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
$$P(X_1+X_3 \leq \alpha \cap X_1< X_2)$$.
Thanks.
Best Answer
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 $P(X_1+X_2\leq\alpha\wedge X_1<X_2)$.
addendum:
$\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} $