STAT 400 Lecture AL1 Examples for 2.3 Spring 2015 Dalpiaz The k th moment of X (the k th moment of X about the origin), µ k , is given by µ k = E ( X k ) = ( )∑⋅xkxfx all The k th central moment of X (the k th moment of X about the mean), µ k' , is given by µ k' = E ( ( X – µ ) k ) = ( ) ()∑⋅−xkxfx all μ The moment-generating function of X, M X ( t ), is given by M X ( t ) = E ( e t X ) = ( )∑⋅xxtxfe all Theorem 1: M X' ( 0 ) = E ( X ) M X" ( 0 ) = E ( X 2 ) M X( k ) ( 0 ) = E ( X k ) Theorem 2: M X 1 ( t ) = M X 2 ( t ) for some interval containing 0 ⇒ f X 1 ( x ) = f X 2 ( x ) Theorem 3: Let Y = a X + b. Then M Y ( t ) = e b t M X ( a t ) 1. Suppose a random variable X has the following probability distribution: x f ( x ) Find the moment-generating function of X, M X ( t ). 10 0.20 11 0.40 M X ( t ) = E ( e t X ) = ( )∑⋅xxtxfe all = 0.20 e 10 t + 0.40 e 11 t + 0.30 e 12 t + 0.10 e 13 t. 12 0.30 13 0.102. Suppose the moment-generating function of a random variable X is M X ( t ) = 0.10 + 0.15 e t + 0.20 e 2 t + 0.25 e – 3 t + 0.30 e 5 t. Find the expected value of X, E(X). M X' ( t ) = 0.15 e t + 0.40 e 2 t – 0.75 e – 3 t + 1.50 e 5 t. E ( X ) = M X' ( 0 ) = 0.15 + 0.40 – 0.75 + 1.50 = 1.30. OR x f ( x ) x ⋅ f ( x ) 0 0.10 0 1 0.15 0.15 2 0.20 0.4 – 3 0.25 – 0.75 E ( X ) = ∑⋅x xfx all)( = 1.30. 5 0.30 1.5 3. Suppose a discrete random variable X has the following probability distribution: f ( 0 ) = P( X = 0 ) = 21 2 e−, f ( k ) = P( X = k ) = !21 kk⋅, k = 1, 2, 3, … a) Find the moment-generating function of X, M X ( t ). M X ( t ) = ∑⋅xxt xf eall)( = 1 ⋅ ( ) 212 e− + ∑∞=⋅⋅1!21 kkktke = ( ) 212 e− + ∑∞=1!2 kktke = ( ) 212 e− + − 1 2 tee = 221 1teee +−.b) Find the expected value of X, E ( X ), and the variance of X, Var ( X ). ( )2 M 2X ' teetet ⋅=, E ( X ) = ( )20M21X ' e=. ( )2 2 M 222X '' tteeeetteet ⋅⋅ +=, E ( X 2 ) = ( )21X 430M'' e⋅=. Var ( X ) = E ( X 2 ) – [ E ( X ) ] 2 = ee⋅⋅−41 4321 . 4. Let X be a Binomial ( n, p ) random variable. Find the moment-generating function of X. M X ( t ) = ( )∑=−−⋅⋅⋅nkknkktppkne0 1 = ()∑=−−⋅⋅⋅nkknktppkne0 1 = [ ( 1 – p ) + p e t ] n. 5. Let X be a geometric random variable with probability of “success” p. a) Find the moment-generating function of X. M X ( t ) = ( )∑∞=−⋅⋅ −11 1 kkktppe = ( )∑∞=−−−⋅⋅⋅111 1)( kkkttpp ee = ( )∑∞=−⋅⋅⋅0 1 nnttee pp = ( )teptep 11 ⋅⋅−−, t < – ln ( 1 – p ).b) Use the moment-generating function of X to find E ( X ). M X' ( t ) = ( )()( )()( )( )2 11 1 11 ttttteeeeeppppp⋅⋅⋅⋅⋅⋅⋅−−−−−−− = ()()2 11 tteepp⋅⋅−−, t < – ln ( 1 – p ). E ( X ) = M X' ( 0 ) = ( )2 pp = p1. 6. a) Find the moment-generating function of a Poisson random variable. M X ( t ) = ∑∞=−⋅⋅0!λ λ kkktkee = ∑∞=−⋅⋅0!λ λ kktkee = teee λλ ⋅⋅− = )(λ 1 −⋅tee. ( ln M X ( t ) ) ' | t = 0 = E ( X ) = µ X ( ln M X ( t ) ) " | t = 0 = E ( X 2 ) – [ E ( X ) ] 2 = σ X2 b) Find E ( X ) and Var ( X ), where X is a Poisson random variable. ln M X ( t ) = λ ( e t – 1 ). ( ln M X ( t ) ) ' = λ e t. ( ln M X ( t ) ) ' | t = 0 = E ( X ) = λ. ( ln M X ( t ) ) " = λ e t. ( ln M X ( t ) ) " | t = 0 = Var ( X ) =
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