Sum of two independent Gaussian r.v.s. given that one is negative

Question

Let $X \sim \mathcal N(0, \sigma_x^2)$ and $Y \sim \mathcal N(0, \sigma_y^2)$ be two independent r.v.s.

Is there a simple solution for $p(X+Y = a ~|~ X \leq 0)$?

I'm aware that without the conditioning on $X \leq 0$ we have that $Z = X+Y \sim \mathcal N(0, \sigma_x^2 + \sigma_y^2)$, but for me it seems that conditioning on $X \leq 0$ makes the problem considerably harder.

Here's what I have so far:

\begin{equation} \begin{split} p(X+Y=a ~|~ X \leq 0) &= \frac{p(X+Y=a, X \leq 0)}{P(X \leq 0)} \\ &= 2\int_{-\infty}^0 p(X+Y=a, X=b) db \\ &= 2 \int_{-\infty}^0 p(Y=a-b) ~ p(X=b) db \\ & \quad\quad\quad\quad \vdots \quad \text{(following the proof for the non-conditioned case)}\\ &= 2 ~ \frac{1}{\sqrt{2 \pi (\sigma_x^2 + \sigma_y^2)}} \exp \left(- \frac{a^2}{2 (\sigma_x^2 + \sigma_y^2)} \right) \int_{-\infty}^0 \frac{1}{\sqrt{2 \pi \frac{\sigma_x^2 \sigma_y^2}{\sigma_x^2 + \sigma_y^2}}} \exp \left(- \frac{\left(b - \frac{\sigma_x^2 a}{\sigma_x^2 + \sigma_y^2}\right)^2}{2 \frac{\sigma_x^2 \sigma_y^2}{\sigma_x^2 + \sigma_y^2}} \right) db \\ &= 2 ~ p(X+Y=a) \int_{-\infty}^0 \frac{1}{\sqrt{2 \pi \frac{\sigma_x^2 \sigma_y^2}{\sigma_x^2 + \sigma_y^2}}} \exp \left(- \frac{\left(b - \frac{\sigma_x^2 a}{\sigma_x^2 + \sigma_y^2}\right)^2}{2 \frac{\sigma_x^2 \sigma_y^2}{\sigma_x^2 + \sigma_y^2}} \right) db \\ &= 2 ~ p(X+Y=a)~p \left(W \geq \frac{\sigma_x a}{\sigma_y \sqrt{\sigma_x^2 + \sigma_y^2}} \right), \quad\quad W \sim \mathcal N(0,1)\\ \end{split} \end{equation}

I'm trying to simplify this further such that its entropy can be 'neatly' computed.

Answer 1

In your third equality line, the integrand is zero as these are absolutely continuous random variables. Dont you get an answer of zero just from that?

Answer 2

The approach is mostly correct, as you expand out the conditional probability with Baye’s rule. However you’ve tried to oversimplify into P(X+Y=a) where this isn’t equivalent to what you had in the line above. This probability actually 0 and is not equivalent to substituting the value ‘a’ into the density.

Not sure there is a slicker way of getting the result.