0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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} $

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.