questions probability.pdf

QUESTION 1.

Let W be the set of all nonnegative integers, and G the class of allW

subsets of W ( 2 ). For A e G let P(A) = 1 if A has a finite number of

elements and P(A) = 0 otherwise. Does P define a probability measure on

(W,G) ? Provide a detailed answer with the defintion of a probability

measure.

Solution question 1

Let ( W, G ) be a measurable space. A set function P(W) ( P: G -----L R )

defined on G, is called a probability measure ( or simply

probability) if it satisfies the following conditions:

1) P(A) > 0 for all A e G.

2) P(W) = 1.

3) For every { A , j e N}, A e G, j=1,2,3,..., a sequence of disjoint setsj j

( A nA =o if j $ k) we havej k

8& 8 *

P7 u A 8 = S P(A ) (= P(A )+P(A )+...).k=1 k 1 2

k=1

In our case W = set of all nonegative integers, so it is infinite.

By definition of P P(W) = 0 which is contradictory to the condition 2)

above, so P given is not a probability measure.

QUESTION 2.

Let A ,A ,... be a infinite sequence of events such that1 2

A C A C A C A C ... ( all A e G from fixed (W,G,P)).1 2 3 4 i

Prove the following

8v

P( A ) = lim P(A )u k n

n[-----L 8k=1

Solution question 2.

Let us define

B = A , B = A \A ,...,B = A \A ,...1 1 2 2 1 n n n-1

The sequence of events {B } is a sequence of disjoint events such thatn

8 8 n nv v v v

B = A , B = A = A . n = 1,2,3,...u k u k u k u k n

k=1 k=1 k=1 k=1

By the 3) (countable additivity) from the definition of a probability

measure we can write:

8 8 n8 n

v v vP( A ) = P( B ) = S P(B ) = lim S P(B ) = lim P( B ) =

u k u k k k u kk=1 n[-----L 8 k=1 n[-----L 8

k=1 k=1 k=1

nv

= lim P( A ) = lim P(A ) which ends the proof.u k n

n[-----L 8 n[-----L 8k=1

QUESTION 3.

Formulate and prove the Bayes Rule.


Let (W,G,P) be a probability space.

Theorem (Bayes Rule). Let { H , n = 1,2,...} be a disjoint sequencen

vof events such that P(H ) > 0, n = 1,2,... and H = W. Let B e G with

n u n

nP(B) > 0 Then

P(H )WP(B/H )k k

(1) P(H /B) = ---------------------------------------------------------------------------k

8s

P(H )WP(B/H )t k kk=1

Proof. From the definition of a conditional probability

P(BnH )k

P(BnH ) = P(H )W----------------------------------- = P(H )WP(B/H )k k k k

P(H )k

and it imply that

P(H )WP(B/H )k k

P(H /B) = -----------------------------------------------------------------k

P(B)

vNote now that B = (BnH ) and sets (BnH ) are disjoint, the by 3) from

u n n

n

the definition of a probability measure (countable additivity) we can write

8 8 8P(BnH )

v s s n sP(B) = P( (BnH )) = P(BnH )= ----------------------------------- P(H )= P(H )P(B/H ).

u n t n t n t n nP(H )

nn n=1 n=1 n=1Hence, we get

P(H )WP(B/H )k k

P(H /B) = ---------------------------------------------------------------------------k

8s

P(H )WP(B/H )t k kk=1

which ends the proof.

QUESTION 4

Let W = [0,1], and G be the Borel s-field of subsets of W. Define X

on W as follows: X(w) =w if 0 < w < 1. Verify that X is a random

variable. Find the event { w; X(w) e (1/4,1/2)}.


The function X mapping W into R is a random variable if and only

if for each x e R-1

a) X ( (-8,x]) = { w 1 X(w) < x } = { X < x} e G.

In our case-1

X ( (-8,x]) = o if x < 1-1

X ( (-8,x]) = [ 0,x] if 0 < x < 1-1

X ( (-8,x]) = [0,1] if x > 1

G - is a borel s field which contain all subintervals of [0,1]

and o therefore a) holds and X is a random variable.

{ w; X(w) e (1/4,1/2)} = (1/4,1/2)

QUESTION 5.

Suppose that the density function of a random variable X is as follows:& 2(9 - x )/36 for -3 < x < 3

f(x) = {{7 0 otherwise

Find: a) P(X < 0)b) The distribution function F(x)

c) Var(X).


Question 5.

2f(x) = 0 (9-x )/36 0

[-----------------------------------------------------------------------------------------------k------------------------------------------------------------k----------------------------------------------------------------------k--------------------------------------------------------------------------------------------------------------L

-3 0 30

0 -3 0 2 & 3 1 *P(X < 0) = i f(x)dx = i 0dx + i (9-x )/36dx = 1/36 9x - x /3 = 0.5

7 1 8-8 -8 -3 -3

xF(x) = P( X < x) = i f(t)dt

-8

x xif x < -3 then F(x) = P( X < x) = i f(t)dt = i 0dt = 0

-8 -8x -3 x 2

if -3 < x < 3 then F(x) = P( X < x) = i f(t)dt = i 0dt + i (9-t )/36dtx -8 -8 -3

& 3 1 * 31/36 9t - t /3 = (9x - x /3+18)/36

7 1 8-3 8x -3 3 2 =if x > 3 then (x) = P( X < x) = i f t)dt = i 0dt + i ( -t )/36dt+ i0dt 1

-8 -8 -3 3&

0 if x < -32

3so F(x) = { (9x - x /3+18)/36 if -3 < x < 3

7 1 if x > 3

2 2 2Var(X) = E(X-EX) = E(X ) - (EX) = 324/180

8 8-3 3 2

E(X) = ixf(x)dx = i 0dx + i x(9-x )/36dx + i 0dx =-8 -8 -3 3

3 3 2 4 3= 1/36i 9x - x dx = 1/36( 9x - x /4 1 ) = 0

/2 -3-3

8 82 2 -3 3 2 2

E(X ) = ix f(x)dx = i 0dx + i x (9-x )/36dx + i 0dx =-8 -8 -3 3

3 2 4 3 5 3 324 324= 1/36i 9x - x dx = 1/36( 3x - x /5 1 ) = -------------------- = ---------------

-3-3 5W36 180

QUESTION 6.

The distribution of a random vector (X,Y) is given by

X \ Y 2 4 5[----------i---------------------------------------------------------------------------------------------------------]1 p 0.1 0.1 0.3

p2 p 0.1 0.09 0.01

p3 p 0.15 0.05 0.1

Find: a) P( X = 2) b) P(Y > 1) c) P(X < 2, Y < 4.5)‘ ‘

d) P( 2 2 < 2) e) F(x).X-Y% %


P(X = 2) = 0.1 + 0.09 + 0.01 = 0.2

P(Y > 1) = 1

P(X < 2, Y < 4.5) = 0.1 + 0.1 = 0.2‘ ‘

P(2 2 < 2) = 0.1 + 0.1 + 0.15 + 0.05 = 0.4X-Y% %

F(x) = P( X < x) = 0 if x < 1

= 0.1 + 0.1 + 0.3 = 0.5 if 1 < x < 2

= 0.1 + 0.1 + 0.3 + 0.1 + 0.09 + 0.01 = 0.7 if 2 < x < 3

= 1 if x > 3

so& 0 if x < 1

F(x) = { 0.5 if 1 < x < 22 0.7 if 2 < x < 37 1 if x > 3

QUESTION 7.

Suppose that the joint density function for a random vector (X,Y) isgiven by:

2& c(x + y) for 0 < x <1, 0 < y < 1-x

f(x,y) = {7 0 otherwise

Find: a) The constant c.b) P( Y < X +1)

c) f(y)


2& c(x + y) for 0 < x <1, 0 < y < 1-x

f(x,y) = {7 0 otherwise

2D = { (x,y);0 < x < 1, 0 < y < 1-x }

A = { (x,y); y < x + 1} D C A

Since f(x,y) is a density function then

8 8i i f(x,y)dxdy = 1

-8 -8

8 8i i f(x,y)dxdy = i if(x,y)dxdy + i if(x,y)dxdy =

-8 -8 D cD

2 21 1-x 1 2 1-x

y 1=c i ix+y dydx = c i(xy + ---------- )dx

10 0 0 2 01

3 2 4= ci x-x +1/2-x +x /2 dx

012 4 3 5 1 ) = c31/60

= c( x /2 - x /4 +x/2 - x /3 +x /1010

c31/60 = 1 therefore c = 60/31.

P(Y < X + 1 ) = P((X,Y) e A) > P((X,Y) e D) = 1

so P(Y < X + 1 ) = 1.

8f(y) = yi f(x,y)dx = 0 if y < 0 or y > 1

-8--------------- ---------------

8 r1-y r1-y2 1

f(y) = yi f(x,y)dx = 60/31i x+ydx = 60/30 (x /2 +xy )1

-8 00

( )2 260

= ----------21-y ---------------2 if 0 < y < 1--------------- + yr1-y2 231

9 2 0( )2 260

f(x) = ----------21-y ---------------2 for 0 < y < 1--------------- + yr1-y2 231

9 2 0and f(x) = 0 otherwise

QUESTION 8


-y& 2xe for 0 < x < 1 and 0 < y < 8

f(x,y) = {0 otherwise

7

Using marginal densities of X and Y verify the independence of X and Y ?

Solution question 8

1 1-y1 2xe 1p p

---------------------------------------------------------------------------i-----------------------------------------------------------------i---------------------------------------------p p0 1

8 8f(x) = i f(x,y)dy = i 0dy if x < 0 or x > 0

-8 -88 8 b

-y -y -y bf(x) = i f(x,y)dy = i 2xe dy = 2x lim ie dy = 2xlim ( -e 1 ) =

0-8 0 b[L8 0 b[L8

-b= 2xlim (-e +1) = 2x if 0 < x < 1

b[L8

& 2x for 0 < x < 1so f(x) = {

0 otherwise7

8 8f(y) = i f(x,y)dx = i 0dx if y < 0

-8 -88 0 1 8

-y -y 2 1 -yf(y) = i f(x,y)dx = i 0dx + i2xe dx + i0dx = e (x 1 ) = e for y > 0

0-8 -8 0 1

-y& e for 0 < y < 8

so f(y) = {0 otherwise

7

-yTherefore f(x)Wf(y) = 2xe for 0 < x < 1, 0 < y < 8

= 0 otherwise

and f(x)Wf(y) = f(x,y) so X and Y are independent.

QUESTION 9.

