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?

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?

  • $\begingroup$ 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. $\endgroup$ Commented Sep 11, 2019 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
    Commented Sep 11, 2019 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 $Y, Z$. $\endgroup$ Commented Sep 11, 2019 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
    Commented Sep 11, 2019 at 16:14
  • 1
    $\begingroup$ It doesn't necessarily have the same distribution. We can suppose all of them to be continuous. $\endgroup$ Commented Sep 11, 2019 at 18:44

1 Answer 1


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

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


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