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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?

1 Answers1

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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 $X$, $$E(X) = \int_0^\infty\left(1-F_X(t)\right)dt$$

Lazy Lee
  • 3,631
  • Ah, I completely missed that $F_{X_{(n)}}(x) = 0$ on $[0,2]$. Thank you! –  Jul 13 '17 at 05:18
  • I should thank you for this very interesting problem! I actually didn't know we could calculate expectation by integrating 1-CDF. Very interesting! – Lazy Lee Jul 13 '17 at 05:19
  • For what it's worth, this method of calculating expectation is unofficially called the "Darth Vader" rule. –  Jul 13 '17 at 05:24
  • Sounds very cool, thank you for the info :) – Lazy Lee Jul 13 '17 at 05:25