Let (W,G) be a measurable space and P the probability measure

defined on (W,G). Prove the following statements:

a) A if A c B then P(B\A) = P(B) - P(A).A,BeG

b) A P(AuB) < P(A) + P(B).A,BeG


a) A c B +++++6 B = (B\A) u A and the sets on the right side are

disjoint, so P(B) = P(B\A) + P(A). Solving we get

P(B\A) = P(B) - P(A).

b) AuB = (A\(AnB) u (B\(AnB) u (AnB) and the sets on the right side

are disjoint and A = (A\(AnB) u (AnB), B = (B\(AnB) u (AnB)

hence

P(AuB) = P(A\(AnB) + P(B\(AnB) + P(AnB) =

= [P(A\(AnB) + P(AnB)] + [P(B\(AnB) + P(AnB)] - P(AnB) =

= P(A) + P(B) - P(AnB) < P(A) + P(B) since P(AnB) > 0.

QUESTION 10.

Prove the following theorem:

If the moments of order t exists for a random variable X of continuous

type with density function f(x), t > 0 then the moments of order s, 0 < s < t

exist.

Solution question 10

8s s

We have to show that E1X1 = i 1x1 f(x)dx < 8-8

8t t

We know that E1X1 = i 1x1 f(x)dx < 8-8

Therefore

8 1 1 8s s s s s

E1X1 = i 1x1 f(x)dx = i1x1 f(x)dx + i1x1 f(x)dx + i1x1 f(x)dx <-8 -8 -1 1

1 1 8t t

i1x1 f(x)dx + i 1Wf(x)dx + i1x1 f(x)dx <-8 -1 1

1 1 8 1t t t t

i1x1 f(x)dx + i1x1 f(x)dx + i1x1 f(x)dx + i 1Wf(x)dx = E1X1 + P{1X1 < 1} < 8.-8 -1 1 -1

t sWe used that for 1x1 > 1 the following 1x1 > 1x1 and

sfor 1x1 < 1 the following 1x1 < 1.

QUESTION 11.

Let W be the set of all natural numbers, and G the class of allW

subsets of W ( 2 ).For A e G, let P(A) = 1 if A is a finite set

and P(A) = 0 otherwise. Does P define a probability measure on (W,G) ?


Definition: A set function P(W) ( P: G L R ) defined on G,

is called a probability measure ( or simply



2) P(W) = 1.

3) For every {A , j e N }, A e G, j = 1,2,3,... , a sequence of disjointj j

sets ( A n A = o if j $ k) we havej k

88& *P u A = S P(A ).7 k8 kk=1 k=1

In our case P(A) = 0 or 1 so P(A) > 0 and 1) holds.

Since our W = N - infinite set so by definition of P P(W) = 0

and it contradicts 2).

Since second condition for the probability meaure is not satisfied

therefore P is not a probability measure.

QUESTION 12.

Suppose that the random variable X has the following probability density

function

&c2x2 for -2 < x < 2

2f(x)={{

2 0 otherwise7

Find a) the constant c

b) P{ -0.5 < X < 1.5 }

c) The distribution function


Density function have to satisfy:

a) f(x) > 0 for all x

8b) i f(x)dx = 1

-88 -2 0 2 8

from b) i f(x)dx = i 0dx + ic(-x)dx + icxdx + i0dx =-8 -8 -2 0 0

2 2x 2 0 x 22

= -c ---------- + c ---------- = 4c = 1 so c = 1/42 2

2 -2 2 0

If c = 1/4 a) holds too.

1.5 0 1.5 2 0 2 1.5x 2 x 2

P{ -0.5 < X < 1.5 } = i f(x)dx = i-x/4dx + ix/4dx = ----- ---------- + ---------- =2 2

-0.5 -0.5 0 8 -0.5 8 0

= 10/24

f(x) 0 -x/4 x/4 0------------------------------------------------------------------------------------------k------------------------------k------------------------------k-----------------------------------------------------------------

-2 0 2

xF(x) = if(t)dt

-8 x xif x < -2 then F(x) = if(t)dt = i0dt = 0

-8 -8x -2 x

2 2xi i i t 1 - xif -2 < x < 0 then F(x) = 2 f(t)dt = 2 0dt + 2 -t/4dt = - -------------------- 1 = -----------------------------------

j j j 8 8-2

-8 -8 -2x -2 0 xi i i i

if 0 < x < 2 then F(x) = 2 f(t)dt = 2 0dt + 2 -t/4dt + 2t/4dt =j j j j-8 -8 -2 0

2 0 2 x 2t 2 t 2 1 x

= ----- ---------- + ---------- = ----- + ----------2 2

8 -2 8 0 2 8

if x > 2 then F(x) = 1

&0 if x < -2

2F(x) = { 2

1 x2 --------------- ----- --------------- if -2 < x < 0

2 822 2

1 x2 --------------- + --------------- if 0 < x < 2

2 822

1 if x > 27

QUESTION 13.

Let (X,Y) be a random vector with the joint density function given

(2 2 22 cx y for x < y < 1

by: f(x,y) = {2 0 otherwise29

Find: a) The constant c, b) P( X < 0.5,Y < 0.2 ), c) f(y)


y2

y = x

y = 1

D

-1 1 x

2D = {(x,y); x < y < 1}

Since f(x,y) is a density function then

8 8i i2 2 f(x,y)dxdy = 1j j-8 -8

8 8i i i i i i2 2 f(x,y)dxdy = 2 2f(x,y)dxdy + 2 2f(x,y)dxdy =j j j j j j

c-8 -8 D D1 1 1i i 2 i 2 2 1

= 2 2 cx ydydx = c 2 (x y /21 )dx =j j j 2

x-1 2 -1

x1

1i 2 6 3 7 4= c/2 2 x - x dx = c/2 (x /3 - x /7)1 = c --------------------

j 21-1-1

Therefore c = 21/4

y2

y = x

y = 1

q==========6-e0.2 xA

12

D = {(x,y); x < y < 1}q==========6 q==========6 2

A = {(x,y);-e0.2 < x < e0.2 ,x < y < 0.2}

i i i i 2P(X < 0.5,Y < 0.2) = 2 20dxdy + 2 2x ydxdy 21/4 =

j j j jA A1

q==========6 q==========6e0.2 0.2 e0.2

1i i 2 i 2 2= 2 2 x ydydx = 2 (x y /21 )dx =

j j j 2q==========6 q==========6 x2

-e0.2 x -e0.2-----q==========6 5 6

e0.2r0.2

i 2 4 3 5= 21/4 2 (0.02x - x /2)dx = 21/4 ( 0.02x /3 - x /101 )

j 5 6q==========6 _____----- r0.2-e0.2

= 0.043826

8i

f(y) = 2 f(x,y)dx = 0 if y < 0 or y > 1j-8

q6===== q6=====8 -ey ey 8i i i i

f(y) = 2 f(x,y)dx = 2 f(x,y)dx + 2 f(x,y)dx + 2 f(x,y)dx =j j j j

q6===== q6=====-8 -8-ey ey

q6=====eyi 2 5/2

= 21/4 2 x ydx = 7y /2jq6=====

-ey5/2

f(y) = 7y /2 if 0 < y <1

f(y) = 0 otherwise

QUESTION 14.

Let X be a random variable with density function given by

&x/c for 0 < x < 4

f(x) = {7 0 otherwise

and Y be a random variable independent of X with density function

given by

& y if y e (0,1)f(y) = { 2-y if y e [1,2)

7 0 otherwise

Find a) c

b) f(x,y)


f(x) is a density if and only if

a) A f(x) > 0xeR8i

b) 2 f(x)dx = 1j-88 0 4 8i i 1 i i2 f(x)dx = 2 0dx + --------------- 2 xdx + 2 0dx =j j c j j-8 -8 0 4

2 41 x 8--------------- (--------------------1 ) = --------------- = 1c 2 c

0

Hence c = 8 and a) holds too.

&x/8 for 0 < x < 4

f(x) = {7 0 otherwise

Since X and Y are independent then

f(x,y) = f(x)f(y)

y

0 0 00

A

0 2-y x(2-y)/8 0

B0y0 xy/8

f(x) = 0 x/8 4 0

0

0 0 0@

f(y)

(2 xy/8 if 0 < x < 4, 0 < y < 1222

f(x,y) = { x(2-y)/8 if 0 < x < 4, 1 < y < 22222 0 otherwise9

QUESTION 15.

The distribution of a random vector (X,Y) is given byX \ Y 2 4 5[----------i---------------------------------------------------------------------------------------------------------]1 p 0.1 0.1 0.2

p2 p 0.2 0.3 0.1

pFind: a) P( X = 2), b) P(Y > 1), c) P(X < 2/ Y < 4.5)Solution question 15.

a) P( X = 2) = P( (X,Y) e{(2,2),(2,4),(2,5)} = 0.2 + 0.3 + 0.1 = 0.6

b) P(Y > 1) = P((X,Y)e { (1,2),(1,4),(1,5),(2,2),(2,4),(2,5)}) = 1

c) P(X < 2/Y < 4.5) = P(X < 2, Y < 4.5)/ P( Y< 4.5) =

= P((X,Y) e{(1,2),(1,4)})/P( (X,Y) e {(1,2),(1,4),(2,2),(2,4)})

= 0.2/0.7 = 2/7

QUESTION 16.


a) Prove the following statement:

A, B e G, P(AuB) = P(A) + P(B) - P(AnB).

1 1b) Consider two events C and D such that P(C) = ---------------, P(D) = ---------------.

4 2c

Determine the value P(CnD ) for each of the following conditions

1i) CnD = o ii) C C D iii) P(CnD) = ---------------

8


a) AuB = (A\(AnB) u (B\(AnB) u (AnB) and the sets on the right side


hence



= P(A) + P(B) - P(AnB)

c c 1b) i) CnD = o 6 CnD = C so P(CnD ) = P(C) = ---------------

4c c

ii) CCD 6 CnD = o so P(CnD ) = P(o) = 0

c 1 1 1iii) P((CnD ) = P(C) - P(CnD) = --------------- - --------------- = ---------------

4 8 8

QUESTION 17.

Let W = [0,1], and G be the Borel s-field of subsets of W. Define

X on W as follows X(w) = w for w e W. Verify that X is a random

variable. Find the event {w; X(w) e (1/4,1/2)}


The function X mapping W into R is a random variable if

and only if for each x e R-1

1) X ((-8, x]) ={w; X(w) < x} = {X < x} e G.

In our case-1

X ((-8, x]) = o if x < 0-1

X ((-8, x]) = [0,x] if 0 < x < 1-1

X ((-8, x]) = [0,1] if x > 0

G - is a Borel s-field which contain all subintervals

of [0,1] and o therefore 1) holds and X is a random variable

QUESTION 18.


function

&c2x2 for -2 < x < 2

2f(x)={{

2 0 otherwise7


b) P{ -0.5 < X < 1.5 }



The function X mapping W into R is a random variable if

and only if for each x e R-1

1) X ((-8, x]) ={w; X(w) < x} = {X < x} e G.

