hw7sol - APPM 4/5520 Solutions to Problem Set Seven 1 The...

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APPM 4/5520 Solutions to Problem Set Seven 1. The first population moment is μ 1 = E [ X ] = integraldisplay η x · e - ( x - η ) dx = η + 1 . (Note that since the pdf is that of an exponential with rate 1 that has been shifted by η to the right, the mean is 1 (the mean of the exponential with rate 1) that has been shifted by η to the right!) The first sample moment is M 1 = 1 n n summationdisplay i =1 X i = X. Equating them gives us η + 1 = X. Solving for θ gives us the MME: ˆ η = X 1 . 2. This is a Beta distribution with a = θ and b = θ . The first distribution moment is μ 1 = E [ X ] = a a + b = θ θ + θ = 1 2 . This does not even involve θ so setting it equal to the sample moment M 1 = X will not get us anywhere where we can solve for θ . The second distribution moment is E [ X 2 ] = V ar [ X ] + ( E [ X ]) 2 = ab ( a + b ) 2 ( a + b +1) + parenleftBig 1 2 parenrightBig 2 = 1 4(2 θ +1) + 1 4 Setting the equal to M 2 = 1 n X 2 i and solving for θ gives ˆ θ MME = 1 4 n X 2 i - 1 1 2
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3. (a) You can get full credit here by just assuming a Poisson distribution. However, this problem is slightly different and I will give the solution for the stated problem... The only way we will observe a zero in the sample is if θ = 0. In the case where θ = 0, we are sampling values from the distribution given by f ( x ; 0) = P ( X = x | θ = 0) Since 1 = f (0; 0) = P ( X = 0 | θ = 0) , when θ = 0 we will always observe the sampled X ’s to be zero. So, if we observe ( x 1 , x 2 , . . . , x n ) = (0 , 0 , . . . , 0), we know that θ must have been zero. When θ > 0, we will be sampling X ’s from the distribution with pdf f ( x ; θ ) = θ x e - θ x ! I { 1 , 2 ,... } ( x ) The likelihood is L ( θ ) n productdisplay i =1 f ( x i ; θ ) = n productdisplay i =1 θ x i e - θ x i ! I { 1 , 2 ,... } ( x i ) = θ x i e - bracketleftBigg n productdisplay i =1 1 x i ! I { 1 , 2 ,... } ( x i ) bracketrightBigg Throwing out the constant of proportionality (wrt θ ), we may take the likelihood to be L ( θ ) = θ x i e - . Then ln L ( θ ) = summationdisplay x x ln θ nθ. Taking the derivative wrt θ and setting it equal to zero gives x i θ n = 0 . Solving for θ gives us the MLE ˆ θ = X i n = X (Note that if all the X ’s were zero then X = 0, so this one expression for the MLE works for zero or non-zero X ’s.) (b) f ( x ; θ ) = (1 ) e - x/θ I (0 , ) ( x ) f ( vectorx ; θ ) = 1 θ n
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