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

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:

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

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

• $P$ here denotes the probability density function. Sep 3, 2022 at 20:17
• To be more specific, $p(X=b) = \frac{1}{\sigma_x \sqrt{2 \pi}} \exp(- \frac{b^2}{2 \sigma_x^2})$ and $p(Y=a-b) =\frac{1}{\sigma_y \sqrt{2 \pi}} \exp(- \frac{(a-b)^2}{2 \sigma_y^2})$. Sep 3, 2022 at 20:44