In our case-1

X ((-8, x]) = o if x < 0-1

X ((-8, x]) = [0,x] if 0 < x < 1-1

X ((-8, x]) = [0,1] if x > 0

G - is a Borel s-field which contain all subintervals

of [0,1] and o therefore 1) holds and X is a random variable

QUESTION 19.


function

&c2x2 for -2 < x < 2

2f(x)={{

2 0 otherwise7


b) P{ -0.5 < X < 1.5 }



Density function have to satisfy:

a) f(x) > 0 for all x

8b) i f(x)dx = 1

-88 -2 0 2 8

from b) i f(x)dx = i 0dx + ic(-x)dx + icxdx + i0dx =-8 -8 -2 0 0

2 2x 2 0 x 22 V

= -c ---------- + c ---------- = 4c = 1 so c = 1/42 2

2 -2 2 0

If c = 1/4 a) holds too.

1.5 0 1.5 2 0 2 1.5x 2 x 2

P{ -0.5 < X < 1.5 } = i f(x)dx = i-x/4dx + ix/4dx = ----- ---------- + ---------- =2 2

-0.5 -0.5 0 8 -0.5 8 0

= 10/24

f(x) 0 -x/4 x/4 0------------------------------------------------------------------------------------------k------------------------------k------------------------------k-----------------------------------------------------------------

-2 0 2

F(x) = if(t)dt-8 x x

if x < -2 then F(x) = if(t)dt = i0dt = 0-8 -8

x -2 x2 2xi i i t 1 - x

if -2 < x < 0 then F(x) = 2 f(t)dt = 2 0dt + 2 -t/4dt = - -------------------- 1 = -----------------------------------j j j 8 8

-2-8 -8 -2x -2 0 xi i i i

if 0 < x < 2 then F(x) = 2 f(t)dt = 2 0dt + 2 -t/4dt + 2t/4dt =j j j j-8 -8 -2 0

2 0 2 x 2t 2 t 2 1 x

= ----- ---------- + ---------- = ----- + ----------2 2

8 -2 8 0 2 8

if x > 2 then F(x) = 1&

0 if x < -22

F(x) = { 21 x

2 --------------- ----- --------------- if -2 < x < 02 8

22 2

1 x2 --------------- + --------------- if 0 < x < 2

2 822

1 if x > 27

QUESTION 20.

Suppose that a random variable X is of a discrete type with the following

distribution given by:

P(X = k) = ck for k = 1,2,3,4,5 and P(X e {1,2,3,4,5} ) = 1.

Find : a) the value of c.

b) The distribution function F(x).

c) P( X < 3,5 / X > 0.5)


a) Since P(X e {1,2,3,4,5} ) = 1 then V = {1,2,3,4,5}is a set of all

values. Let us denote p = P(X = k), k = 1,2,3,4,5.k

{P } constitute an assignment of a probability if and only ifk

1) each p > 0k

2) S p = 1k

By 2) we have S p = c + 2c + 3c + 4c + 5c = 15c = 1 so c = 1/15.k

If c = 1/15 then a) also holds so indeed c = 1/15 gives a probability

distribution.

b) F(x) = P(X < x)

if x < 1 then F(x) = P(X < x) = P(o) = 0

if 1 < x < 2 then F(x) = P(X < x) = P(X = 1) = 1/15

if 2 < x < 3 then F(x) = P(X < x) = P(X e {1,2}) = 3/15

if 3 < x < 4 then F(x) = P(X < x) = P(X e {1,2,3}) = 6/15

if 4 < x < 5 then F(x) = P(X < x) = P(X e {1,2,3,4}) =10/15

if 5 < x then F(x) = P(X < x) = P(X e {1,2,3,4,5}) = 1

Then & 0 if x < 12

1/15 if 1 < x < 22

F(x) = { 3/15 if 2 < x < 32

6/15 if 3 < x < 422 10/15 if 4 < x < 52

1 if 5 < x then7

P(X < 3.5, X > 0.5) P(X e {1,2,3})c) P(X < 3.5 / X > 0.5) = --------------------------------------------------------------------------------------------------------- = ----------------------------------------------------------------------------------------------- =

P(X > 0.5) P(X e {1,2,3,4,5})

= (6/15)/(1) = 6/15QUESTION 21.

The distribution of a random vector (X,Y) is given byX \ Y 2 4 5[----------i---------------------------------------------------------------------------------------------------------]1 p 0.1 0.1 0.2

p2 p 0.2 0.3 0.1

pFind: a) P( X = 2), b) P(Y > 1), c) P(X < 2/ Y < 4.5)


a) P( X = 2) = P( (X,Y) e{(2,2),(2,4),(2,5)} = 0.2 + 0.3 + 0.1 = 0.6

b) P(Y > 1) = P((X,Y)e { (1,2),(1,4),(1,5),(2,2),(2,4),(2,5)}) = 1

c) P(X < 2/Y < 4.5) = P(X < 2, Y < 4.5)/ P( Y< 4.5) =

= P((X,Y) e{(1,2),(1,4)})/P( (X,Y) e {(1,2),(1,4),(2,2),(2,4)})

= 0.2/0.7 = 2/7

sWe use the following P((X,Y)eA) = p

t i,ji,j;(x ,y )eA

i j

QUESTION 22.

Let W be the set of all natural numbers, and G the class of allW

subsets of W ( 2 ).For A e G, let P(A) = 1 if A is a finite set

and P(A) = 0 otherwise. Does P define a probability measure on (W,G) ?


Definition: A set function P(W) ( P: G L R ) defined on G,

is called a probability measure ( or simply



2) P(W) = 1.



88& *P u A = S P(A ).7 k8 kk=1 k=1

In our case P(A) = 0 or 1 so P(A) > 0 and 1) holds.

Since our W = N - infinite set so by definition of P P(W) = 0

and it contradicts 2).

Since second condition for the probability meaure is not satisfied

therefore P is not a probability measure.

QUESTION 23.

Let W be the set of all nonnegative integers and G the class of allW

subsets of W ( G = 2 ). For A e G let

s kP(A) = p(1-p) 0 < p < 1.

tkeA

Does P define a probability measure on (W,G) ?

Give detail answer with all definition required.

nn+1

s k 1 - tHint: t = -------------------------------------------------- ( t =/ 1)

t 1 - tk=0


Let ( W, G ) be a measurable space. A real set function P(W)

( P: G L R ) defined on G, is called a probability measure ( or simply



2) P(W) = 1.



88& *P u A = S P(A ).7 k8 kk=1 k=1

s kIn our case P(A) = p(1-p) 0 < p < 1

tkeA

and each term in the sum is > 0 so P(A) > 0 and 1) holds.

8 ns k s k

P(W) = P({0,1,2,3,...}) = p(1-P) = p lim (1-p) =t t

n-----L8k=0 k=0

n+11 - (1-p) 1

= p lim ---------------------------------------------------------------------- = p ------------------------- = 11 - (1-p) 1-1+p

n-----L8

so 2) holds.

Let {A , j e N } be as in 3).j

( )2 8 22 2

v s kP2 A 2 = p(1-p) = { series converges absolutely can be rearranged} =

u j t2 22j=1 2 k eUuA

j9 0

8s k s k s k

= p(1-p) + p(1-p) + ... + p(1-p) + ... = S P(A )t t t j

j=1keA keA keA

1 2 2

so 3) holds. Since 1), 2) and 3) are satisfied therefore P is a probability

measure.

QUESTION 24.



a) A if A c B then P(B\A) = P(B) - P(A).A,BeG

b) A P(AuB) < P(A) + P(B).A,BeG


a) A c B +++++6 B = (B\A) u A and the sets on the right side are

disjoint, so P(B) = P(B\A) + P(A). Solving we get

P(B\A) = P(B) - P(A).

b) AuB = (A\(AnB) u (B\(AnB) u (AnB) and the sets on the right side


hence



= P(A) + P(B) - P(AnB) < P(A) + P(B) since P(AnB) > 0.

QUESTION 25.


Prove the following statements:c

a) If A and B are independent events then A and B

are independent events.

b) If A and B are independent events and P(B) > 0 then

P(A/B) = P(A)


a) A and B are independent if P(AnB) = P(A)P(B)c c

We have to show that P(BnA ) = P(B)P(A ).c

B = (BnA) u (BnA ) and the sets on the right side are disjoint

hencec

P(B) = P(BnA) + P(BnA ), solving we getc

P(BnA ) = P(B) - P(BnA) = P(B) - P(B)P(A) =c

= P(B)[1 - P(A)] = P(B)P(A )

P(AnB) P(A)P(B)b) P(A/B) = ---------------------------------------- = -------------------------------------------------- = P(A).

P(B) P(B)

QUESTION 26.

Does the following function define distribution function(2 0 if x < 0

F(x) = {-x22 1 - e if x > 0.

9Solution question 26.

A real-valued function F defined on R ( (-8,8) ) that is

a) nondecreasing, b) right continuous and

c) lim F(x) = 0 d) lim F(x) = 1x-----L -8 x-----L8

is called a distribution function.

y = F(x)y

x

Our function is continuous so b) holds.

if x < y < 0 then F(x) = 0 = F(y)-y

if x < 0 < y then 0 = F(x) and F(y) = 1 - e > 0

so F(x) < F(y)-x -y -x -y

if 0 < x < y then e > e so 1 - e < 1 - e

so F(x) < F(y)

Therefore if x < y +++++6 F(x) < F(y) so F(x) is nondecreasing

function and a) holds

lim F(x) = lim 0 = 0 so c) holdsxL -8 xL -8

-xlim F(x) = lim (1 - e ) = 1 so d) holds.

xL 8 xL 8

Since all conditions are satisfied therefore F(x) is

a distribution function.

QUESTION 27.

