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### Quiz 1 - Solution

Course: ECON 501, Spring 2011
School: Istanbul Technical...
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Summer EC489.02 2011 Quiz 1 - Solution You have 8 minutes to solve the following questions. 1. For a random variable X with E[X ] = c; answer the following questions. and V ar(X ) = 2 ; and two constants b and (a) (1.5 pts)E [aX + b] =? Answer: E [aX + b] = aE[X ] + b = a + b: (b) (1.5 pts)V ar (aX + b) =? Answer: V ar(aX + b) = a2 V ar(X ) = a2 2 : (c) (2 pts)E[aX + bX 2 ] =? (Hint: remember the denition...

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Summer EC489.02 2011 Quiz 1 - Solution You have 8 minutes to solve the following questions. 1. For a random variable X with E[X ] = c; answer the following questions. and V ar(X ) = 2 ; and two constants b and (a) (1.5 pts)E [aX + b] =? Answer: E [aX + b] = aE[X ] + b = a + b: (b) (1.5 pts)V ar (aX + b) =? Answer: V ar(aX + b) = a2 V ar(X ) = a2 2 : (c) (2 pts)E[aX + bX 2 ] =? (Hint: remember the denition of V ar(X )) Answer: Since V ar(X ) = E[X 2 ] fE[X ]g2 ; we have E[X 2 ] = V ar(X ) + fE[X ]g2 = 2 + 2 : Then E[aX + bX 2 ] = E[aX ] + E[bX 2 ] = aE[X ] + bE[X 2 a ] = + 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 MX (t) = 1 (1 t) : Use this information to nd (a) (2.5 pts)E[X ], Answer: Remember that @MX (t) @t t=0 = E[X ]: Then, @ 1 = (1 t) 2 ( @t (1 t) @MX (t) = E[X ] = : and @t t=0 1 )= (1 t)2 (b) (2.5 pts)E[X 2 ]: Answer: Remember that @ 2 MX (t) @t2 t=0 = E[X 2 ]: Then, 1 @2 2 (1 @t t) = = and @ 2 MX (t) @t2 t=0 @ @t (1 2 (1 t)2 t) ( = E[X 2 ] = 2 2 : 2 3 )= 2 (1 2 t)3
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