Random Variables and Discrete probability Distributions

Random Variables and Discrete probability Distributions

Chapter 7

Random Variables and Probability Distributions

• A random variable is a function or rule that assigns a numerical value to each simple event in a sample space.

• A random variable reflects the aspect of a random experiment that is of interest for us.

• There are two types of random variables:– Discrete random variable– Continuous random variable.

• A table, formula, or graph that lists all possible values a discrete random variable can assume, together with associated probabilities, is called a discrete probability distribution.

• To calculate the probability that the random variable X assumes the value x, P(X = x), – add the probabilities of all the simple events for which

X is equal to x, or– Use probability calculation tools (tree diagram),– Apply probability definitions

Discrete Probability Distribution

• Example – The number of cars a dealer is selling daily were recorded

in the last 100 days. This data was summarized in the table below.

– Estimate the probability distribution, and determine the probability of selling more than 2 cars a day.

Daily sales Frequency0 51 152 353 254 20

100

Developing Probability Distribution and Finding the Probability of an Event

0 1 2 3 4

• Solution– From the table of frequencies we can calculate the

relative frequencies, which becomes our estimated probability distribution

Daily sales Relative Frequency0 5/100=.051 15/100=.152 35/100=.353 25/100=.254 20/100=.20

1.00

.05

.15

.35.25

.20

X

P(X>2) = P(X=3) + P(X=4) =.25 + .20 = .45

Developing Probability Distribution and Finding the Probability of an Event

Describing the Population/ Probability Distribution

• The probability distribution represents a population

• We’re interested in describing the population by computing various parameters.

• Specifically, we calculate the population mean and population variance.

Population Mean (Expected Value)

• Given a discrete random variable X with values xi, that occur with probabilities p(xi), the population mean of X is.

ixall

ii )x(px)X(E ixall

ii )x(px)X(E

– Let X be a discrete random variable with possible values xi that occur with probabilities p(xi), and let E(xi) = The variance of X is defined by

ixall

i2

i22 )x(p)x()X(E)X(V

ixalli

2i

22 )x(p)x()X(E)X(V

Population Variance

2

isdeviationdardtansThe

2

isdeviationdardtansThe

• Solution – continued– The variance can also be calculated as

follows:

The Mean and the Variance

107.1084.2...)2(p2)1(p1)0(p0

)x(px)X(E)X(V

2222

2

xalli

2i

222

i

107.1084.2...)2(p2)1(p1)0(p0

)x(px)X(E)X(V

2222

2

xalli

2i

222

i

Laws of Expected Value E(c) = c E(X + c) = E(X) + c E(cX) = cE(X)

Laws of Variance V(c) = 0 V(X + c) = V(X) V(cX) = c2V(X)

Laws of Expected Value and Variance

Example 1

• The random variable X has the following distribution:

• a. Find the mean and variance for the probability distribution• b. Determine the probability distributions of Y where Y=5X• c. Use the probability distribution in part (b) to compute the mean

and variance of Y• d. Use the laws of expected value and variance to find the expected

value and variance of Y from the parameters of X

x -4 5 7 8

P(x)

.59 .15 .25 .01

Solution

• a. = E(X) = –4(.59) +5(.15) + 7(.25) +8(.01) = .22

• V(X) = (–4–.22)^2(.59) + (5–.22)^2(.15) + (7–.22)^2(.25) + (8–.22)^2(.01) = 26.03

• Also, you can use V(X)=E(X2)-{E(X)}2

=(-4)2(.59)+(5)2(.15)+(7)2(.25)+(8)2(.01)-(.22)2 =26.03

Solution

b. x –4 5 7 8 y –20 25 35 40 P(y) .59 .15 .25 .01

c. E(Y) = –20(.59) + 25(.15) + 35(.25) + 40(.01) = 1.10

V(Y) = (–20–1.10)^2(.59) + (25–1.10)^2(.15) + (35–1.1)^2(.25) + (40–1.1)^2(.01) = 650.8

Solution

d.

E(Y) = E(5X) = 5E(X) = 5(.22) = 1.10

V(Y) = V(5X) = 25V(X) = 25(26.03) = 650.8

• The bivariate (or joint) distribution is used when the relationship between two random variables is studied.

• The probability that X assumes the value x, and Y assumes the value y is denoted

p(x,y) = P(X=x and Y = y)

Bivariate Distributions

Bivariate Distributions

1y)p(x, 2.1y)p(x,0 1.