Suppose that a joint density function for a random vector (X,Y) is given by(2 c(x+y) for 0 < x < y < 1

f(x,y) = {2 0 otherwise9

Find: a) The constant c.

b) F(y)

c) P( Y < X + 0.5)


8 8i i

f(x,y) - joint density function 46 a) 2 2 f(x,y)dxdy = 1j j-8 -8

b) A f(x,y) > 0.x,y

D = { (x,y); 0 < y < 1, 0 < x < y}


c-8 -8 D D1 y 1

2( ) ( y )i i i x= 2 {2 2 c(x,y)dx }2dy = c 2 {2 -------------------- + xy 1 }2 dy =

j j j 29 0 9 000 0 0

1 12 2 3 1i y 2 i 3y y c

= c 2 -------------------- + y dy = c 2 ------------------------- dy = c --------------------1 = --------------- = 1j 2 j 2 2 200 0

so c = 2.yi

F(y) = P( Y < y) = 2 f (t)dtj 2-8

where8i

f (y) = 2 f(x,y)dx2 j

-8

8 0 y 8i i i i

y e (0,1) +++++6 f (y) = 2 f(x,y)dx = 2 0 dx + 22(x+y)dx + 20 dx2 j j j j

-8 -8 0 y

2 yx 2= 2 ( -------------------- + xy 1 ) = 3y

20

8i

y e/ (0,1) +++++6 f (y) = 2 0 dx = 02 j

-8(2 2

3y if 0 < y < 1f (y) = {2

22 0 otherwise9yi

F(y) = 2 f (t)dtj 2-8

yi

y < 0 +++++6 F(y) = 2 0dt = 0j-8

0 yi i 2 3 y 3

0 < y < 1 +++++6 F(y) = 2 0dt + 23t dt = t 1 = yj j 0-8 0

0 1 yi i 2 i 3 1

y > 1 +++++6 F(y) = 2 0dt + + 23t dt + 20dt = t 1 = 1j j j 0-8 0 1

(2 0 if y < 022 3

F(y) = { y if 0 < y < 12222 1 if y > 19

P(Y < X + 0.5) = P((X,Y) e C) + P((X,Y) e C ) =1

i i i i= 2 2f(x,y)dxdy + 2 2f(x,y)dxdy =

j j j jC C

1

C C1

C = {(x,y); 0 < y < 0.5,0 < x < y } u {(x,y); 0.5 < y < 1,y - 0.5 < x < y}

i i i i2 22(x+y)dxdy + 2 20dxdy =j j j jC C

10.5 y 1 yi i i i

= 2 22(x+y)dxdy + 2 2 2(x+y)dxdy =j j j j0 0 0.5 y-0.50.5 1

y yi 2 i 2= 2 (x +xy1 )dy + 2 (x +xy1 )dy =

j j0 y-0.5

0 0.50.5 1i 2 i 2 2

= 2 3y dy + 2 3y - (y-0.5) - 2(y-0.5)ydy =j j0 0.5

0.5 10.5 1i 2 i 3 2

= 2 3y dy + 2 2y -0.25dy = y 1 + (y - 0.25y)1 = 0.75.j j 0 0.50 0.5

P(Y < X +0.25) = 0.75.

QUESTION 28.

Let X,Y be independent identically distributed random variables

with common density function(2 -z2 e if z > 0

f(z) = {222 0 otherwise9

Find the distribution function of M = max{X,Y}.


X and Y are independent so for all x,y P(X < x,Y< y) = P(X < x)P(Y < y)

and also F(x,y) = F(x)F(y)

Distribution function for X and Y is the same and is equal to

z zi i

F(z) = P(X < z) = P( Y < z) = 2 f(t)dt = 2 0dt = 0 if z < 0j j

-8 -80 zi i -t -t z -z

= 2 0dt + 2e dt = - e 1 = 1 - e if z > 0j j

0-8 0

(2 0 if x < 02

F (x) = P( max{X,Y} < x) = P(X < x,Y < x) = F(x)F(x) ={M 2 -x 222 (1 - e ) if x > 1

9QUESTION 29.

Let (X,Y) be a random vector with joint density given by(2 22 3x

2xy + ------------------------------ if 0 < x < 1, 0 < y < 1f(x,y) = { 2

222 0 otherwise9

Find the line of regression of Y on X


cov(X,Y)y - EY = ----------------------------------------(x - E)

Var(X)

I1j---------------------------------------------o1 11 2 1

3x12xy+--------------- 1 D = {(x,y);0<x<1,0<y<1}

21 1---------------------------------------------k---------------------------------------------k---------------------------------------------L

1 1

8i

f(x) = 2 f(x,y)dy = 0 if x e/ (0,1)j

-88 0 1 8

2 2i i i 3x i 2 3yx 1

f(x) = 2 f(x,y)dy = 2 0dy + 2 2xy+---------------dy + 20dy = xy + --------------------1 =j j j 2 j 2

0-8 -8 0 1

23x

= x + --------------- if x e (0,1)2

8 0 1 82

i i i 3x iEX = 2 xf(x)dx = 2 0dx + 2x(x + --------------- )dx + 20dx =

j j j 2 j-8 -8 0 1

3 4 1 17= x /3 + 3x /8 1 = -------------------- = 0.7083

2408 0 1 8

22 i 2 i i 2 3x i

EX = 2 x f(x)dx = 2 0dx + 2x (x + --------------- )dx + 20dx =j j j 2 j

-8 -8 0 1

4 5 1 11= x /4 + 3x /10 1 = --------------------

200

2 2 11 17 2Var(X) = EX - (EX) = -------------------- - (--------------------) = 0.048

20 24

8i

f(y) = 2 f(x,y)dx = 0 if y e/ (0,1)j

-88 0 1 8

2i i i 3x i

f(y) = 2 f(x,y)dx = 2 0dx + 2 2xy+---------------dx + 20dx =j j j 2 j

-8 -8 0 12 3 1

= x y + x /2 1 = y + 1/2 if y e (0,1)0

8 0 1 8i i i 2 y i 3 2 1 7

EY = 2 yf(y)dy = 2 0dy + 2y + ----- dy + 20dy = y /3 + y /4 1 = -------------------- = 0.5833j j j 2 j 120

-8 -8 0 1

cov(X,Y) = E(X - EX)(Y - EY) = EXY - EXEY

8 8 11 12 3 4

i i ii 3x i 2 x 3yx 1EXY = 2 2 xyf(x,y)dxdy = 22xy(2xy+---------------)dxdy = 2(2y -------------------- + -------------------- 1 )dy =

j j jj 2 j 3 8 0-8 -8 00 01i 2 3 2 1

= 2 2y /3 + 3y/8 dy = 2y /9 - 3y /16 1 = 0.4097j 00

cov(X,Y) = E(X - EX)(Y - EY) = EXY - EXEY = 0.4097 - 0.7083Q0.5833 = -0.0034

cov(X,Y)y - EY = ----------------------------------------(x - E)

Var(X)

y - 0.5833 = -0.0034/0.048(x - 0.7083)

y = -0.0708x +0.6334

QUESTION 30.



1) A, B e G, P(AuB) = P(A) + P(B) - P(AnB)c

2) A, B e G, P(AnB ) = P(A) - P(AnB)


1) AuB = (A\(AnB) u (B\(AnB) u (AnB) and the sets on the right side


hence



= P(A) + P(B) - P(AnB) < P(A) + P(B)c

2) A = (AnB)u(AnB ) and the sets on the right side are disjoint thereforec

P(A) = P(AnB) + P(AnB ) solving we getc

P(AnB ) = P(A) - P(AnB)

QUESTION 31.

Let X be an r.v. Is Z =1X1 a random variable ?



Definition: A real function X : W -------------------------L R is called a random variable (r.v.)

if the inverse images under X of all Borel sets in R are events, that is,-1

1) X (B) = { w; X(w) e B } e G for all B e B(R)

This condition is equivalent to the following:

X is an r.v. if and only if for each x e R-1

2) X ((-8,x]) = { w; X(w) < x} = { X < x } e G

We have to show that-1

Z ((-8.x]) e G .

We have-1

Z ((-8.x]) = { w;Z(w) < x} = { w; 1X(w)1 < x } = {w; -x < X(w) < x } =-1

X ( [-x,x]) J-------------------- this set is in G since interval [-x,x]

is a borel set and 1) holds (because X is a random variable).

Therefore Z = 1X1 is a random variable.

QUESTION 32.

Suppose that the density function of a random variable X is

as follows:(2 22 (9 - x )/36 for -3 < x < 3

f(x) = {22 0 otherwise9

Find: a) P(X < 0)

b) The distribution function F(x)

c) P( X > -0.5/ X < 2)


2f(x) = 0 (9-x )/36 0

[-----------------------------------------------------------------------------------------------k------------------------------------------------------------k----------------------------------------------------------------------k--------------------------------------------------------------------------------------------------------------L

-3 0 3

00 -3 0 2 & 3 1 * 1

P(X < 0) = i f(x)dx = i 0dx + i (9-x )/36dx = 1/36 9x - x /3 = --------------------7 1 8 2

-8 -8 -3 -3

xF(x) = P( X < x) = i f(t)dt

-8x x

if x < -3 then F(x) = P( X < x) = i f(t)dt = i 0dt = 0-8 -8

x -3 x 2if -3 < x < 3 then F(x) = P( X < x) = i f(t)dt = i 0dt + i (9-t )/36dt

x -8 -8 -3& 3 1 * 3

1/36 9t - t /3 = (9x - x /3+18)/367 1 8

-3 8x -3 3 2

if x > 3 then F(x) = P( X < x) = i f(t)dt = i 0dt + i (9-t )/36dt+ i0dt=1-8 -8 -3 3

&0 if x < -3

23

so F(x) = { (9x - x /3+18)/36 if -3 < x < 37 1 if x > 3

P(-0.5 < X < 2 ) 0.549P( X > -0.5 / X < 2) = -------------------------------------------------------------------------------- = ------------------------- = 0.593

P( X < 2 ) 0.9252 2

2 3P(-0.5 < X < 2) = i (9 - x )/36 dx = 1/36 ( 9x - x /3)1 = 0.549

-0.5-0.5

P( X < 2) = P( X < 2) = F(2) = 0.925

QUESTION 33.

Suppose that a random variable X is of a discrete type with the following

distribution given by:

P(X = k) = ck for k = 1,2,3,4,5 and P(X e {1,2,3,4,5} ) = 1.

Find : a) the value of c.

b) The distribution function F(x).

c) P( X < 3,5 / X > 0.5)


a) Since P(X e {1,2,3,4,5} ) = 1 then V = {1,2,3,4,5}is a set of all

values. Let us denote p = P(X = k), k = 1,2,3,4,5.k

{P } constitute an assignment of a probability if and only ifk

1) each p > 0k

2) S p = 1k

By 2) we have S p = c + 2c + 3c + 4c + 5c = 15c = 1 so c = 1/15.k

If c = 1/15 then a) also holds so indeed c = 1/15 gives a probability

distribution.

b) F(x) = P(X < x)

if x < 1 then F(x) = P(X < x) = P(o) = 0

if 1 < x < 2 then F(x) = P(X < x) = P(X = 1) = 1/15

if 2 < x < 3 then F(x) = P(X < x) = P(X e {1,2}) = 3/15

if 3 < x < 4 then F(x) = P(X < x) = P(X e {1,2,3}) = 6/15

if 4 < x < 5 then F(x) = P(X < x) = P(X e {1,2,3,4}) =10/15

if 5 < x then F(x) = P(X < x) = P(X e {1,2,3,4,5}) = 1

Then & 0 if x < 12

1/15 if 1 < x < 22

F(x) = { 3/15 if 2 < x < 32

6/15 if 3 < x < 422 10/15 if 4 < x < 52

1 if 5 < x then7

P(X < 3.5, X > 0.5) P(X e {1,2,3})c) P(X < 3.5 / X > 0.5) = --------------------------------------------------------------------------------------------------------- = ----------------------------------------------------------------------------------------------- =

P(X > 0.5) P(X e {1,2,3,4,5})

= (6/15)/(1) = 6/15QUESTION 34.


