# Find the probability that $P(X_1\leq \alpha \cap X_2\leq X_1)$ and $P(X_1+X_3 \leq \alpha \cap X_1< X_2)$

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 $$0\leq X_3 \leq \alpha -X_1$$ $$0\leq X_1 $$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)$$. Thanks. • Are$X_1,X_2,X_3independent? – drhab Mar 7 at 12:55 • yes they are indpendent – Monir Mar 7 at 13:21 • Then edit your question and give that (essential) extra info. It is not enough to mention it only in a comment. Also it seems that the rv's are supposed to be nonnegative. – drhab Mar 7 at 13:24 ## 1 Answer If $$X_1,X_2$$ are rv's having a joint PDF $$f_{X_1,X_2}$$ then: $$P(X_2 If moreover $$X_1,X_2$$ are independent then this can be rewritten as:$$\cdots=\int\int\mathbf1_{x_2 Further we can change the order of integration and make use of equality $$\mathbf1_{x_2. This can for instance lead to:$$\cdots=\int\mathbf1_{x_1\leq\alpha}f_{X_1}(x_1)\int\mathbf1_{x_2$$\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. addendum: \begin{aligned}\int\int\int\mathbf{1}_{x_{3}\leq\alpha-x_{1}}\mathbf{1}_{x_{1} • What you mean By\mathbf{I}_x$and$\mathbb{E}$? – Monir Mar 7 at 13:40 • For$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? – Monir Mar 7 at 13:52
• Do you mean that $$0 \leq X_3 \leq \alpha- X_1$$ and $$0\leq X_2\leq X_2$$ and $$0\leq X_2\leq \infty$$ – Monir Mar 7 at 13:57
• #drhab Thanks so much I will try to get the second probability? – Monir Mar 7 at 14:21
• Can any one help me for the last part in my quetion? – Monir Mar 8 at 0:09