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?


  • 2
    $\begingroup$ 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 $P(Z_3)$. $\endgroup$ – 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 $x_1,x_2$.

  • 1
    $\begingroup$ 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. $\endgroup$ – Robert Israel Jul 30 '12 at 18:51
  • $\begingroup$ Maybe a naive question, but are there clear conditions for the existence of closed-form solutions to the given problem? $\endgroup$ – Libra Jul 31 '12 at 7:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.