:conditionsfollowingthesatisfies functiony probabilitjoint The

xall y all

• Example 7.5– Xavier and Yvette are two real estate agents.

Let X and Y denote the number of houses that Xavier and Yvette will sell next week, respectively.

– The bivariate probability distribution is presented next.

Bivariate Distributions

p(x,y)

Bivariate Distributions

X

YX=0 X=2X=1

y=1

y=2

y=0

0.42

0.120.21

0.070.06

0.02

0.06

0.03

0.01

Example 7.5 – continued X

Y 0 1 20 .12 .42 .061 .21 .06 .032 .07 .02 .01

Marginal Probabilities

• Example 7.5 – continued– Sum across rows and down columns

XY 0 1 2 p(y)0 .12 .42 .06 .601 .21 .06 .03 .302 .07 .02 .01 .10p(x) .40 .50 .10 1.00

p(0,0)p(0,1)p(0,2)

The marginal probability P(X=0)

P(Y=1), the marginal probability.

Describing the Bivariate Distribution

• The joint distribution can be described by the mean, variance, and standard deviation of each variable.

• This is done using the marginal distributions. x p(x) y

p(y)0 .4 0 .61 .5 1 .32 .1 2 .1

E(X) = .7 E(Y) = .5V(X) = .41 V(Y) = .45

=

Describing the Bivariate Distribution • To describe the relationship between the two variables

we compute the covariance and the coefficient of correlation– Covariance:

COV(X,Y) = (X – x)(Y- y)p(x,y)

– Coefficient of Correlation

– COV(X,Y) xy

• To describe the relationship between the two variables we compute the covariance and the coefficient of correlation– Covariance:

COV(X,Y) = (X – x)(Y- y)p(x,y)

– Coefficient of Correlation

– COV(X,Y) xy

• Example 7.6– Calculate the covariance and coefficient of c

orrelation between the number of houses sold by the two agents in Example 7.5

– Solution• COV(X,Y) = (x-x)(y-y)p(x,y) =

(0-.7)(0-.5)p(0,0)+…(2-.7)(2-.5)p(2,2) = -.15

• =COV(X,Y)/xy = - .15/(.64)(.67) = -.35

Describing the Bivariate Distribution

Sum of Two Variables

• The probability distribution of X + Y is determined by– Determining all the possible values that X+Y can assume– For every possible value C of X+Y, adding the probabilities of all

the combinations of X and Y for which X+Y = C

• Example 7.5 - continued– Find the probability distribution of the total number of houses

sold per week by Xavier and Yvette.– Solution

• X+Y is the total number of houses sold. X+Y can have the values 0, 1, 2, 3, 4.

P(X+Y=0) = P(X=0 and Y=0) = .12

P(X+Y=1) = P(X=0 and Y=1)+ P(X=1 and Y=0) =.21 + .42 = .63

P(X+Y=2) = P(X=0 and Y=2)+ P(X=1 and Y=1)+ P(X=2 and Y=0) = .07 + .06 + .06 = .19

The probabilities P(X+Y)=3 and P(X+Y) =4 are calculated the same way. The distribution follows

XY 0 1 2 p(y)0 .12 .42 .06 .601 .21 .06 .03 .302 .07 .02 .01 .10p(x) .40 .50 .10 1.00

The Probability Distribution of X+Y

• The distribution of X+Y

• The expected value and variance of X+Y can be calculated from the distribution of X+Y.– E(X+Y)=0(.12)+ 1(63)+2(.19)+3(.05)+4(.01)=1.2– V(X+Y)=(0-1.2)2(.12)+(1-1.2)2(.63)+… =.56

x + y 0 1 2 3 4p(x+y) .12 .63 .19 .05 .01

The Expected Value and Variance of X+Y

– The following relationship can assist in calculating E(X+Y) and V(X+Y)

• E(X+Y) =E(X) + E(Y);

• V(X+Y) = V(X) +V(Y) +2COV(X,Y)

• When X and Y are independent COV(X,Y) = 0, and V(X+Y) = V(X)+V(Y).

The Expected Value and Variance of X+Y

Example2 (Text 7.47&48)

• The bivariate distribution of X and Y is described here.