X \ Y 2 4 5[----------i---------------------------------------------------------------------------------------------------------]1 p 0.1 0.1 0.2

p2 p 0.2 0.3 0.1

p

Find: a) P( X = 2)

b) P(Y > 1)

c) P(X < 2/ Y < 4.5)


a) P( X = 2) = P( (X,Y) e{(2,2),(2,4),(2,5)} = 0.2 + 0.3 + 0.1 = 0.6

b) P(Y > 1) = P((X,Y)e { (1,2),(1,4),(1,5),(2,2),(2,4),(2,5)}) = 1

c) P(X < 2/Y < 4.5) = P(X < 2, Y < 4.5)/ P( Y< 4.5) =

= P((X,Y) e{(1,2),(1,4)})/P( (X,Y) e {(1,2),(1,4),(2,2),(2,4)})

= 0.2/0.7 = 2/7

QUESTION 35.


a) Prove the following statement:

A, B e G, P(AuB) = P(A) + P(B) - P(AnB).

1 1b) Consider two events C and D such that P(C) = ---------------, P(D) = ---------------.

4 2c

Determine the value P(CnD ) for each of the following conditions

1i) CnD = o ii) C C D iii) P(CnD) = ---------------

8


a) AuB = (A\(AnB) u (B\(AnB) u (AnB) and the sets on the right side


hence



= P(A) + P(B) - P(AnB)

c c 1b) i) CnD = o 6 CnD = C so P(CnD ) = P(C) = ---------------

4c c

ii) CCD 6 CnD = o so P(CnD ) = P(o) = 0

c 1 1 1iii) P((CnD ) = P(C) - P(CnD) = --------------- - --------------- = ---------------

4 8 8

QUESTION 36.

Let (W,G,P) be a probability space. A , B e G such that P(A) > 0,

P(B) > 0 and AnB = o. Define X(w) = -I (w) + 2I (w)A B

& &1 if w e A 1 if w e B

where I (w) = { and I (w) = { .A B

7 0 if w m A 7 0 if w m B

Verify that X is a random variable. Find the event { w; X(w) e (-2,1)}.


The function X mapping W into R is a random variable if and only

if for each x e R-1

a) X ( (-8,x]) = { w 1 X(w) < x } = { X < x} e G.

X(w) = -I (w) + 2I (w)A B

I (w) = 0 ! 0 ! 0 ! 1 ! 0B

I (w) = 0 ! 1 ! 0 ! 0 ! 0A

-----------------------------------------------------------------k--------------------------------------------------k-------------------------k-----------------------------------k-------------------------------------------------------A B (A and B disjoint)

Therefore(2 -1 if w e A2 c

X(w) = { 0 if w e (AuB)22 2 if w e B9

In our case-1

X ( (-8,x]) = o if x < -1-1

X ( (-8,x]) = A if -1 < x < 0-1 c

X ( (-8,x]) = A u (AuB) if 0 < x < 2-1

X ( (-8,x]) = W if x > 2

Since G is a s-field so o e G and by our assumption A,B eG soc

AuB e G and (AuB) , therefore a) holds and X is a random variable.c

{ w; X(w) e (-2,1)} = {w: X(w) = -1 or 0} = A u (AuB)

QUESTION 37.


Let X be a random variable on a probability space (W,G,P) of thek

continuous type with the density function f(x). Let E1X1 < 8 for

some k > 0. Thenk

n P(1X1 >n) ----------L 0 as n ----------L 8.


We have

8 nk k

8 > i 1x1 f(x)dx = lim i 1x1 f(x)dx .-8 n[-----L 8 -n

It follows that-n 8

& k k *lim i 1x1 f(x)dx + i 1x1 f(x)dx = 0

7 8n[-----L 8 -8 n

But-n 8 -n 8

k k k ki1x1 f(x)dx + i1x1 f(x)dx > i n f(x)dx + i n f(x)dx

-8 n -8 n-n 8 -n 8

k k k k> i n f(x)dx + i n f(x)dx = n ( i f(x)dx + i f(x)dx) = n P{1X1 > n}

-8 n -8 ncompleting the proof.

QUESTION 38.

Does the following function define distribution function

(2 0 if x < 0

F(x) = {-x22 1 - e if x > 0.

9If yes find: EX, P(-1 < X < 4)


A real-valued function F defined on R ( (-8,8) ) that is

a) nondecreasing, b) right continuous and

c) lim F(x) = 0 d) lim F(x) = 1x-----L -8 x-----L8

is called a distribution function.

y y = F(x)

x

Our function is continuous so b) holds.

if x < y < 0 then F(x) = 0 = F(y)-y

if x < 0 < y then 0 = F(x) and F(y) = 1 - e > 0

so F(x) < F(y)-x -y -x -y

if 0 < x < y then e > e so 1 - e < 1 - e

so F(x) < F(y)

Therefore if x < y +++++6 F(x) < F(y) so F(x) is nondecreasing

function and a) holds-x

Or since F’(x) = e or 0 so is > 0 therefore F(x) is nondecreasing

and a) holds.

lim F(x) = lim 0 = 0 so c) holdsxL -8 xL -8

-xlim F(x) = lim (1 - e ) = 1 so d) holds.

xL 8 xL 8

Since all conditions are satisfied therefore F(x) is

a distribution function.

8i

EX = 2 xf(x)dxj-8

(2 0 if x < 0

f(x) = F’(x) = {-x22 e if x > 0.

98 0 8i i i -x

EX = 2 xf(x)dx = 2 0dx + 2xe dx =j j j-8 -8 0

q===============================================================================================eb b2 -x 2

2v’ = e u = x 2 bi -x -x i -x= lim 2xe dx = 2 2 = lim [-xe 1 + 2e dx]

j jbL8 2 -x 2 bL8 02v = -e 20 2 u’= 0 2 0z===============================================================================================c

b-b -x -b -b= lim [-be + (-e 1 )] = lim [-be -e + 1] = 1

bL8 bL80

-4P(-1 < X < 4) = F(4) - F(-1) = 1 - e

QUESTION 39.


as follows:(2 x--------------- if 0 < x < 4

cf(x) = {22 0 otherwise9

Find: a) c2

b) distribution of Y = X


a) f(x) is a density function if and only if f(x) > 0 for all x

8i

and 2 f(x)dx = 1j-8

8 0 4 82 4i i i x i 1 x

2 f(x)dx = 2 0dx + 2--------------- dx + 2 0dx = --------------- [ -------------------- 1 ] =j j j c j c 2

0-8 -8 0 4

8= --------------- = 1 so c = 8.

c(2 x--------------- if 0 < x < 4

8f(x) = {22 0 otherwise9

b) f(x) > 0 if x e (0,4) , so (0,4) - set of all values taken by X.

2Y = g(X) = X .g’(x) = 2x > 0 for x e (0,4).

Therefore Y takes values from (g(0), g(4)) = (0,16)q6===== q6=====2 -1

y = x so x = ey for x e (0,4) so g (y) = ey.

Y is a continuous random variable with density function given by:

-1( -1 dg (y)2 f(g (y))1----------------------------------1 if y e (0,16)

dyh(y) = {2 0 otherwise9

-1dg (y) 1

------------------------------------------------ = ---------------q6=====dy2ey

(2 62 ry 12 --------------- -------------------- if y e (0,16)6h(y) = { 8 2ry222 0 otherwise9

( 12 --------------- if 0 < y < 162 16

Then h(y) = {22 0 otherwise9

QUESTION 40.

Formulate and prove the Chebychev’s inequality.

Solution question 40.2

Chebychev’s inequality If EX = m, var(X) = s < 8, then for

any e > 0

2s

P(1X - m1 > e ) < ---------2e

Proof.2 2 2s = var(X) = E(X - m) = E((X - m) I ) =

W2 2

E((X - m) (I + I )) > E((X-m) I ))[w;1X-m1<e] [w;1X-m1>e] [w;1X-m1>e]

2 2> Ee I = e P(1X-m1 > e).

[w;1X-e1>e]

Hence

2s

P(1X - m1 > e) < -------------------2e

QUESTION 41.

Suppose that a joint density function for a random vector (X,Y)

is given by:

(2 c(x+y) for 0 < x < y < 1

f(x,y) = {2 0 otherwise9

Find: a) The constant c.

b) F(y)

c) P( Y < X + 0.5)


8 8i i

f(x,y) - joint density function 46 a) 2 2 f(x,y)dxdy = 1j j-8 -8

b) A f(x,y) > 0.x,y

D = { (x,y); 0 < y < 1, 0 < x < y}


c-8 -8 D D1 y 1

2( ) ( y )i i i x= 2 {2 2 c(x,y)dx }2dy = c 2 {2 -------------------- + xy 1 }2 dy =

j j j 29 0 9 000 0 0

1 12 2 3 1i y 2 i 3y y c

= c 2 -------------------- + y dy = c 2 ------------------------- dy = c --------------------1 = --------------- = 1j 2 j 2 2 200 0

so c = 2.

yi

F(y) = P( Y < y) = 2 f (t)dtj 2-8

where8i

f (y) = 2 f(x,y)dx2 j

-8

8 0 y 8i i i i

y e (0,1) +++++6 f (y) = 2 f(x,y)dx = 2 0 dx + 22(x+y)dx + 20 dx2 j j j j

-8 -8 0 y

2 yx 2= 2 ( -------------------- + xy 1 ) = 3y

20

8i

y e/ (0,1) +++++6 f (y) = 2 0 dx = 02 j

-8

(2 2

3y if 0 < y < 1f (y) = {2

22 0 otherwise9

yi

F(y) = 2 f (t)dtj 2-8

yi

y < 0 +++++6 F(y) = 2 0dt = 0j-8

0 yi i 2 3 y 3

0 < y < 1 +++++6 F(y) = 2 0dt + 23t dt = t 1 = yj j 0-8 0

0 1 yi i 2 i 3 1

y > 1 +++++6 F(y) = 2 0dt + + 23t dt + 20dt = t 1 = 1j j j 0-8 0 1

(2 0 if y < 022 3

F(y) = { y if 0 < y < 12222 1 if y > 19

P(Y < X + 0.5) = P((X,Y) e C) + P((X,Y) e C ) =1

i i i i= 2 2f(x,y)dxdy + 2 2f(x,y)dxdy =

j j j jC C

1

C C1

C = {(x,y); 0 < y < 0.5,0 < x < y } u {(x,y); 0.5 < y < 1,y - 0.5 < x < y}

i i i i2 22(x+y)dxdy + 2 20dxdy =j j j jC C

1

0.5 y 1 yi i i i

= 2 22(x+y)dxdy + 2 2 2(x+y)dxdy =j j j j0 0 0.5 y-0.5

0.5 1y yi 2 i 2

= 2 (x +xy1 )dy + 2 (x +xy1 )dy =j j

0 y-0.50 0.5

0.5 1i 2 i 2 2

= 2 3y dy + 2 3y - (y-0.5) - 2(y-0.5)ydy =j j0 0.5

0.5 10.5 1i 2 i 3 2

= 2 3y dy + 2 2y -0.25dy = y 1 + (y - 0.25y)1 = 0.75.j j 0 0.50 0.5

P(Y < X +0.25) = 0.75.

