Let $X_1, ... X_n \stackrel{i.i.d} \sim Unif(0,2).$ Find $P(Y_1 < \frac12 < Y_n),$ where $Y_1 = \min \{X_1, ..., X_n\} $ and $ \ Y_n = \max\{X_1, ..., X_n\}$

When I attempted this problem, I decided that $n>2 \implies Y_1 $ and $ Y_n$ are independent, which leads me to the answer

$$P(Y_1 < \frac12 < Y_n) = \left(1-\left(\frac14\right)^n\right) \cdot \left(1-\left(\frac34\right)^n\right)$$

I have since decided is untrue. Clearly, if I know $Y_n = 1,$ then $Y_1 \neq 1.5.$ I have also discovered that in my notes there is a joint pdf of two order statistics of a given sample:

$$f_{Y_i,Y_j}(u,v) = \cfrac{n!} {(i-1)!(j-i-1)!(n-j)!} [F_X(u)]^{i-1} [F_X(v)-F_X(u)]^{j-i-1} [1-F_X(v)]^{n-j} f_X(u)f_X(v), \forall \ u<v$$

I am looking first of all for the derivation and perhaps some intuition for this equation. For context, I am familiar with the derivation of the cdf/pdf of a single order statistic, but this one is throwing me for a loop. Second of all, I would like to know if I am using this equation correctly:

$$P(Y_1<\frac12<Y_n) = P(Y_1 < \frac12 \ \bigcap \ Y_n > \frac12) \\ = \int_0^\frac12 \int_\frac12^2 \cfrac{n!}{0!(n-2)!0!} \left[\frac u2\right]^{0} \left[\frac v2 - \frac u2 \right]^{n-2} \left[1- \frac v2 \right]^{0} \left(\frac12 \right) \left(\frac12 \right)\mathrm{d}v\mathrm{d}u \\ = \frac14 \int_0^\frac12 \int_\frac12^2 n(n-1) \left[\cfrac{v-u}2 \right]^{n-2} \mathrm{d}v\mathrm{d}u$$

which I don't know how to solve explicitly, but is still useful as it can be solved given a value of $n$.

  • 2
    $\begingroup$ Clearly $Y_1$ and $Y_n$ are not independent, since $Y_1$ cannot exceed $Y_n$. $\endgroup$
    – Glen_b
    Apr 22, 2020 at 5:40
  • 1
    $\begingroup$ The integral you have set up in the end misses the $u<v$ restriction. If this is some sort of homework, please add self-study tag. $\endgroup$ Apr 22, 2020 at 6:31
  • $\begingroup$ See stats.stackexchange.com/questions/225867 for a more complicated version of this problem, because its solution easily applies here. Concerning the issue of independence, see stats.stackexchange.com/questions/208658. A formula for the joint pdf is given at stats.stackexchange.com/a/69582/919 and, in the special case of a uniform distribution, at stats.stackexchange.com/a/207596/919. $\endgroup$
    – whuber
    Apr 22, 2020 at 13:46
  • $\begingroup$ @StubbornAtom How does my integral miss the restriction? No $u$ value is above any $v$ value in the integration. Also, this is a former exam question not assigned to me in any capacity - let me know if that still qualifies it as self-study $\endgroup$
    – jeremy909
    Apr 22, 2020 at 15:42
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    $\begingroup$ Yes, all exam questions etc. qualify as self-study: link through to the tag wiki for details. As far as computing your integral goes, it's far easier than it might look: you're just integrating polynomials, after all. $\endgroup$
    – whuber
    Apr 22, 2020 at 16:25

1 Answer 1


I don't care to evaluate the integral mechanically (although Wolfram Alpha will). And nor should anyone else, because the pedagogical usefulness of that exercise is doubtful.

Instead, considering that the event $Y_{1} \le 1/2 \le Y_{n}$ is the complement of the union of events

$$\left(Y_{1}\gt 1/2\right)\ \cup\ \left(Y_{n}\lt 1/2\right)$$

and, because $Y_1$ cannot exceed $Y_n$ those events are disjoint, it follows that

$$\Pr(Y_{1}\le 1/2\le Y_{n}) = 1 - \Pr(Y_{1} \gt 1/2) - \Pr(Y_{n} \lt 1/2).$$

These probabilities are simple to evaluate because the first is the chance that all the $X_i$ exceed $1/2,$ the second is the chance that all the $X_i$ are less than $1/2,$ and the $X_i$ are independent.

Even if your interest is only on whether you are doing the integration correctly, this approach provides a way to check your answer.


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