This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ....
View
Full
Document
This note was uploaded on 02/07/2012 for the course APPM 4520 taught by Professor Manuel during the Fall '11 term at University of Colorado Denver.
 Fall '11
 Manuel

Click to edit the document details