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Homework for 1/8

1. [4-6] Let X be a continuous random variable with probability densityfunction f(x) = 2x, 0 x 1.(a) Find E[X].(b) Find E[X2].(c) Find Var[X].

(a) We have

E[X] =

xf(x) dx =

10

x 2x dx =(

2x3

3

) 10

=2

3.

(b) We have

E[X2] =

x2f(x) dx =

10

x2 2x dx =(x4

2

) 10

=1

2.

(c) We have

Var[X] = E[X2] E[X]2 = 12(

2

3

)2=

1

18.

2. [4-31] Let X be uniformly distributed on the interval [1, 2]. Find E[1/X].Is E[1/X] = 1/E[X]?

Since X is uniformly distributed on [1, 2], the pdf of X is

f(x) =

{1, 1 x 20, otherwise

.

On one hand, we have

E[X] =

xf(x) dx =

21

x 1 dx =(x2

2

) 21

=3

2.

On the other hand, we have

E[

1

X

]=

1

xf(x) dx =

21

1

x 1 dx = (lnx)

21

= ln 2 ln 1.

It follows immediately that E[1/X] 6= 1/E[X].

3. [4-42] Let X be an exponential random variable with standard deviation. Find P(|X E[X]| > k) for k = 2, 3, 4, and compare the results tothe bounds from Chebyshevs inequality.

Since X is exponentially distributed and has variance 2, the pdf of X is

f(x) =

1

ex/, x > 0

0, otherwise.

It follows immediately that E[X] = . Thus we have

P(|X E[X]| > k) = P(X > (k + 1) or X < (1 k))

=

(k+1)

1

ex/ dx we only consider k = 2, 3, 4

=

k+1

ey dy (change of variable y = x/)

= e(k+1).

Therefore, we have

P(|X E[X]| > 2) = e3 = .0498, P(|X E[X]| > 3) = e4 = .0183,P(|X E[X]| > 4) = e5 = .0067.On the other hand, the Chebyshevs inequality gives us

P(|X E[X]| > k) Var[X]k22

=1

k2.

In particular, we have

P(|X E[X]| > 2) 122

= 0.25, P(|X E[X]| > 3) 132

= 0.1111,

P(|X E[X]| > 4) 142

= 0.0625.

4. [4-80] Let X be a continuous random variable with density functionf(x) = 2x, 0 x 1. Find the moment-generating function of X,M(t), and verify that E[X] = M (0) and that E[X2] = M (0).

2

The mgf of X is

M(t) = E[etX ] =

etxf(x) dx =

10

etx 2x dx

= 2

[1

tetxx

10

1t

10

etx dx

]

= 2

[et

t 1t2(et 1)] (integration by parts)

= 2

(et

t e

t

t2+

1

t2

).

Note that by LHospital rule, we have

limt0

M(t) = 2 limt0

tet et + 1t2

= 2 limt0

tet + et et2t

= limt0

et = 1.

Furthermore, we have

M (t) = 2(et

t 2e

t

t2+

2et

t3 2t3

), and

limt0

M (t) = 2 limt0

t2et 2tet + 2et 2t3

= 2 limt0

t2et + 2tet 2tet 2et + 2et3t2

=2

3limt0

et =2

3,

and

M (t) = 2(et

t 3e

t

t2+

6et

t3 6e

t

t4+

6

t4

), and

limt0

M (t) = 2 limt0

t3et 3t2et + 6tet 6et + 6t4

= 2 limt0

t3et + 3t2et 3t2et 6tet + 6tet + 6et 6et4t3

=1

2limt0

et =1

2.

Combing with the first problem (4-6), we see that E[X] = M (0) andE[X2] = M (0).

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5. [4-91] Use the mgf to show that if X follows an exponential distribution,cX(c > 0) does also.

Let the pdf of X be

f(X) =

{ex, x > 00, otherwise

.

Then the mgf of X is

MX(t) =

0

etx ex dx = t

0

( t)e(t) dx = t ,

for all t < . (The trick here is to identify a pdf for Exp( t)). LetY = cX. Then the mgf of Y is

MY (t) = E[etY ] = E[et(cX)] = E[e(ct)X ] = MX(ct) =

ct =/c

/c t ,

for all t with ct < , that is, t < /c. Compare MY (t) to MX(t), we seethat Y is also exponentially distributed and the parameter is /c.

Note that the resultMcX(t) = MX(ct)

is generally true as long as the mgf are defined (for example, we need tomake sure ct < ). More generally, we have

MaX+b(t) = ebtMX(t).

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Homework for 1/10

1. [5-16] Suppose that X1, . . . , X20 are independent random variables withdensity functions

f(x) = 2x, 0 x 1Let S = X1 + + X20. Use the central limit theorem to approximateP(S 10).

From Problem 1 of Homeowrk 1/8, we have

= E[X] =2

3, and 2 = Var[X] =

1

18.

Therefore, according the central limit theorem, we have

P(S 10) = P(S nn 10 n

n

)= P

S 20 23118

20 10 20

23

118

20

P(Z 3.16) = 0.0008,

where Z N(0, 1).

2. [5-17] Suppose that a measurement has mean and variance 2 = 25.Let X be the average of n such independent measurements. How largeshould n be so that P(|X | < 1) = .95?

By applying central limit theorem, we have

0.95 = P(|X | < 1) = P( |X |/n

1700 100

100

)= P

(S100 100 15

10 100 >1700 100 15

10 100

) P(Z > 2) = 0.0228.

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