The difference between stupidity and genius is that genius has its limits.
"Albert Einstein"
Heres what I have:
First of all I note that (X,Z) are both linear combinations of (X,Y) that are independent so their distribution is bivariate normal.
It then holds that is normal with mean:
and the variance is given as
to find the terms we do:
and using that and independence of X and Y:
however since
we have:
and then
I know that this does not prove what you are assked to prove but the place where you are stuck you probably have to collect terms and complete the square in order to get the conditional distribution on the form of a normal. The above gives you what I believe you have to end up with .....
askazy (06-10-2015)
ok so working with the density form youre first equation but dropping constant factors .. here is what I get ...
first expand the square:
remove any additive terms in exp function not containing x since can be seperated out as factors not including x. Hence the expression is proportional to:
adding and substracting the constant C needed to complete the square:
where
and then quickly removing the constant to get the proportional expression:
conditional on z this is a normal kernel with the variance and conditional mean:
and since we can add and substract:
to get:
to check that the result is the same as we get if we use the sentence I wrote up in the previous post
askazy (06-10-2015)
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