QUESTION 42.


X \ Y 2 4 5[----------i---------------------------------------------------------------------------------------------------------]1 p 0.1 0.1 0.2

p2 p 0.2 0.3 0.1

p

Find: a) P( X = 2), b) P(Y > 1), c) P(X < 2/ Y < 4.5)


a) P( X = 2) = P( (X,Y) e{(2,2),(2,4),(2,5)} = 0.2 + 0.3 + 0.1 = 0.6

b) P(Y > 1) = P((X,Y)e { (1,2),(1,4),(1,5),(2,2),(2,4),(2,5)}) = 1

c) P(X < 2/Y < 4.5) = P(X < 2, Y < 4.5)/ P( Y< 4.5) =

= P((X,Y) e{(1,2),(1,4)})/P( (X,Y) e {(1,2),(1,4),(2,2),(2,4)})

= 0.2/0.7 = 2/7

QUESTION 43.

Let X, Y be two independent identically distributed random variables with

common density function(2 -t

e if t > 0f(t) = {

22 0 otherwise9

Find the distribution function of Z = max {X,Y}.


X and Y are identically distributed so thay have the same distribution function F(

x 2 0 if2 x < 0

i xF(x) = P(X < x) = 2 f(t)dt = { -t -t x -xj i e dt = -e 1 = 1 - e if x > 0

2-8 2 0 0

9X and Y are independent so for all x,y

P( X < x,Y < y) = P(X < x)P(Y < y) = F (x)F (y) =F(x)F(y)X X

Therefore

F (z) = P(Z < z) = P( max{X,Y} < z) = P(X < z, Y < z) = P(X < z)P(Y < z) =Z

(2 0 if z < o

2= F (z)F (z) = [F(z)] = {

X Y -z 22 (1 - e ) if z > 09

QUESTION 44.

A point (x,y) is to be selected from the square S containing all points

(x,y) such that 0 < x < 1 and 0 < y < 1. Suppose that each point has the

same chance to be selected. Find the probability that the selected point

will be from the area given by 1/2 < x + y < 3/2.


1

0.5 DD = {(x,y);0.5 < x + y < 1.5}

0.5 1 x

Applying geometric probability we have

area of D 1 - 2W0.5W0.5W0.5 3P(D) = --------------------------------------------- = ------------------------------------------------------------------------------------------ = ---------------

area of W 1 4

QUESTION 45.

Let A and B be two independent events such that P(A) = 1/3, P(B) = 1/2.

Find P(AuB).


P(AuB) = P(A) + P(B) - P(AnB)

Since A and B are independent

P(AnB) = P(A)P(B)

Therefore

P(AuB) = P(A) + P(B) - P(A)P(B) =

1 1 1 1 2= --------------- + --------------- - --------------- --------------- = ---------------

3 2 3 2 3

QUESTION 46.

Suppose that the distribution function of a random variable X is given by:

& 0 for x < -12 0.25 for -1 < x < 0

F(x) = { 0.45 for 0 < x < 22 0.75 for 2 < x < 42 0.95 for 4 < x < 67 1 for x > 6

Find: a) The distribution of X

b) P( -1 < X < 3.5)

c) P( X > 1.5)

d) EX and Var(X)


Since F(x) has a jumps at certain points and is constant in between

therefore X is a random variable of discrete type with values equal

to the points where F has jumps and corresponding probabilities equal

to the size of jumps.

Hence distribution is given by

x 1-1 1 0 1 2 1 4 1 6i-------------------------k--------------------k--------------------k--------------------k--------------------k-------------------------

p 10.2510.2010.3010.2010.05i

P(a < X < b) = F(a) - F(b)

P(-1 < X < 3.5) = P(Xe{-1,0,2}) = 0.75

P(X > 1.5) = 1 - P(X < 1.5) = 1 - F(1.5) = 1 - 0.45 = 0.55

EX = S x p = 1.45i i

2 2 2 2Var(X) = E(X - EX) = E(X ) - (EX) = 6.45 - (1.45) = 4.39

2 2E(X ) = S x p = 6.45

i iQUESTION 47.

Let (X,Y) be a random vector with a common density function given by:

1 + xy& ------------------------------ if -1 < x < 1, -1 < y < 1f(x,y) = { 4

7 0 otherwise

Find a) P( X < Y/ Y > 0.1 )

b) Var(X)


P( X < Y, Y > 0.1 )P( X < Y/ Y > 0.1 ) = ---------------------------------------------------------------------------------------------------------

P(Y > 0.1)

yC

y = xy = 1

BA

y = 0.1

1 +xy 1 +xyP(X < Y, Y > 0.1) = Ii i -----------------------------------dxdy + i i -----------------------------------dxdy + i i0dxdy =

4 4A B C

0.1 1 1 1i i 1 +xy i i 1 +xy

= 2 2 -----------------------------------dydx + 2 2 -----------------------------------dydx =j j 4 j j 4

-1 0.1 0.1 x0.1 1

11 i 2 i 2 1= ---------------[ 2 {y + xy /21 }dx + 2 {y + xy /21 }dx] =

4 j j x0.1-1 0.10.1 1

1 i i 3= --------------- [ 2 0.9 + 0.99 x/2 dx + 2 1 - x/2 - x /2dx] =

4 j j-1 0.1

0.1 11 2 2 4= ---------------[(0.9 x + 0.99 x /41 ) + ( x - x /4 - x /81 )]=

4 -10.1

= 0.3181

C

D

1 11 + xy i i 1 + xy

P(Y > 0.1 ) = Ii i----------------------------------------dxdy + i i0dxdy = 2 2 ---------------------------------------- dydx =4 j j 4

D C-1 0.1

1 11 i 2 1 1 i

= --------------- 2 { y + xy /21 }dx = --------------- 2 {0.9 + 0.99 x/2 }dx =4 j 4 j

0.1-1 -1

11 2= --------------- (0.9 x + 0.99 x /41 ) = 0.45

4 -1

0.3181P( X < Y/ Y > 0.1 ) = ----------------------------------- = 0.7068

0.452 2 2

Var(x) = E(X - EX) = EX - (EX)

8i

EX = 2 xf(x)dxj-8

8i

f(x) = 2 f(x,y)dyj-8

y

x

8i

if -1 < x < 1 then f(x) = 2 f(x,y)dy =j-8

-1 1 81i i 1 + xy i 1 2

= 2 0dy + 2 ----------------------------------------dy + 20dy = ---------------( y + xy /21 ) =j j 4 j 4 -1-8 -1 1

1= --------------- ( 1 + x/2 +1 -x/2) = 0.5

4

8 8i i

if x < -1 or x > 1 then f(x) = 2 f(x,y)dy = 2 0dy = 0j j-8 -8

Hence(2 0.5 if -1 < x < 1

f(x) = {2 0 otherwise98 1i i 2 1

EX = 2 xf(x)dx = 2 0.5xdx = 0.5 (x /21 ) = 0j j -1-8 -18 1

2 i 2 i 2 3 1EX = 2 x f(x)dx = 2 0.5x dx = 0.5 ( x /31 )= 0.33

j j -1-8 -1

Hence Var(X) = 0.33 - 0 = 0.33

QUESTION 48.


as follows:(2 22 (9 - x )/36 for -3 < x < 3


Find: a) P(X < 0)

b) The distribution function F(x)

c) P( X > -0.5/ X < 2)


2f(x) = 0 (9-x )/36 0

[-----------------------------------------------------------------------------------------------k------------------------------------------------------------k----------------------------------------------------------------------k--------------------------------------------------------------------------------------------------------------L

-3 0 3

00 -3 0 2 & 3 1 * 1

P(X < 0) = i f(x)dx = i 0dx + i (9-x )/36dx = 1/36 9x - x /3 = --------------------7 1 8 2

-8 -8 -3 -3

xF(x) = P( X < x) = i f(t)dt

-8x x

if x < -3 then F(x) = P( X < x) = i f(t)dt = i 0dt = 0-8 -8

x -3 x 2if -3 < x < 3 then F(x) = P( X < x) = i f(t)dt = i 0dt + i (9-t )/36dt

x -8 -8 -3& 3 1 * 3

1/36 9t - t /3 = (9x - x /3+18)/367 1 8

-3 8x -3 3 2

if x > 3 then F(x) = P( X < x) = i f(t)dt = i 0dt + i (9-t )/36dt+ i0dt=1-8 -8 -3 3

&0 if x < -3

23

so F(x) = { (9x - x /3+18)/36 if -3 < x < 37 1 if x > 3

P(-0.5 < X < 2 ) 0.549P( X > -0.5 / X < 2) = -------------------------------------------------------------------------------- = ------------------------- = 0.593

P( X < 2 ) 0.9252 2

2 3P(-0.5 < X < 2) = i (9 - x )/36 dx = 1/36 ( 9x - x /3)1 = 0.549

-0.5-0.5

P( X < 2) = P( X < 2) = F(2) = 0.925

QUESTION 49.

Let X be a random variable. Is Z =1X1 a random variable ?



Definition: A real function X : W -------------------------L R is called a random variable (r.v.)

if the inverse images under X of all Borel sets in R are events, that is,-1

1) X (B) = { w; X(w) e B } e G for all B e B(R)

This condition is equivalent to the following:

X is an r.v. if and only if for each x e R-1

2) X ((-8,x]) = { w; X(w) < x} = { X < x } e G


Z ((-8.x]) e G .

We have-1

Z ((-8.x]) = { w;Z(w) < x} = { w; 1X(w)1 < x } = o if x < 0

For x > 0 we have-1

Z ((-8.x]) = { w;Z(w) < x} = { w; 1X(w)1 < x } = {w; -x < X(w) < x } =-1

X ( [-x,x]) J-------------------- this set is in G since interval [-x,x]

is a borel set and 1) holds (because X is a random variable).

Therefore Z = 1X1 is a random variable.

QUESTION 50.

Let X, Y be two independent identically distributed random variables with

common density function(2 -t

e if t > 0f(t) = {

22 0 otherwise9

Find the distribution function of Z = max {X,Y}.


X and Y are identically distributed so they have the same distribution function F(

x 2 0 if2 x < 0

i xF(x) = P(X < x) = 2 f(t)dt = { -t -t x -xj i e dt = -e 1 = 1 - e if x > 0

2-8 2 0 0

9X and Y are independent so for all x,y

P( X < x,Y < y) = P(X < x)P(Y < y) = F (x)F (y) =F(x)F(y)X X

Therefore

F (z) = P(Z < z) = P( max{X,Y} < z) = P(X < z, Y < z) = P(X < z)P(Y < z) =Z

(2 0 if z < o

2= F (z)F (z) = [F(z)] = {

X Y -z 22 (1 - e ) if z > 09

QUESTION 60.


