1
$\begingroup$

I have three random variables $X$, $Y$, $Z$. $X$ is independent from the two others. On the other hand, $Y$ and $Z$ may be dependent, and of different distributions, for example consecutive order statistics for a given sampling. How can I estimate $\mathbb P (X \in [Y, Z))$ ?

I know that this probability is $\mathbb P (\{X - Y \geq 0\} \cap \{Z-X > 0\})$, but these are not independent. Where do I go from here?

EDIT
Following @Dilip Sarwate's and @whuber's comments below. For a given $x$,

\begin{align} \mathbb P(X\in [Y, Z) | X=x) &= \mathbb E[1\{X\in [Y, Z) \} | X=x] \\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}1\{x\in [y, z) \}f_{Y, Z}(y, z)dydz \\ &= \int_{-\infty}^{x}\int_{x}^{\infty}f_{Y, Z}(y, z)dydz \end{align}

Then returning to the original problem:

\begin{align} \mathbb P (X \in [Y, Z)) = \int_{-\infty}^\infty\int_{-\infty}^{x}\int_{x}^{\infty}f_{Y, Z}(y, z)f_X(x)dydzdx \end{align}

Similarly, we would then have

\begin{align} \mathbb E (X 1\{X \in [Y, Z)\}) = \int_{-\infty}^\infty\int_{-\infty}^{x}\int_{x}^{\infty}xf_{Y, Z}(y, z)f_X(x)dydzdx \end{align} ?

Are the two previous equations correct?

$\endgroup$
  • $\begingroup$ Try finding the conditional probability that $Y \leq X$ and $Z > X$ conditioned on $X$ having taken on the value $x$. Then, remove the conditioning on $X$ by multiplying your result by the pdf $f_X(x)$ of $X$ and integrating. $\endgroup$ – Dilip Sarwate Sep 11 at 3:23
  • $\begingroup$ (1) How might $X$ be related to $Y$ and $Z$? Same distribution? Independent? (2) What information or data would you consider using to estimate this probability? Or do you really mean to ask how to compute its value? $\endgroup$ – whuber Sep 11 at 15:57
  • $\begingroup$ @whuber, I edited the question: $X$ is independent from the others. The idea would be to have some bound on some distance in probability space between the distribution of $X$ and $Y, Z$. $\endgroup$ – geoffn91 Sep 11 at 16:12
  • $\begingroup$ Does $X$ have the same distribution as $Y$ and $Z$ or not? It would also be useful to indicate whether any of these distributions might not be continuous. $\endgroup$ – whuber Sep 11 at 16:14
  • 1
    $\begingroup$ It doesn't necessarily have the same distribution. We can suppose all of them to be continuous. $\endgroup$ – geoffn91 Sep 11 at 18:44
0
$\begingroup$

$\mathbb P (X \in [Y, Z)) = \int_{-\infty}^{\infty} \mathbb P (x \in [Y, Z)) f_X(x) dx = \int_{-\infty}^{\infty} (\underset{(x, \infty) \times (-\infty, x)}{\iint} f_{YZ}(y, z) dydz) f_X(x) dx$

since $Y, Z$ are dependent can you substitute double integral with two integrals? also it may be easier to compute this integral

$\endgroup$

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.