Quiz 1 - Solution - (a) (2.5 pts) E [ X ] , Answer :...

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EC489.02 Summer 2011 Quiz 1 - Solution You have 8 minutes to solve the following questions. 1. For a random variable X with E [ X ] = and V ar ( X ) = ± 2 ; and two constants b and c; answer the following questions. (a) (1.5 pts) E [ aX + b ] =? Answer : E [ aX + b ] = a E [ X ] + b = + b: (b) (1.5 pts) V ar ( aX + b ) =? Answer : V ar ( aX + b ) = a 2 V ar ( X ) = a 2 ± 2 : (c) (2 pts) E [ aX + bX 2 ] =? ( Hint V ar ( X ) ) Answer : Since V ar ( X ) = E [ X 2 ] E [ X ] g 2 ; we have E [ X 2 ] = V ar ( X ) + f E [ X ] g 2 = ± 2 + 2 : Then E [ aX + bX 2 ] = E [ aX ] + E [ bX 2 ] = a E [ X ] + b E [ X 2 ] = + b ( ± 2 + 2 ) : 2. Let X be a random variable distributed with the Exponential distribution. Remember that the moment generating function of the Exponential distribution is given by M X ( t ) = 1 (1 ²t ) :
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Unformatted text preview: (a) (2.5 pts) E [ X ] , Answer : Remember that @M X ( t ) @t & & & t =0 = E [ X ] : Then, @ @t ± 1 (1 & ²t ) ² = ³ & (1 & ²t ) & 2 ( & ² ) ´ = ² (1 & ²t ) 2 and @M X ( t ) @t & & & t =0 = E [ X ] = ²: 1 (b) (2.5 pts) E [ X 2 ] : Answer : Remember that @ 2 M X ( t ) @t 2 & & & t =0 = E [ X 2 ] : Then, @ 2 @t 2 ± 1 (1 & &t ) ² = @ @t ± & (1 & &t ) 2 ² = & 2 & (1 & &t ) & 3 ( & & ) = 2 & 2 (1 & &t ) 3 and @ 2 M X ( t ) @t 2 & & & t =0 = E [ X 2 ] = 2 & 2 : 2...
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Quiz 1 - Solution - (a) (2.5 pts) E [ X ] , Answer :...

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