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

For example, for exponential random variables we have

$$f_{X_i}(x)=\beta_i e^{-\beta_i x}$$

$$F_{X_i}(x)=1-e^{-\beta_i x}$$

where $\beta_i$ is the parametre of $X_i$.

My question is, if there is any formula using the PDF and CDF of $X_i$ to get the following probabilities $$P(X_1\leq \alpha \cap X_2\leq X_1)$$ and $$P(X_1+X_3 \leq \alpha \cap X_1< X_2)$$. For the first, $drhab help me to understanding

$$P(X_1\leq \alpha \cap X_2\leq X_1)=\int_{x_1=0}^{\alpha}\left(\int_{x_2=0}^{x1}f_{X_2}(x_2)dx_2\right)f_{x_1}(x_1) dx_1$$

But it sill form me the second part where I have three random variables. I was try to use the same as #drhab as follow $$ 0\leq X_3 \leq \alpha -X_1 $$ $$ 0\leq X_1<X_2 $$ $$ 0\leq X_2\leq \infty $$ But what then or how we get

$$P(X_1+X_3 \leq \alpha \cap X_1< X_2)$$.


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    $\begingroup$ Are $X_1,X_2,X_3$ independent? $\endgroup$ – drhab Mar 7 at 12:55
  • $\begingroup$ yes they are indpendent $\endgroup$ – Monir Mar 7 at 13:21
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    $\begingroup$ Then edit your question and give that (essential) extra info. It is not enough to mention it only in a comment. Also it seems that the rv's are supposed to be nonnegative. $\endgroup$ – drhab Mar 7 at 13:24

If $X_1,X_2$ are rv's having a joint PDF $f_{X_1,X_2}$ then:

$$P(X_2<X_1\leq\alpha)=\mathbb E\mathbf1_{X_2<X_1\leq\alpha}=\int\int\mathbf1_{x_2<x_1\leq\alpha}f_{X_1,X_2}(x_1,x_2)\;dx_1\;dx_2$$

If moreover $X_1,X_2$ are independent then this can be rewritten as:$$\cdots=\int\int\mathbf1_{x_2<x_1\leq\alpha}f_{X_1}(x_1)f_{X_2}(x_2)\;dx_1\;dx_2$$

Further we can change the order of integration and make use of equality $\mathbf1_{x_2<x_1\leq\alpha}=\mathbf1_{x_2<x_1}\mathbf1_{x_1\leq\alpha}$.

This can for instance lead to:$$\cdots=\int\mathbf1_{x_1\leq\alpha}f_{X_1}(x_1)\int\mathbf1_{x_2<x_1}f_{X_2}(x_2)\;dx_2\;dx_1=$$$$\int_{-\infty}^\alpha f_{X_1}(x_1)\int^{x_1}_{-\infty}f_{X_2}(x_2)\;dx_2\;dx_1=\int_{-\infty}^\alpha f_{X_1}(x_1)P(X_2\leq x_1)\;dx_1=\int_{-\infty}^\alpha f_{X_1}(x_1)F_{X_2}(x_1))\;dx_1$$

The same technique: $$P(\text{condition on }X_1,X_2,X_3)=\mathbb E\mathbf1_{\text{condition on }X_1,X_2,X_3}$$ can be applied to find an integral expression for $P(X_1+X_2\leq\alpha\wedge X_1<X_2)$.


$\begin{aligned}\int\int\int\mathbf{1}_{x_{3}\leq\alpha-x_{1}}\mathbf{1}_{x_{1}<x_{2}}f_{X_{1}}\left(x_{1}\right)f_{X_{2}}\left(x_{2}\right)f_{X_{3}}\left(x_{3}\right)dx_{3}dx_{1}dx_{2} & =\int f_{X_{2}}\left(x_{2}\right)\int\mathbf{1}_{x_{1}<x_{2}}f_{X_{1}}\left(x_{1}\right)\int\mathbf{1}_{x_{3}\leq\alpha-x_{1}}f_{X_{3}}\left(x_{3}\right)dx_{3}dx_{1}dx_{2}\\ & =\int f_{X_{2}}\left(x_{2}\right)\int\mathbf{1}_{x_{1}<x_{2}}f_{X_{1}}\left(x_{1}\right)F_{X_{3}}\left(\alpha-x_{1}\right)dx_{1}dx_{2}\\ & =\int f_{X_{2}}\left(x_{2}\right)\int_{-\infty}^{x_{2}}f_{X_{1}}\left(x_{1}\right)F_{X_{3}}\left(\alpha-x_{1}\right)dx_{1}dx_{2} \end{aligned} $

  • $\begingroup$ What you mean By $\mathbf{I}_x$ and $\mathbb{E}$? $\endgroup$ – Monir Mar 7 at 13:40
  • $\begingroup$ For $P(X_1+X_3 \wedge X_1<X_2)$ can we say $X_1\leq \alpha-X_3$ and $X_1<X_2$. But I didn't understand how we use your technique for three random variable? $\endgroup$ – Monir Mar 7 at 13:52
  • $\begingroup$ Do you mean that $$0 \leq X_3 \leq \alpha- X_1$$ and $$0\leq X_2\leq X_2$$ and $$0\leq X_2\leq \infty$$ $\endgroup$ – Monir Mar 7 at 13:57
  • $\begingroup$ #drhab Thanks so much I will try to get the second probability? $\endgroup$ – Monir Mar 7 at 14:21
  • $\begingroup$ Can any one help me for the last part in my quetion? $\endgroup$ – Monir Mar 8 at 0:09

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