# Joint pdf of Order Statistics

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

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:

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

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

• Clearly $Y_1$ and $Y_n$ are not independent, since $Y_1$ cannot exceed $Y_n$. Apr 22, 2020 at 5:40
• 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. Apr 22, 2020 at 6:31
• 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.
– whuber
Apr 22, 2020 at 13:46
• @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 Apr 22, 2020 at 15:42
• 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.
– whuber
Apr 22, 2020 at 16:25

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.