Distribution of the sum of independent r.v.

Assume that $X_1$ and $X_2$ are independent random variables with given distribution $f(.)$ (say Normal distribution with $\mu_i$ and $\sigma_i$). I am stuck with the calculation of:

$P(\{X_1 \leq a\} \; \cap\; \{X_2 \leq b\} \; \cap\; \{c \leq X_1 + X_2 \leq a+b\} )$

where $c < a+b$. For brevity, let's say $Z_1 = \{X_1 \leq a\}$, $Z_2 = \{X_2 \leq b\}$ and $Z_3 = \{c \leq X_1 + X_2 \leq a+b\}$.

Using the product rule I can write:

$P(Z_1 \cap Z_2 \cap Z_3) = P(Z_1)\cdot P(Z_2|Z_1) \cdot P(Z_3|Z_1\cap Z_2)$

and since $X_1$ and $X_2$ are independent, then $P(Z_2|Z_1) = P(Z_2)$ and $P(Z_1 \cap Z_2) = P(Z_1)\cdot P(Z_2)$. But now I don't know how to calculate the third factor $P(Z_3|Z_1\cap Z_2)$ since, as far as I understand, $Z_3$ is not independent of $Z_1$ and $Z_2$. Is there any other way to solve this problem?

Thanks.

• You can remove the condition that $X_1 + X_2 \leq a+b$ in the definition of $Z_3$ as it follows from conditions for $Z_1$ and $Z_2$. If you know the densities of distributions, you can express the desired probability as an integral over the region $\{(x,y): x\leq a, y \leq b, x + y \geq c\}$. However, you cannot express it in terms of $P(Z_1)$, $P(Z_2)$ and $P(Z_3)$. – Yury Jul 30 '12 at 15:33

If your random variables each have a probability density (like in the case of a normal distribution )and we call these densities $f_1,f_2$, you can just write down your probability as an integral over the density functions on a suitably chosen domain. \begin{equation} \int_{A} f_1(x_1)f_2(x_2)dx_1dx_2. \end{equation} The domain $A$ should be given by the conditions $x_1 \le a$, $x_2 \le b$ and $c \le x_1+x_2 \le a + b$, where the last inequality is in fact redundant because $x_1 + x_2 \le a+b$ follows from the individual conditions on $x_1,x_2$.
• Of course, there's no guarantee that the integral can be found in closed form, even in the case of normal distributions. Thus if $X_1$ and $X_2$ have standard normal distributions, the probability is $$\frac{\sqrt{2}}{4 \sqrt{\pi}} \int_{c-a}^b e^{-y^2/2} \left( \text{erf}(a/\sqrt{2}) - \text{erf}((c-y)/\sqrt{2})\right)\ dy$$ which, as far as I know, does not have a closed form in general. – Robert Israel Jul 30 '12 at 18:51