• a. Find the marginal probability distribution of X.• b. Find the marginal probability distribution of Y• c. Compute the mean and variance of X• d. Compute the mean and variance of Y• e. Compute the covariance and variance of Y

x

y 1 2

1 0.28 0.42

2 0.12 0.18

Solution

• a x P(x) 1 .4 2 .6• b y P(y) 1 .7 2 .3• c E(X) = 1(.4) + 2(.6) = 1.6 V(X) = (1–1.6)^2*(.4) + (2–1.6)^2*(.6) = .24 or (1^2)*0.4+(2^2)*0.6-(1.6)^2=.24

Solution

• d E(Y) = 1(.7) + 2(.3) = 1.3 V(Y) = (1–1.3)^2(.7) + (2–1.3)^2(.3) = .21• e = (1)(1)(.28) + (1)(2)(.12) + (2)(1)(.42) + (2)(2)(.18) = 2.08

• COV(X, Y) = – = 2.08 – (1.6)(1.3) = 0 • = 0

yallxall

)y,x(xyP

yallxall

)y,x(xyPyx

yx

)Y,X(COV

The Binomial Distribution

• The binomial experiment can result in only one of two possible outcomes.

• Typical cases where the binomial experiment applies:– A coin flipped results in heads or tails– An election candidate wins or loses– An employee is male or female– A car uses 87octane gasoline, or another

gasoline.

– There are n trials (n is finite and fixed).– Each trial can result in a success or a failure.– The probability p of success is the same for all the

trials.– All the trials of the experiment are independent.

• Binomial Random Variable– The binomial random variable counts the number

of successes in n trials of the binomial experiment.– By definition, this is a discrete random variable.

Binomial Experiment

Calculating the Binomial Probability

xnxnx )p1(pC)x(p)xX(P xnxn

x )p1(pC)x(p)xX(P

In general, The binomial probability is calculated by:

)!xn(!x!n

Cwhere nx

E(X) = = npV(X) = 2 = np(1-p)

Mean and Variance of Binomial Variable

Example 3(Text 7.102)

• In the game of blackjack as played in casinos in Las Vegas, Atlantic City, Niagara Falls, as well as many other cities, the dealer has the advantages. Most players do not play very well. As a result, the probability that the average player wins a about 45%.Find the probability that an average player wins

a. twice in 5 hands

b. ten or more times in 25 hands

Solution

• a P(X = 2) = = .3369

• b Excel with n = 25 and p = .45:

P(X 10) = 1 – P(X 9) = 1 – .2424 = .7576

)!25(!2

!5

252 )45.1()45(.

The Binomial Distribution

• [插入 ][函數 ][選取類別 (統計 )]BINOMDIST– Number_s 表示二項分配中成功的次數– Trials 總共的實驗次數– Probability_s 成功的機率– Cumulative 打入 true則為累加分配函數，打入 false則為機率函數

• The Poisson experiment typically fits cases of rare events that occur over a fixed amount of time or within a specified region

• Typical cases– The number of errors a typist makes per page– The number of customers entering a service

station per hour– The number of telephone calls received by a

switchboard per hour.

Poisson Distribution

• The number of successes (events) that occur in a certain time interval is independent of the number of successes that occur in another time interval.

• The probability of a success in a certain time interval is– the same for all time intervals of the same size,– proportional to the length of the interval.

• The probability that two or more successes will occur in an interval approaches zero as the interval becomes smaller.

Properties of the Poisson Experiment

• The Poisson Random Variable– The Poisson variable indicates the number of

successes that occur during a given time interval or in a specific region in a Poisson experiment

• Probability Distribution of the Poisson Random Variable.

)X(V)X(E

...2,1,0x!x

e)x(p)xX(P

x

The Poisson Variable and Distribution

Example 4

• The number of students who seek assistance with their statistics assignments is Poisson distributed with a mean of three per day.

a. What is the probability that no student seek assistance tomorrow?

b. Find the probability that 10 students seek assistance in a week.

Solution

• a. P(X = 0 with = 3) = =

= .0498

• b. P(X = 10 with = 21) = =

= .0035

!x

e x

!0

)3(e 03

!x

e x

!10

)21(e 1021

The Poisson Distribution

• [插入 ][函數 ][選取類別 (統計 )]POISSON– X 表示平均時間內發生的次數– Mean 平均值– Cumulative 打入 true為累加分配函數，打入 false則為機率函數