Let X be a random variable on a probability space (W,G,P) of thek

continuous type with the density function f(x). Let E1X1 < 8 for

some k > 0. Thenk

n P(1X1 >n) ----------L 0 as n ----------L 8.


We have

8 nk k

8 > i 1x1 f(x)dx = lim i 1x1 f(x)dx .-8 n[-----L 8 -n

It follows that-n 8

& k k *lim i 1x1 f(x)dx + i 1x1 f(x)dx = 0

7 8n[-----L 8 -8 n

Let us notice that for x e (-8,-n) or x e (n,8)k k

1x1 > nUsing it we get

-n 8 -n 8k k k k

i1x1 f(x)dx + i1x1 f(x)dx > i n f(x)dx + i n f(x)dx-8 n -8 n

-n 8 -n 8k k k k

> i n f(x)dx + i n f(x)dx = n ( i f(x)dx + i f(x)dx) = n P{1X1 > n}-8 n -8 n

completing the proof.

QUESTION 61.

Let (X,Y) be the random vector with the joint density function given by:

&1 for 0 < x < 1, 0 < y < 1

f(x,y)= {7 0 otherwise

Find the distribution of U(X,Y) = ( X + Y, X - Y).


Let U(X,Y) = ( X + Y, X - Y) ( u (x,y) = x + y, u (x,y) = x - y)1 2

& u = x + y u + u u - u1 1 2 1 2

Solving { we have x(u ,u ) = ---------------------------------------------, y(u ,u )= --------------------------------------------- .1 2 2 1 2 2u = x - y7 2

Then the Jacobian of the inverse function is given by ( h = x(u ,u ),1 1 2

h = y(u ,u ))2 1 2

q e2 2dh dh q e2 21 12 2 2 2--------------- --------------- 1 12 2 2 2du du --------------- ---------------2 2 2 21 2 2 2 1

J = det2 2 = det 2 2 = - ---------------2dh dh 1 12 2 2 22 2 --------------- ----------------2 2 2 2--------------- --------------- 2 22 2 2 2du du2 21 2 z c2 2

z cand the joint density of U = (U ,U ) is given by

1 2f (u ,u ) = 1J1Wf (h (u ,u ),h (u ,u )) =U 1 2 X 1 1 2 2 1 2

u + u u - u1 2 1 2& 1/2 if 0 < ---------------------------------------- < 1, 0 < ---------------------------------------- < 12 2= {

7 0 otherwise

QUESTION 62.

Let X be a random variable with the moment generating function-n/2

given by M (t) = (1 - 2t)X

3Find variance of X.

Solution question 62.2 2 2

Var(X) = E(X - EX) = EX - (EX)

-n/2 n -n/2 -1EX = (M (t))’ = ((1 2t) )’ = (------(1-2t) (-2)) = n1X 10 10 2 0

2 -n/2 -1EX = (M (t))" = (n(1-2t) )’ =

X 10 10n -n/2 - 2 2

= n(------ - 1)(1-2t) (-2) = n + 2n2 102 2

Var(X) = n + 2n - n = 2n

QUESTION 63.

Let (X,Y) be a random vector with joint density given by(2 22 3x

2xy + ------------------------------ if 0 < x < 1, 0 < y < 1f(x,y) = { 2

222 0 otherwise9

Find the line of regression of Y on X


cov(X,Y)y - EY = ----------------------------------------(x - EX)

Var(X)

I1j---------------------------------------------o1 11 2 1

3x12xy+--------------- 1 D = {(x,y);0<x<1,0<y<1}

21 1---------------------------------------------k---------------------------------------------k---------------------------------------------L

1 18i

f(x) = 2 f(x,y)dy = 0 if x e/ (0,1)j

-88 0 1 8

2 2i i i 3x i 2 3yx 1

f(x) = 2 f(x,y)dy = 2 0dy + 2 2xy+---------------dy + 20dy = xy + --------------------1 =j j j 2 j 2

0-8 -8 0 1

23x

= x + --------------- if x e (0,1)2

8 0 1 82

i i i 3x iEX = 2 xf(x)dx = 2 0dx + 2x(x + --------------- )dx + 20dx =

j j j 2 j-8 -8 0 1

3 4 1 17= x /3 + 3x /8 1 = -------------------- = 0.7083

2408 0 1 8

22 i 2 i i 2 3x i

EX = 2 x f(x)dx = 2 0dx + 2x (x + --------------- )dx + 20dx =j j j 2 j

-8 -8 0 1

4 5 1 11= x /4 + 3x /10 1 = --------------------

200

2 2 11 17 2Var(X) = EX - (EX) = -------------------- - (--------------------) = 0.048

20 24

8i

f(y) = 2 f(x,y)dx = 0 if y e/ (0,1)j

-88 0 1 8

2i i i 3x i

f(y) = 2 f(x,y)dx = 2 0dx + 2 2xy+---------------dx + 20dx =j j j 2 j

-8 -8 0 12 3 1

= x y + x /2 1 = y + 1/2 if y e (0,1)0

8 0 1 8i i i 2 y i 3 2 1 7

EY = 2 yf(y)dy = 2 0dy + 2y + ----- dy + 20dy = y /3 + y /4 1 = -------------------- = 0.5833j j j 2 j 120

-8 -8 0 1

cov(X,Y) = E(X - EX)(Y - EY) = EXY - EXEY

8 8 11 12 3 4

i i ii 3x i 2 x 3yx 1EXY = 2 2 xyf(x,y)dxdy = 22xy(2xy+---------------)dxdy = 2(2y -------------------- + -------------------- 1 )dy =

j j jj 2 j 3 8 0-8 -8 00 01i 2 3 2 1

= 2 2y /3 + 3y/8 dy = 2y /9 - 3y /16 1 = 0.4097j 00

cov(X,Y) = E(X - EX)(Y - EY) = EXY - EXEY = 0.4097 - 0.7083Q0.5833 = -0.0034

cov(X,Y)y - EY = ----------------------------------------(x - E)

Var(X)

y - 0.5833 = -0.0034/0.048(x - 0.7083)

y = -0.0708x +0.6334

QUESTION 64.

Let X ,X ,...,X ,... be i.i.d. random variable with joint1 2 n

density function given by(2 12--------------- if 0 < x < 42

4f(x) = {

22 0 otherwise29

PLet M = max {X ,X ,...,X ). Prove that M ------------------------------L 4 as n --------------------L 8.

n 1 2 n nSolution question 64.

We have to show that

A lim P(1M - 41 > e) = 0ne>0 nL8

Let e > 0 ( e < 4) be fixed

P(1M - 41 > e) = 1 - P(1M - 41 < e) = 1 - P(4-e < M < 4 + e) =n n n

1 - P( 4 - e < M ) { since values taken by each X are in (0,4)n i

so values taken by M ) = P( M < 4 - e) =n n

P(max{X ,X ,...,X } < 4 - e) = P(X < 4-e,X < 4-e,...,X < 4 - e)1 2 n 1 2 n

= P(X < 4-e)P(X < 4-e)...,P(X < 4 - e) ={since identical dist.)1 2 n

4-e 0 4-en i n i i 1 n

= P(X < 4-e) = ( 2 f(x)dx) = ( 2 0dx + 2 ---------------dx) =1 j j j 4

-8 -8 0n4 - e 4 - e

= (-------------------------) ------------------------------L 0 as n --------------------L 8 since 1-------------------------1 < 1.4 4

PHence M ------------------------------L 4 as n --------------------L 8.

n

QUESTION 65.

Let X be a random variable with the density function f(x) given by(2 1/2 for -1 < x < 1


2Find the distribution of Z = X

Solution question 65.Our random variables take values from (-1,1). Z has the values [0,1)Our function can be restricted to two 1:1 functions

2 2Let Y = X ( g(x) = x ). D = (-1,0), D = [0,1)

1 2-1 -1 6 -1 -1 6

g (y) = g (y) = - ry, g (y) = g (y) = ry , y e [0,1) (C = [0,1)).1 1D 2 1D

1 2& -1 d -1 -1 d -1

f(g (y))W1--------------- g (y)1 + f(g (y))W1--------------- g (y)1 if y e [0,1)2 1 dy 1 2 dy 2h(y) = {

7 0 otherwise&

1 1 1 12 --------------- 1--------------------1 + --------------- 1---------------1 if y e (0,1]2 q6===== q6=====2 2-2ey 2ey= {

27 0 otherwise

1&------------------------- if y e (0,1]q==========2 e2 y= {

7 0 otherwise

QUESTION 66.

Let X be a random variable with the moment generating function( )-12 2t

given by M (t) = 21 - ---------------2 t < l.X l2 2

9 0Find variance of X.



( ) ( )( ) ’ ( )2 2 2 2-1 -22 2 2 2 2 2 2 2t t 1 1

EX = (M (t))’ = 2 21 - ---------------2 2 = 2 21 - ---------------2 --------------- 2 = ---------------X 10 l l l l2 2 2 2 2 2 2 2

2 2 2 29 0 9 09 010 9 010

( ) ( )( ) ’ ( )2 2 2 2-2 -32 2 2 2 2 2 2 22 t 1 t 1 2

EX = (M (t))" = 2 21 - ---------------2 --------------- 2 = 2 221 - ---------------2 ------------------------- 2 = -------------------------X 10 l l l 2 22 2 2 2 2 2 2 2l l2 2 2 29 0 9 0

9 010 9 010( )

22 22 1 1Var(X) = ------------------------- - 2 --------------- 2 = -------------------------

2 l 22 2l l9 0

QUESTION 67.

Let X ,X ,...,X ,... be i.i.d. random variable with joint1 2 n

density function given by

(2 12--------------- if 0 < x < 42

4f(x) = {

22 0 otherwise29

PLet M = max {X ,X ,...,X ). Prove that M ------------------------------L 4 as n --------------------L 8.

n 1 2 n nSolution question 67.


A lim P(1M - 41 > e) = 0ne>0 nL8

Let e > 0 ( e < 4) be fixed

P(1M - 41 > e) = 1 - P(1M - 41 < e) = 1 - P(4-e < M < 4 + e) =n n n

1 - P( 4 - e < M ) { since values taken by each X are in (0,4)n i

so values taken by M ) = P( M < 4 - e) =n n

P(max{X ,X ,...,X } < 4 - e) = P(X < 4-e,X < 4-e,...,X < 4 - e)1 2 n 1 2 n

= P(X < 4-e)P(X < 4-e)...,P(X < 4 - e) ={since identical dist.)1 2 n

4-e 0 4-en i n i i 1 n

= P(X < 4-e) = ( 2 f(x)dx) = ( 2 0dx + 2 ---------------dx) =1 j j j 4

-8 -8 0n4 - e 4 - e

= (-------------------------) ------------------------------L 0 as n --------------------L 8 since 1-------------------------1 < 1.4 4

QUESTION 68.


