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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

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

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  • $\begingroup$ Once again you have specific distribution =s in mind but you don't mention it in the question. What start the intergals from $0$? $\endgroup$
    – Kavi Rama Murthy
    Jun 16, 2019 at 23:40
  • $\begingroup$ For example exponential random varaible the important for me are all positive. Just I would like the true representation using PDF and CDFs. $\endgroup$
    – Monir
    Jun 16, 2019 at 23:42

1 Answer 1

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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.

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  • $\begingroup$ Ok Thanks. I hop it will work with me. $\endgroup$
    – Monir
    Jun 16, 2019 at 23:46
  • $\begingroup$ 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. $\endgroup$
    – Monir
    Jun 18, 2019 at 10:44
  • $\begingroup$ 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. $\endgroup$
    – Monir
    Jun 18, 2019 at 10:54
  • $\begingroup$ @Monir I have added some explanation to my answer. $\endgroup$
    – Kavi Rama Murthy
    Jun 18, 2019 at 11:44
  • $\begingroup$ Ok because I tried for integral from $0$ to $\infty$ it dose not work for me. I will try from $0$ to $\alpha$. Thanks. $\endgroup$
    – Monir
    Jun 18, 2019 at 11:47

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