Conditional density of bivariate normal and normal

Question

Let $Z=X+Y$ where $X \sim N(\mu,\sigma^2)$ and $Y \sim N(0,1)$ are independents. 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,Is there any easier way to find the conditional?

Answer 1

$X$ and $Z =X+Y$ enjoy a bivariate normal density where $X\sim N(\mu,\sigma^2)$, $Z \sim (\mu,\sigma^2+1)$, and $$\operatorname{cov}(X,Z)=\operatorname{cov}(X+Y,X) = \operatorname{cov}(X,X) + \operatorname{cov}(Y,X) = \operatorname{cov}(X,X)+0 = \sigma^2.$$ If you are allowed to apply the well-known result that the conditional density of $X$ given $Z = z$ is a normal density with mean $$\mu_X + \left.\left. \frac{\operatorname{cov}(X,Z)}{\operatorname{var}(Z)}\right(z-\mu_Z\right) = \mu+\left.\left.\frac{\sigma^2}{\sigma^2+1}\right(z-\mu \right) = \frac{\sigma^2}{\sigma^2+1}z +\frac{\mu}{\sigma^2+1}\tag{1}$$ and variance $$\operatorname{var}(X)\left(1-\frac{(\operatorname{cov}(X,Z))^2}{\operatorname{var}(X)\operatorname{var}(Z)}\right) = \sigma^2\left(1-\frac{\sigma^4}{\sigma^2(\sigma^2+1)}\right)= \frac{\sigma^2}{\sigma^2+1}\tag{2}$$ then you can just write down the density. Just be careful with the algebra; it is not hard, just tediously lengthy.

If you are not allowed to use this result, you can proceed as you described in your question and just try to massage the result into the final form that you now know it must have. I note, for example, that the multiplicative factor is correct in your "final answer" since $$\frac{\sqrt{\sigma^2+1}}{\sqrt{2\pi}\sigma} = \frac{1}{\sqrt{2\pi}\sqrt{\frac{\sigma^2}{\sigma^2+1}}}$$ is consistent with the variance given in $(2)$, but I am less sanguine about the multiplicative factor $\displaystyle\frac{1}{2\sigma^2(\sigma^2+1)}$ in the exponent: unless something comes out after simplifying the rest of the exponent that modifies the multiplicative factor, the factor should be $\displaystyle\frac{\sigma^2+1}{2\sigma^2}$ to be consistent with $(2)$.