will have a high-school degree. Let be the number who respond and have a X high-school degree, and let be the total number who respond. Note that Y X Y. Find E[ = ]. Y X x The trick clearly comes in getting the joint law (x, We do know that pX,Y y). ~ Poisson() and = ~ Bin(y, ). Thus Y X Y y (x, = (y) (x | pX,Y y) pY pXY y) = I(0 x y) Using this, we can sum out to get the marginal probability law for . y X = = pX(x) = = = = [change index to = - ] k y x = = = Thus, we see that (marginally) ~ Poisson(). Thus, we can complete the derivation of X the conditional law of given = Y, X x. = = = = = In this form, it appears that, given = it happens that is Poisson with parameter X x, Y x (1 ). This can be made completely explicit by writing = The probability law for Z Y x. (given will be Z x) = pZ(x) Since E = (1 ), we have E[ | = ] = + (1 ). Z Y X x x
will have a high-school degree. Let X be the number who respond and have a high-school degree, and let Y be the total number who respond. Note that X Y. Find E[ Y X = x ]. The trick clearly comes in getting the joint law pX,Y (x, y). We do know that Y ~ Poisson() and X Y = y ~ Bin(y, ). Thus pX,Y (x, y) = pY (y) pXY (x | y) = I(0 x y) Using this, we can sum out y to get the marginal probability law for X . pX(x) = = = = = = [change index to k = y - x ] = = = Thus, we see that (marginally) X ~ Poisson(). Thus, we can complete the derivation of the conditional law of Y, given X = x. = = = = = In this form, it appears that, given X = x, it happens that Y x is Poisson with parameter (1 ). This can be made completely explicit by writing Z = Y x. The probability law for Z (given x) will be pZ(x) = Since E Z = (1 ), we have E[ Y | X = x ] = x + (1 ).
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