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# hw3sol - APPM 4/5520 Solutions to Problem Set Three 1(a f x...

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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 ) &amp;lt; 1. (Note that it is already positive since e t &amp;gt; 0 and since 0 p 1 for the geometric pdf.) So, we have convergence if e t &amp;lt; (1- p ) 1 which implies that t &amp;lt; ln(1- p ) 1 =- ln(1- p ). Using the fact that x =0 r n = 1 1 r for | r | &amp;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 &amp;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 &amp;lt;- ln(1- p ). 2. The moment generating function for X ( , ) is M ( t ) = bracketleftbigg - t bracketrightbigg for t &amp;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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