# P(X + Y < a | Y < b) where X and Y are independent continuous random variables

I’m looking for an online source/article/lecture notes/ text book that would contain a detailed/rigorous discussion/explanation/proof of the following result, which was used in Conditional normal distribution

$P(X+Y<a,\,Y<b)=\int\limits_{-\infty}^bg_Y(y)\int\limits_{-\infty}^{a-y}g_X(x)\mathrm dx\,\mathrm dy$

Many thanks

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Using indicator function $\chi$: $$\begin{eqnarray} \mathbb{P}(X+Y<a, Y<b) &=& \int_{-\infty}^{\infty} \int_{-\infty}^\infty \chi_{y < b} (x,y) \chi_{x+y < a}(x,y) \, g_Y(y) \, g_X(x) \, \mathrm{d} x \, \mathrm{d} y \\ &=& \int_{-\infty}^{b} \int_{-\infty}^\infty \chi_{x+y < a}(x,y) \, g_Y(y) \, g_X(x) \, \mathrm{d} x \, \mathrm{d} y \\ &=& \int_{-\infty}^{b} \int_{-\infty}^{a-y} g_Y(y) \, g_X(x) \, \mathrm{d} x \, \mathrm{d} y \end{eqnarray}$$

The first line is the definition of probability. Since $\chi_{y<b}(x,y)=0$ in the region $y>b$, we replaced the upper bound of integration w.r.t. $y$ variable with $b$, similarly, $\chi_{x+y<a}(x,y)$ is zero for $x>a-y$.

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just fixed a little typo in your answer – Ilya Sep 29 '11 at 16:04

First of all, how this formula can be derived. Suppose, distributions of both $X,Y$ has continuous densities $g_X,g_Y$. Then $$P(X+Y<a,Y<b) = \int\limits_\mathbb R P(X+Y<a,Y<b|Y = y) g_Y(y)\mathrm d y$$ by the Law of total probability. Note that $P(X+Y<a,Y<b|Y = y) = 0$ if $y\geq b$ and $$P(X+Y<a,Y<b|Y = y) = P(X+Y<a|Y=y) = P(X<a-y) = \int\limits_{-\infty}^{a-y}g_X(x)\mathrm dx$$ if $y<b$ so $$\int\limits_\mathbb R P(X+Y<a,Y<b|Y = y) g_Y(y)\mathrm d y = \int\limits_{-\infty}^b \left(\int\limits_{-\infty}^{a-y}g_X(x)\mathrm dx g_Y(y)\right)\mathrm d y$$ which is the formula in your post.

Note: although I've cited the Law of total probability from Wikipedia you may be interested in the proof of it. I've found it in these lecture notes. They are in *.ps so you may need some software to read it. Also, I would advise you to read the serious book in probability about this topic. Durrett's book is very nice.

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Thanks you for your reply, but for continuous r.v. the event Y=y has probabilty 0, so how can we then condition on this event? Thanks – Confounded Sep 29 '11 at 15:41
That's why we don't multiply by its probability but multiply by the density of $Y$. I would say that for this case the notion of conditional probability is define through the joint distribution rather than through a fraction of two probabilities. You can think about it like a limit case, but for the better understanding I still recommend you to read at least lecture notes I've cited. – Ilya Sep 29 '11 at 15:50