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Expectation

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Expectation

Let X denote a discrete random variable with probability function p(x) (probability density function f(x) if X is continuous) then the expected value of X, E(X) is defined to be:and if X is continuous with probability density function f(x)

Example: Suppose we are observing a seven game series where the teams are evenly matched and the games are independent. Let X denote the length of the series. Find:The distribution of X. the expected value of X, E(X).

Solution: Let A denote the event that team A, wins and B denote the event that team B wins. Then the sample space for this experiment (together with probabilities and values of X) would be (next slide):

continued

At this stage it is recognized that it might be easier to determine the distribution of X using counting techniques

outcomeAAAABBBBBAAAAABAAAAABAAAAABAProb( )4( )4( )5( )5( )5( )5X445555outcomeABBBBBABBBBBABBBBBABABBBBBBAAAAProb( )5( )5( )5( )5( )6( )6X555566outcomeBABAAABAABAABAAABAABBAAAABABAAABAABAProb( )6( )6( )6( )6( )6( )6X666666outcomeAABBAAAABABAAAABBAAABBBBABABBBABBABBProb( )6( )6( )6( )6( )6( )6X666666

The possible values of X are {4, 5, 6, 7}The probability of a sequence of length x is ()xThe series can either be won by A or B.If the series is of length x and won by one of the teams (A say) then the number of such series is:

In a series of that lasts x games, the winning team wins 4 games and the losing team wins x - 4 games. The winning team has to win the last games. The no. of ways of choosing the games that the losing team wins is:

Thus

The no. of ways of choosing the games that the losing team wins

The no. of ways of choosing the winning team

The probability of a series of length x.

x4567p(x)

Interpretation of E(X)The expected value of X, E(X), is the centre of gravity of the probability distribution of X.The expected value of X, E(X), is the long-run average value of X. (shown later Law of Large Numbers)E(X)

Example: The Binomal distributionLet X be a discrete random variable having the Binomial distribution. i. e. X = the number of successes in n independent repetitions of a Bernoulli trial. Find the expected value of X, E(X).

Solution:

Example: A continuous random variableThe Exponential distributionLet X have an exponential distribution with parameter l.This will be the case if: P[X 0] = 1, and P[ x X x + dx| X x] = ldx.The probability density function of X is:The expected value of X is:

We will determineusing integration by parts.

Summary:If X has an exponential distribution with parameter l then:

Example:The Uniform distributionSuppose X has a uniform distribution from a to b. Then:The expected value of X is:

Example:The Normal distributionSuppose X has a Normal distribution with parameters m and s. Then:The expected value of X is:Make the substitution:

HenceNow

Example:The Gamma distributionSuppose X has a Gamma distribution with parameters a and l. Then:Note:This is a very useful formula when working with the Gamma distribution.

The expected value of X is:This is now equal to 1.

Thus if X has a Gamma (a ,l) distribution then the expected value of X is:Special Cases: (a ,l) distribution then the expected value of X is: Exponential (l) distribution: a = 1, l arbitrary Chi-square (n) distribution: a = n/2, l = .

The Gamma distribution

The Exponential distribution

The Chi-square (c2) distribution

Expectation of functions of Random Variables

DefinitionLet X denote a discrete random variable with probability function p(x) (probability density function f(x) if X is continuous) then the expected value of g(X), E[g(X)] is defined to be:and if X is continuous with probability density function f(x)

Example:The Uniform distributionSuppose X has a uniform distribution from 0 to b. Then:Find the expected value of A = X2 . If X is the length of a side of a square (chosen at random form 0 to b) then A is the area of the square= 1/3 the maximum area of the square

Example:The Geometric distributionSuppose X (discrete) has a geometric distribution with parameter p. Then:Find the expected value of X A and the expected value of X2.

Recall: The sum of a geometric SeriesDifferentiating both sides with respect to r we get: Differentiating both sides with respect to r we get:

ThusThis formula could also be developed by noting:

This formula can be used to calculate:

To compute the expected value of X2. we need to find a formula forNoteDifferentiating with respect to r we get

Differentiating again with respect to r we getThus

impliesThus

Thus

Moments of Random Variables

DefinitionLet X be a random variable (discrete or continuous), then the kth moment of X is defined to be:The first moment of X , m = m1 = E(X) is the center of gravity of the distribution of X. The higher moments give different information regarding the distribution of X.

DefinitionLet X be a random variable (discrete or continuous), then the kth central moment of X is defined to be:where m = m1 = E(X) = the first moment of X .

The central moments describe how the probability distribution is distributed about the centre of gravity, m. and is denoted by the symbol var(X).= 2nd central moment.depends on the spread of the probability distribution of X about m.is called the variance of X.

is called the standarddeviation of X and is denoted by the symbol s.The third central momentcontains information about the skewness of a distribution.

The third central momentcontains information about the skewness of a distribution.Measure of skewness

Positively skewed distribution

Negatively skewed distribution

Symmetric distribution

The fourth central momentAlso contains information about the shape of a distribution. The property of shape that is measured by the fourth central moment is called kurtosisThe measure of kurtosis

Mesokurtic distribution

Platykurtic distribution

leptokurtic distribution

Example: The uniform distribution from 0 to 1Finding the moments

Finding the central moments:

ThusThe standard deviationThe measure of skewnessThe measure of kurtosis

Rules for expectation

Rules:ProofThe proof for discrete random variables is similar.

ProofThe proof for discrete random variables is similar.

ProofThe proof for discrete random variables is similar.

Proof

Moment generating functions

DefinitionLet X denote a random variable, Then the moment generating function of X , mX(t) is defined by:Recall

ExamplesThe moment generating function of X , mX(t) is:The Binomial distribution (parameters p, n)

The moment generating function of X , mX(t) is:The Poisson distribution (parameter l)

The moment generating function of X , mX(t) is:The Exponential distribution (parameter l)

The moment generating function of X , mX(t) is:The Standard Normal distribution (m = 0, s = 1)

We will now use the fact thatWe have completed the squareThis is 1

The moment generating function of X , mX(t) is:The Gamma distribution (parameters a, l)

We use the factEqual to 1

Properties of Moment Generating Functions

mX(0) = 1Note: the moment generating functions of the following distributions satisfy the property mX(0) = 1

We use the expansion of the exponential function:

Now

Property 3 is very useful in determining the moments of a random variable X.Examples

To find the moments we set t = 0.

The moments for the exponential distribution can be calculated in an alternative way. This is note by expanding mX(t) in powers of t and equating the coefficients of tk to the coefficients in:Equating the coefficients of tk we get:

The moments for the standard normal distributionWe use the expansion of eu.We now equate the coefficients tk in:

If k is odd:mk = 0.For even 2k:

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