13.7.25!1!Summer 2013! Stat 100A/Sanchez-Introduction to Probability!0!The Moment Generating Function, review and new results."Announcements • Homework 6 due on Wednesday Feb 27th!Summer 2013! Stat 100A/Sanchez-Introduction to Probability!1!Outline"Ross, Section 7.7, example 7i. Include tables 7.1 and 7.2, • Moment generating functions for discrete and continuous random variables !• Properties of a Moment generating function.!• Using the moment generating function to identify the distribution of a sum of random variables and of a linear combination of a random variable. !• The Mgf of the normal distribution!Summer 2013! Stat 100A/Sanchez-Introduction to Probability!2!13.7.25!2!Introduction!A very important function of a random variable X is . The expectation of this function of X, , is called the moment generating function of X. The moment generating function is unique for each random variable. It helps us generate expectations of other nonlinear functions of X, such as !Summer 2013! Stat 100A/Sanchez-Introduction to Probability!3!€ etx€ E etx( )€ E X3( ), E X4( ),etcSummer 2013! Stat 100A/Sanchez-Introduction to Probability!4!1. Moment generating function of a discrete random variable!The moment generating function M(t) of any discrete random variable X is defined for all real values of t by:!€ M(t) = E(etx) = etxx∑p(x)Remember the definition of an expected value of a function of a r.v. g(x): !E(g(x)) = Σ ig(xi)p(xi)!Where P(x) is the probability mass function of X.!Summer 2013! Stat 100A/Sanchez-Introduction to Probability!5!We call M(t) the moment generating function !because all the moments of X can be obtained by successively differentiating M(t) and then evaluating the result at t=0.!Define moments of X:!!E(x) =1st moment!!E(x2) =2nd moment!!E(x3) =3rd moment!13.7.25!3!Summer 2013! Stat 100A/Sanchez-Introduction to Probability!6!€ ʹ′ M (t) =ddtE etx( )[ ]= Eddtetx( )⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = E xetx( )ʹ′ M (0) = E(x)Example: the first moment of a r.v. X!Interchange of the differentiation and expectation operators is legitimate: !€ ddtetxp(x)x∑⎡ ⎣ ⎢ ⎤ ⎦ ⎥ =ddtetxp(x)[ ]x∑Summer 2013! Stat 100A/Sanchez-Introduction to Probability!7!Example: the second moment of a r.v. X!€ ʹ′ ʹ′ M (t) =ddtʹ′ M (t)=ddtE (xetx)= Eddtxetx( )= E (x2etx)ʹ′ ʹ′ M (0) = E (x2)Summer 2013! Stat 100A/Sanchez-Introduction to Probability!8!In general, the nth derivative of M(t) is given by:!€ Mn(t) = E(xnetx)Mn(0) = E (xn)⎫ ⎬ ⎭ n ≥113.7.25!4!Summer 2013! Stat 100A/Sanchez-Introduction to Probability!9!Example 1.1 The moment generating function for the binomial distribution: X ~ Bin(n, p)!€ M(t) = E(etx)= etknk⎛ ⎝ ⎜ ⎞ ⎠ ⎟ pk(1− p)n−kk= 0n∑=nk⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ( pet)k(1− p)n−kk= 0n∑= ( pet+ 1− p)nBinomial Theorem: !€ (x + y)n=nk⎛ ⎝ ⎜ ⎞ ⎠ ⎟ xkyn−kk= 0n∑Summer 2013! Stat 100A/Sanchez-Introduction to Probability!10!Differentiation yields:!€ ʹ′ M (t) = n( pet+ 1− p)n−1petE(x) =ʹ′ M (0) = npand thus:!Summer 2013! Stat 100A/Sanchez-Introduction to Probability!11!2nd moment and variance for X~Bin:!