Let { X } be a sequence of random variables. Thenn

P -1 Pmax 1X 1---------------L 0 ++++++++++6 1n S 1---------------L 0

k n1<k<n

where S = X + ...+ Xn 1 n

Solution question 68.P

Let Z = max 1X 1. Since Z ---------------L 0 then we haven k n1<k<n

A P(1Z 1 > e) --------------------L 0 as n ---------------L 8ne>0


A P(1n S 1 > e) ---------------L 0 as n ----------L 8ne>0

-1Let us notice that if each 1X 1 < d then 1n S 1 < d

i nand als that P(AnB) < P(A)

Let e,d such that 0 < d < e < 8 be fixed.-1 -1

P(1n S 1 > e) = 1 - P(1n S 1 < e) =n n

-1 -1= 1 - P(1n S 1 < e, max 1X 1 < d ) - P(1n S 1 < e, max 1X 1 < d )

n k n k1<k<n 1<k<n

Let us notice-1

P(1n S 1 < e, max 1X 1 < d ) = P( max 1X 1 < d ) ----------L 1 as n ----------L 8n k k1<k<n 1<k<n

and-1

0 < P(1n S 1 < e, max 1X 1 > d ) < P( max 1X 1 > d ) ----------L 0 as n ----------L 8.n k k1<k<n 1<k<n

Therefore we get-1

P(1n S 1 > e) ---------------L 0 as n ----------L 8 which end the proof.n

QUESTION 69.

Let X , X , X , ... be a sequence of random variables defined on the1 2 3

probability space (W,G,P) and F (x), F (x), F (x), ... be a sequence1 2 3

of corresponding distribution functions.D P

Prove that X -----------------------------------L a ++++++++++6 X -----------------------------------L a, as n -----------------------------------L 8, wheren n

a is a constant.


The limit random variable X is a constant with the distribution function

F(x) = 0 if x < a and F(x) = 1 if x > a. F(x) is continuous for all

x $ a. By the definition of convergence in distribution we have

lim F (x) = F(x) for all x $ a.n[L8 n

PIn order to prove the X -----------------------------------L a we have to show that

nA lim P(1X - X1 > e) = 0

n[L8 ne > 0

Let e > 0 be fixed. Let d > 0 be fixed such that 0 < d < e

P(1X - X1 > e) = 1 - P(1X - a1 < e) = 1 - P( -e < X - a < e)n n n

= 1 - P( a - e < X < a + e) < 1 - P( a - e < X < a + e - d) =n

= 1 - F (a + e - d) - F (a - e) [------------------------------L 1 - 1 + 0 as n [--------------------L 8 whichn n

ends the proof.

QUESTION 70.

Let X be a r.v. with density

& 0 if x < 0f(x) = {

1/2 if 0 < x < 12

27 1/(2x ) if 1 < x

Find the density of Y = U(X) = 1/X


1

1

U(x) is 1:1 for x > 0 and all values taken by X are greater than 0

(since f(x) > 0 for x > 0).

(0,1] is transferred into [1,8) by U and (1,8) is transferred into (0,1)

by U.

-1 -1 1u = 1/x so x = 1/u therefore U (u) = 1/u and (U (u))’ = ----- ----------

2u

In this case

& -1 -1f(U (u))1(U (u))’1 if 0 < u < 8

2h(u) = {

2 0 otherwise7

& 1/2 if u > 1-1

f(U (u) = f(1/u) = {{ 2 21/(2(1/u) ) = u /2 for 0 < u < 1

7

Finally we get

& 0 if u < 0f(u) = {

1/2 if 0 < u < 12

27 1/(2u ) if 1 < u

QUESTION 71.


& 2(x+y) for 0 < x < y < 1f(x,y) = {

0 otherwise7



& 2(x+y) for 0 < x < y < 1f(x,y) = {

0 otherwise7

U(X,Y) = ( X + Y, X - Y) ( u (x,y) = x + y, u (x,y) = x - y)1 2

(2 u = x + y2 1 u + u u - u

1 2 1 2solving { we have x = ---------------------------------------------, y = ---------------------------------------------

2 22 u = x - y22 29

Then the Jacobian of the inverse function is given by ( h = x(u ,u ), h = y(u ,u ))1 1 2 2 1 2

q e2 2dh dh q e2 21 12 2 2 2--------------- --------------- 1 12 2 2 2du du --------------- ---------------2 2 2 21 2 2 2 1



1 2f (u ,u ) = 1J1Wf (h (u ,u ),h (u ,u )) =U 1 2 X 1 1 2 2 1 2

& u for (u ,u ) e C1 1 2

= {7 0 otherwise

u +u u -u1 2 1 2

0 < ------------------------- < ------------------------- < 12 2

C = { (u ,u ): 0 < u + u < u - u < 2}1 2 1 2 1 2

C = { (u ,u ): 0 < u < 1, -u < u < 0}u{(u ,u ): 1< u < 2,u -2 < u < 0}1 2 1 1 2 1 2 1 1 2

u2

u = 02

2 uC 1

u u = u - 2= -u2 2 11

QUESTION 72.


given by M (t) = (1 - 2t) .Find variance of X.X



-n/2 n -n/2 -1EX = (M (t))’ = ((1 2t) )’ = (------(1-2t) (-2)) = n1X 10 10 2 0

2 -n/2 -1EX = (M (t))" = (n(1-2t) )’ =

X 10 10n n -n/2 - 2 2

= ------(------ - 1)(1-2t) (-2) = n + 2n2 2 10

2 2Var(X) = n + 2n - n = 2n

QUESTION 73.

Let X be a random variable with the density function f(x) given by(2 1/2 for -1 < x < 1


2Find the distribution of Y = X

Solution question 73.2

Our random variables take values from (-1,1). Y = X has the values2 2

in [0,1), Y = X ( g(x) = x )

Our function g(x) can be restricted to two 1:1 functions

by using the partition of R into D = (-1,0), D = [0,1)1 2

-1 -1 6 -1 -1 6g (y) = g (y) = - ry, g (y) = g (y) = ry , y e [0,1) (C = [0,1)).1 1D 2 1D

1 2& -1 d -1 -1 d -1

f(g (y))W1--------------- g (y)1 + f(g (y))W1--------------- g (y)1 if y e [0,1)2 1 dy 1 2 dy 2h(y) = {

7 0 otherwise&

1 1 1 12 --------------- 1--------------------1 + --------------- 1---------------1 if y e (0,1]2 q6===== q6=====2 2-2ey 2ey= {

27 0 otherwise

1&------------------------- if y e (0,1]q==========2 e2 y= {

7 0 otherwise

QUESTION 74.


given by M (t) = (1 - 2t)X

Find variance of X.



-n/2 n -n/2 -1EX = (M (t))’ = ((1 2t) )’ = (------(1-2t) (-2)) = n1X 10 10 2 0

2 -n/2 -1EX = (M (t))" = (n(1-2t) )’ =

X 10 10

n -n/2 - 2 2= n(------ - 1)(1-2t) (-2) = n + 2n

2 102 2

Var(X) = n + 2n - n = 2n

QUESTION 75.

Let (X,Y) be the random vector with the joint density function given by:

&1 for 0 < x < 1, 0 < y < 1

f(x,y)= {7 0 otherwise



Let U(X,Y) = ( X + Y, X - Y) ( u (x,y) = x + y, u (x,y) = x - y)1 2

& u = x + y u + u u - u1 1 2 1 2

Solving { we have x(u ,u ) = ---------------------------------------------, y(u ,u )= --------------------------------------------- .1 2 2 1 2 2u = x - y7 2

Then the Jacobian of the inverse function is given by ( h = x(u ,u ),1 1 2

h = y(u ,u ))2 1 2

q e2 2dh dh q e2 21 12 2 2 2--------------- --------------- 1 12 2 2 2du du --------------- ---------------2 2 2 21 2 2 2 1



1 2f (u ,u ) = 1J1Wf (h (u ,u ),h (u ,u )) =U 1 2 X 1 1 2 2 1 2

u + u u - u1 2 1 2& 1/2 if 0 < ---------------------------------------- < 1, 0 < ---------------------------------------- < 12 2= {

7 0 otherwise

QUESTION 76.

Let {X } be a sequence of independent identically distributed randomn

n2 s

variables with EX = 0 and EX = 1. Let S = X , n = 1,2,3,...1 n t k

k=1

Sn P

Prove that -------------------- ----------------------------------------L 0 as n ----------------------------------------L 8.n



A limP(1S /n - 01 > e) = 0ne>0 n-----L8

Let e > 0 be fixed but arbitrary. ES = E(X +X +...+X ) = 0n 1 2 n

Var(X ) = 1 so Var(S ) = Var(X +X +...+X ) = Var(X )+Var(X )+...+Var(X ) = ni n 1 2 n 1 2 n

We have to evaluate P(1S - 01 > ne) = P(1S - ES 1 > ne)n n n

If in the Chebychev’s inequality

2s

P(1Z - m1 > e) < --------------------2e

we substitute S in place of Z we will getn

Var(S )n n 1

P(1S - 01 > ne) = P(1S - ES 1 > ne) < ----------------------------------- = -------------------- = --------------- ----------L 0n n n 2 2 2 2ne(ne) n e

as n ---------------L 8 which completes the proof.

QUESTION 77.

Let (X,Y) be a random vector with distribution given by

Y \ X1 0 1 11 1----------------------------------------------------------------------------------------------------1 11 0.25 0.251 1

2 1 0 1 0.5

Find the line of regression of Y on X.


cov(X,Y)y - EY = --------------------------------------------------W(x - EX)

var(X)

Y \ X1 0 1 1 pqj1 1

--------------------------------------------------------------------------------------------------------------------------------------------1 11 0.25 0.25 0.51 12 1 0 1 0.5 0.5

p 0.25 0.75iq

Cov(X,Y) = E(X - EX)(Y - EY) = EXY - EXEY

sEXY = x y p = 0*1*0.25 + 0*2*0 + 1*1*0.25 + 1*2*0.5 =

t i j iji,j

= 0.25 + 1 = 1.25

sEX = x p = 0*0.25 + 1*0.75 = 0.75

t i iqi

sEY = y p = 1*0.5 + 2*0.5 = 1.5

t j jqj

Cov(X,y) = 1.25 - 0.75*1.5 = 0.1252 2 2


2 s 2 2 2EX = x p = 0 *0.25 + 1 *0.75 = 0.75

t i iqi

2Var(X) = 0.75 - 0.75 = 0.75 - 0.5625 = 0.1875

0.125y - 1.5 = ------------------------------(x - 0.75)

0.1875

y - 1.5 = 0.66(x - 0.75)

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