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 Seven 1. The first population moment is 1 = E [ X ] = integraldisplay x e ( x ) dx = + 1 . (Note that since the pdf is that of an exponential with rate 1 that has been shifted by to the right, the mean is 1 (the mean of the exponential with rate 1) that has been shifted by to the right!) The first sample moment is M 1 = 1 n n summationdisplay i =1 X i = X. Equating them gives us + 1 = X. Solving for gives us the MME: = X 1 . 2. This is a Beta distribution with a = and b = . The first distribution moment is 1 = E [ X ] = a a + b = + = 1 2 . This does not even involve so setting it equal to the sample moment M 1 = X will not get us anywhere where we can solve for . 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 +1) + 1 4 Setting the equal to M 2 = 1 n X 2 i and solving for gives MME = 1 4 n X 2 i 1 1 2 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 = 0. In the case where = 0, we are sampling values from the distribution given by f ( x ; 0) = P ( X = x  = 0) Since 1 = f (0; 0) = P ( X = 0  = 0) , when = 0 we will always observe the sampled X s to be zero. So, if we observe ( x 1 ,x 2 ,... ,x n ) = (0 , ,... , 0), we know that must have been zero. When > 0, we will be sampling X s from the distribution with pdf f ( x ; ) = x e x ! I { 1 , 2 ,... } ( x ) The likelihood is L ( ) n productdisplay i =1 f ( x i ; ) = n productdisplay i =1 x i e x i ! I { 1 , 2 ,... } ( x i ) = x i e n bracketleftBigg n productdisplay i =1 1 x i ! I { 1 , 2 ,... } ( x i ) bracketrightBigg Throwing out the constant of proportionality (wrt ), we may take the likelihood to be L ( ) = x i e n ....
View Full
Document
 Fall '11
 Manuel

Click to edit the document details