1
$\begingroup$

Suppose we have a random sample $(X_1,..,X_n)$ from a Uniform$(2,2\theta)$ distribution. We're interested in $\mathbb{E}[X_{(n)}]$, the expectation of the maximum order statistic. I found that the CDF of $X_{(n)}$ is,

$$ F_{X_{(n)}}(x) = \left(\frac{x-2}{2\theta-2}\right)^n \;\;\;\; 2 < x < 2\theta$$

The corresponding pdf is then,

$$ f_{X_{(n)}}(x) = \frac{n(x-2)^{n-1}}{(2\theta-2)^n} \;\;\;\; 2 < x < 2\theta $$

I've been trying to calculate the expectation using two methods,

$$ \mathbb{E}[X_{(n)}] = \int\limits_2^{2\theta}xf_{X_{(n)}}(x)dx \;\;\;\;\;\; \& \;\;\;\;\;\; \mathbb{E}[X_{(n)}] = \int\limits_2^{2\theta}(1-F_{X_{(n)}}(x))dx $$

However, I'm getting different results,

$$ \int\limits_2^{2\theta}xf_{X_{(n)}}(x)dx = \int\limits_2^{2\theta}x\frac{n(x-2)^{n-1}}{(2\theta-2)^n}dx = \frac{2\theta n + 2}{n+1} $$ $$ \int\limits_2^{2\theta}(1-F_{X_{(n)}}(x))dx = \int\limits_2^{2\theta}\left(1-\left(\frac{x-2}{2\theta-2}\right)^n\right)dx = \frac{2\theta n - 2n}{n+1}$$

Am I not applying the formula correctly? Under what conditions (other than $X$ being nonnegative) does the formula, $\mathbb{E}[X] = \int\limits_0^\infty(1-F_X(x))dx$, apply?

$\endgroup$
1
$\begingroup$

For the second integral, you should start integrating from $0$ rather than $2$, hence $$\int_0^\infty(1-F_{X_n}(x)dx = \int_0^21dx + \int_2^{2\theta}(1-F_{X_n}(x))dx = 2+\frac{2\theta n -2n}{n+1} = \frac{2\theta n+2}{n+1}$$ as desired.

Note: You can see this post for a proof that for non-negative random variable $X$, $$E(X) = \int_0^\infty\left(1-F_X(t)\right)dt$$

$\endgroup$
  • $\begingroup$ Ah, I completely missed that $F_{X_{(n)}}(x) = 0$ on $[0,2]$. Thank you! $\endgroup$ – Flowsnake Jul 13 '17 at 5:18
  • $\begingroup$ I should thank you for this very interesting problem! I actually didn't know we could calculate expectation by integrating 1-CDF. Very interesting! $\endgroup$ – Lazy Lee Jul 13 '17 at 5:19
  • $\begingroup$ For what it's worth, this method of calculating expectation is unofficially called the "Darth Vader" rule. $\endgroup$ – Flowsnake Jul 13 '17 at 5:24
  • $\begingroup$ Sounds very cool, thank you for the info :) $\endgroup$ – Lazy Lee Jul 13 '17 at 5:25

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.