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Unformatted text preview: APPM 4/5520 Solutions to Problem Set Three 1. (a) f ( x ) = p (1 p ) x I { , 1 , 2 ,... } ( x ) M X ( t ) = E [ e tX ] = x =0 e tx p (1 p ) x = p x =0 bracketleftbig e t (1 p ) bracketrightbig x This geometric sum converges if e t (1 p ) &lt; 1. (Note that it is already positive since e t &gt; 0 and since 0 p 1 for the geometric pdf.) So, we have convergence if e t &lt; (1 p ) 1 which implies that t &lt; ln(1 p ) 1 = ln(1 p ). Using the fact that x =0 r n = 1 1 r for  r  &lt; 1, we have that M X ( t ) = p summationdisplay x =0 bracketleftBig e t (1 p ) bracketrightBig x = p 1 1 e t (1 p ) = p 1 e t (1 p ) for t &lt; ln(1 p ). (b) Y = X + 1 So, M Y ( t ) = E [ e tY ] = E [ e t ( X + 1)] = e t E [ e tX ] = e t M X ( t ) = pe t 1 e t (1 p ) for t &lt; ln(1 p ). 2. The moment generating function for X ( , ) is M ( t ) = bracketleftbigg  t bracketrightbigg for t &lt; . So, M ( t ) = bracketleftBig t bracketrightBig 1 d dt bracketleftBig t bracketrightBig = bracketleftBig t bracketrightBig 1 ( t ) 2 and M ( t ) = d dt braceleftbigg bracketleftBig t bracketrightBig 1 ( t ) 2 bracerightbigg = bracketleftBig t bracketrightBig 1 2 ( t ) 3 + ( t ) 2 (  1) bracketleftBig t bracketrightBig 2 ( t ) 2 . Now, E [ X ] = M (0) = bracketleftbigg  bracketrightbigg 1 (  0) 2 = , as desired, and E [ X 2 ] = M (0) = bracketleftBig bracketrightBig 1 2 ( 0) 3 + ( 0) 2 (  1) bracketleftBig bracketrightBig 2 ( 0) 2 = 2 + 2 ....
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 Fall '11
 Manuel
 Probability theory, 2 j, 1 degree

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