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Expectation

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

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Text of Expectation. Let X denote a discrete random variable with probability function p(x) (probability...

• 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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