Conditional density, bivariate normal

Question

Let $Z=X+Y$ where $X \sim N(\mu,\sigma^2)$ and $Y \sim N(0,1)$ are independent. What is the conditional density of X given Z, $f_{X|Z}(x|z)$?

I already found that $f_{X,Z}(x,z)=\frac{1}{2\pi\sigma}e^{-\frac{1}{2\sigma^2}[(x-\mu)^2+\sigma^2(x-z)^2]}$ and $Z \sim N(\mu,\sigma^2+1)$

$$f_{X|Z}(x|z)=\frac{f_{X,Z}(x,z)}{f_Z(z)}=\frac{\frac{1}{2\pi\sigma}e^{-\frac{1}{2\sigma^2}[(x-\mu)^2+\sigma^2(x-z)^2]}}{\frac{1}{\sqrt{2\pi}\sqrt{\sigma^2+1}}e^{-\frac{1}{2(\sigma^2+1)}(z-u)^2}}$$ $$f_{X|Z}(x|z)=\frac{\sqrt{\sigma^2+1}}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2\sigma^2}[(x-\mu)^2+\sigma^2(x-z)^2]+\frac{1}{(2\sigma^2+1)}(z-\mu)^2}$$ After some simplifications I get $$f_{X|Z}(x|z)=\frac{\sqrt{\sigma^2+1}}{\sqrt{2\pi}\sigma}e^{\frac{1}{2\sigma^2(\sigma^2+1)}[-(\sigma^2+1)(x^2-2\mu x+\sigma^2(x^2-2zx)-2\sigma^2\mu z]}$$

from here I'm stuck.I can not further simplify things and not get in a know density. Is there any easier way to find the conditional?

Answer 1

$$X = Z - Y\text{ and }Y \sim N(0,1) \implies X|Z \sim N(Z,1)$$

Here I've just treated $Z$ as a constant because when you're conditioning on it being known, that's pretty much what it is.

Then you know the density $f_{X|Z}(x|z) = \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(x-z)^2}$.

This may not be as rigorous as you want it to be but I'm sure if you did enough manipulation of the pdf's and didn't make any mistakes, then this is what you should get.