# $u$~$N(0,A)$ and z$|u$~$N(u,1)$ how to show that $u|z$~$N(Bz,B)$ where $B=A/(A+1)$?

$u$~$N(0,A)$ and $z|u$~$N(u,1)$ how to show that $u|z$~$N(Bz,B)$ where $B=A/(A+1)$ ?

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Knowing $f_U$ and $f_{Z\mid U}$ yields $f_{U,Z}$, which yields $f_{U\mid Z}$. Which part of this program have you trouble with? –  Did Jan 28 '13 at 16:11
what is $f(U,Z)$ ? $f(U,Z)=?$ –  Qbik Jan 28 '13 at 16:26
Never wrote $f(U,Z)$, but $f_{U,Z}$ the joint density of $(U,Z)$. –  Did Jan 28 '13 at 16:53
Which part of this program have you trouble with? –  Did Jan 28 '13 at 16:53
Somebody is erasing their footprints... –  Did Jan 30 '13 at 22:18

$$p(\mu|z) \propto p(z|\mu)p(\mu)$$
After you plug it $p(z|\mu)$ and $p(\mu)$ and you will find $p(\mu|z)$ has a pdf that is normal. And you can easily derive its mean and variance.
but what is $p(z)$ ? $N(0,1)$ ? –  Qbik Jan 28 '13 at 16:22
@Qbik There is no $p(z)$ involved here. You only need the conditional distribution of $p(z|\mu)$. –  Patrick Li Jan 28 '13 at 16:55