Conditional probability against sum of two normal random variables I am trying to solve the following problem:
Let $X$, $Y$ be two variables with the distribution $\mathcal{N}(0,\lambda^2)$. Find the distribution of $X$ under the condtion $X+Y=t$.
So what we are looking for is, for $B$ a Borel set, the following thing:
$$
P(X\in B\:|\:X+Y=t)
$$
But $$P(X\in B\:|\:X+Y=t)=P(Y\in t-B)=\int_{t-B}\frac{1}{\lambda\sqrt{2\pi}}e^\frac{-x^2}{2\lambda^2}dx.$$
By the change of variables, the last integral is equal to
$$
-\int_{B}\frac{1}{\lambda\sqrt{2\pi}}e^\frac{-(y-t)^2}{2\lambda^2}dy.
$$
This would mean that the distribution of $X|X+Y=t$ is $\mathcal{N}(t,\lambda^2)$ BUT:


*

*There is unexpected minus sign, which makes a measure of the set negative. I suppose it comes from wrong change of variables?

*The answer in the textbook is that this distribution is $\mathcal{N}(t/2, \lambda^2/2)$. So where is the mistake?


Here's another way.  I'm assuming $X$and $Y$ are independent. 
Since $\begin{pmatrix}X\\X+Y\end{pmatrix} = \begin{pmatrix}1 & 0\\ 1 & 1\end{pmatrix} \begin{pmatrix}X\\Y\end{pmatrix}$, the joint distribution of $(X,X+Y)$ is  $\mathcal N_2\left(0,\lambda^2 \begin{pmatrix}1 & 1\\ 1 & 2\end{pmatrix}\right)$.
Note also that the distribution of $X+Y$ is $\mathcal N(0,2\lambda^2)$.
By Bayes' theorem, the conditional density of $X$ given $X+Y=t$ is given by
$$\begin{aligned}f_{X|X+Y=t}(x) 
= \frac{f_{(X,X+Y)}(x,t)}{f_{X+Y}(t)}
&\propto\exp\left(-\frac 1{2\lambda^2} (x,t)\begin{pmatrix}2 & -1\\ -1 & 1\end{pmatrix} \begin{pmatrix}x\\t\end{pmatrix}\right)\\
&\propto \exp\left(-\frac{1}{\lambda^2} (x^2-tx)\right) \\
&\propto \exp\left(-\frac{1}{2 \frac{\lambda^2}2} (x-\frac t2)^2\right) \end{aligned}$$
$\propto$ means proportional, which allows me to drop multiplicative factors that do not depend on $x$.
Given the final expression, $f_{X|X+Y=t}(x) $ is the density of $\mathcal N(\frac t2, \frac{\lambda^2}2)$, as claimed in your textbook.