€ ʹ′ M (t) = n( pet+ 1− p)n−1petʹ′ ʹ′ M (t) = n(n −1)( pet+ 1− p)n− 2( pet)2+n( pet+ 1− p)n−1( pet)E(x2) =ʹ′ ʹ′ M (0) = n(n −1)p2+ npVar(x) = E( x2) − E(x)[ ]2= n(n −1) p2+ np − n2p2= np(1− p)And:!13.7.25!5!Summer 2013! Stat 100A/Sanchez-Introduction to Probability!12!€ M(t) = E (etx)= etxe−λλxx!x= 0∞∑= e−λ(λet)xx!x= 0∞∑= e−λeλet= eλ(et−1)Example 1.2 For X a Poisson r.v. with mean λ:!This Result uses the Maclarin polynomial for ek:! € f (x) = f (0) +ʹ′ f (0)x +ʹ′ ʹ′ f (0)x22!+ +fn(0)xnn!Pn=knn!n= 0∑= ekSummer 2013! Stat 100A/Sanchez-Introduction to Probability!13!€ ʹ′ M (t) =λeteλ(et−1)ʹ′ ʹ′ M (t) =λeteλ(et−1)+ (λet)2eλ(et−1)E (x) =ʹ′ M (0) =λE (x2) =ʹ′ ʹ′ M (0) =λ2+λVar(x) = E (x2) − E (x)[ ]2=λ2+λ−λ2=λ1st and 2nd moments for a Poisson r.v.!Summer 2013! Stat 100A/Sanchez-Introduction to Probability!14!Example 1.3 Suppose that the moment generating function of a r.v. X is given by "M(t) = e3(et-1). What is the P(x = 0)?!13.7.25!6!Summer 2013! Stat 100A/Sanchez-Introduction to Probability!15!Example 1.3 Suppose that the moment generating function of a r.v. X is given by "M(t) = e3(et-1). What is the P(x = 0)?!First, identify the probability distribution by looking at the M(t) formula. It looks like this is the mgf of ?!Summer 2013! Stat 100A/Sanchez-Introduction to Probability!16!Example 1.3 Suppose that the moment generating function of a r.v. X is given by "M(t) = e3(et-1). What is the P(x = 0)?!First, identify the probability distribution by looking at the M(t) formula. It looks like this is the mgf of a Poisson distribution with parameter λ=3. !So use the Poisson formula to compute the probability!€ P(X = 0) =30e−30!= e−3= 0.0497Summer 2013! Stat 100A/Sanchez-Introduction to Probability!17!2. Moment generating functions of a continuous random variable!The moment generating function M(t) of the continuous random variable X is defined for all real values of t by:!€ M(t) = E(etx)= etx−∞∞∫f (x)dx}If X is continuous with density f(x)!Remember the definition of an expected value of a function of a r.v. g(x): !E(g(x)) = ∫ g(x)f(x)dx!13.7.25!7!Summer 2013! Stat 100A/Sanchez-Introduction to Probability!18!As before, we call M(t) the moment!generating function because all the moments of X can be obtained by successively differentiating M(t) and then evaluating the result at t = 0.!€ ʹ′ M (t) =ddtE (etx) = Eddt(etx) = E (xetx)€ ʹ′ M (t)t= 0= E (x)ʹ′ ʹ′ M (t)t= 0= E (x2)ʹ′ ʹ′ ʹ′ M (t)t= 0= E (x3)So:!Summer 2013! Stat 100A/Sanchez-Introduction to Probability!19!Example 2.1. The moment generating function of the exponential distribution with parameter λ!€ M(t) = E (etx)= etx0∞∫λe−λxdx=λetx0∞∫e−λxdx =λe−(λ−t )x0∞∫dx=λλ− tFor t < λ!Summer 2013! Stat 100A/Sanchez-Introduction to Probability!20!E(x) and Var(x) for the exponential distribution with parameter λ!€ ʹ′ M (t) =λ(λ− t)2ʹ′ ʹ′ M (t) =2λ(λ− t)3Var(x) = E(x2) − E (x)[ ]2=1λ2At t = 0, M’(t) = 1/ lambda = E(x)!At t = 0, M’’(t) = 2/ lambda2 =
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