Gaussian conditional probability $X$ given $X+Y$

Question

Let $X$ and $Y$ be independent Gaussian random variables with means $\mu_1 = 1$ and $\mu_2 = 0$ and variances $\sigma^2_1 = 1$ and $\sigma^2_2 = 1$ respectively. How to find the conditional density of $X$ given $X+Y = 2$: $\mathbf{P}(X|X+Y = 2)$?

Answer 1

In the following I am abusing notation slightly. I write $\mathbb{P}(X=x)$ instead of $p_X(x)$ etc.

Recall that $$ \mathbb{P}(X=x|X+Y=2)=\frac{\mathbb{P}(X=x,X+Y=2)}{\mathbb{P}(X+Y=2)}. $$ Now by independence, $$ \mathbb{P}(X=x,X+Y=2) = \mathbb{P}(X=x,Y=2-x) = \mathbb{P}(X=x)\mathbb{P}(Y=2-x). $$ For the denominator, recall that for independent, normally random variables, $$ X+Y \sim \mathcal{N}(\mu_1+\mu_2, \sigma_1² + \sigma_2²). $$ Put everything together to get

\begin{align*} \mathbb{P}(X=x|X+Y=2) &= \frac{\frac{1}{\sqrt{2\pi}}e^{-\frac{(x-1)^{2}}{2}} \frac{1}{\sqrt{2\pi}}e^{-\frac{(2-x)^{2}}{2}}}{\frac{1}{\sqrt{2\pi}\sqrt{2}} e^{-\frac{(2-1)^{2}}{4}}} \\\\ &= \frac{\sqrt{2}}{\sqrt{2\pi}}e^{-x^2+3x-\frac{9}{4}} \\\\ &= \frac{\sqrt{2}}{\sqrt{2 \pi}}e^{-(x-\frac{3}{2})^{2}}, \end{align*} which means that $$ X | (X+Y=2) \sim \mathcal{N}\bigg(\frac{3}{2},\frac{1}{2}\bigg). $$