# Express $P\{\{X_1+X_2\leq \alpha \}\cap\{X_3\leq X_2\}\}$ using CDF and PDF

Let $$X_1$$ $$X_2$$ and $$X_3$$ be three positive independent random variables. The PDF and CDF of $$X_i$$ are denoted by $$f_{X_i}(x_i)$$ and $$F_{X_i}(x_i)$$ respectively.

I would like to compute the following probability using CDF and PDF
$$P\{\{X_1+X_2\leq \alpha \}\cap\{X_3\leq X_2\}\}$$

I use the following domain $$X_3\leq X_2 \leq \alpha -X_1.$$ So is it $$P\{\{X_1+X_2\leq \alpha \}\cap\{X_3\leq X_2\}\}=P\{\{X_2\leq \alpha-X_1 \}\cap\{X_3\leq X_2\}\}$$ $$= \int_{x_1=0}^{\infty}f_{X_1}(x_1)F_{X_2}(\alpha-x_1)dx_1- \int_{x_3=0}^{\infty}f_{X_3}(x_3)F_{X_2}(x_3)dx_3.$$

or

$$P\{\{X_1+X_2\leq \alpha \}\cap\{X_3\leq X_2\}\}=P\{\{X_2\leq \alpha-X_1 \}\cap\{X_3\leq X_2\}\}$$ $$= \int_{x_1=0}^{\infty}f_{X_1}(x_1)\left( \int_{x_2=0}^{\alpha-x_1}f_{X_2}(x_2)F_{X_3}(x_2)dx_2\right)dx_1.$$

• Once again you have specific distribution =s in mind but you don't mention it in the question. What start the intergals from $0$? – Kavi Rama Murthy Jun 16 at 23:40
• For example exponential random varaible the important for me are all positive. Just I would like the true representation using PDF and CDFs. – Monir Jun 16 at 23:42

In the the second one if you change the outer integral to $$\int_0^{\alpha}$$ it becomes correct provided the random variables are positive. Alternatively you can make it correct as stated if you interpret the inside integral as $$0$$ when $$\alpha -x_1 <0$$. The first one seems to be wrong.

• Ok Thanks. I hop it will work with me. – Monir Jun 16 at 23:46
• I dont think the second is correct the last integral should evaluate from 0 to $\alpha$. Since $X_3\leq X_2$ is positive, and $X_1$ is positive. – Monir Jun 18 at 10:44
• I dont think the second is correct the last integral should evaluate from 0 to $\alpha$. Since $X_3\leq X_2$ is positive random variable , and $X_1$ is also positive random varble. – Monir Jun 18 at 10:54
• @Monir I have added some explanation to my answer. – Kavi Rama Murthy Jun 18 at 11:44
• Ok because I tried for integral from $0$ to $\infty$ it dose not work for me. I will try from $0$ to $\alpha$. Thanks. – Monir Jun 18 at 11:47