Conditional distribution of $X$ given $Y(X)$ Let $X$ and $Z$ be random variables that independently follow $N(\mu,1)$ and $N(0,1)$, respectively, and define $Y = cX + Z$, where c is a const. How to find that the conditional distribution of $X$ given $Y$ is a normal distribution?
I tried to solve this by:
$$F_{X|Y}=P\left(X\leq x|Y\right)=P\left(\frac{Y-Z}{c}\leq x|Y\right)=P\left(Y-Z\leq cx|Y\right)$$
but get stuck. I'm not sure it is the right way.
 A: Start by finding the joint distributions of $(X,Y)$.
$$(X,Y) \sim \mathcal N\left((\mu,c\mu), \begin{pmatrix}1 & c\\ 
c & c^2+1\end{pmatrix}\right)$$
So using the conditional distributions $\left(\text{with $\mu_1 = \mu$, $\mu_2 = c\mu$, $\sigma^2_1 = 1$, $\sigma^2_2 = 1+c^2$ and $\rho = \frac{c}{\sqrt{1+c^2}}$}\right)$
$$X_{|Y} \sim \mathcal N \left(\mu + \frac{1}{1+c^2}\frac{c}{\sqrt{1+c^2}}(Y-c\mu), \frac{1}{1+c^2}\right)$$
A: $\newcommand{\prob}{\mathrm{P}}$I will try to give an answer without using the normality assumption. You can use the law of total probability. We have
\begin{align}
F_{X|Y}(x|y)
{}={}&
\prob[X \leq x | Y = y]
{}={}
\prob[X \leq x | cX + Z = y] 
\\
{}={}&
\int_{-\infty}^{\infty} \prob[X \leq x | cX + Z = y, Z = z] p_Z(z) \mathrm{d}z
\\
{}={}&
\int_{-\infty}^{\infty} \prob[X \leq x | cX + z = y] p_Z(z) \mathrm{d}z
\end{align}
Now we need to determine a conditional probability involving $X$ given $cX$ (we assume $c\neq 0$).
If $c>0$, then
\begin{align}
\prob[X \leq x | cX + z = y] 
{}={}&
\prob[cX \leq cx | cX  = y - z] 
\\
{}={}&
\prob[y - z \leq cx]
\\
{}={}&
\prob[z \geq y - cx]
\\
{}={}&
\begin{cases}
1, &\text{ if } z \geq y-cx
\\
0, &\text{ otherwise}
\end{cases}
\\
{}=:{}& 1_{\geq y-cx}(z)
\end{align}
so,
$$
F_{X | Y}(x|y)
{}={}
\int_{-\infty}^{\infty} 1_{\geq y-cx}(z) p_Z(z) \mathrm{d}z
{}={}
\int_{y-cx}^{\infty}p_Z(z) \mathrm{d}z,
$$
so, if $c>0$, $F_{X | Y}(x|y) = 1 - F_Z(y-cx)$.
