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Monte Carlo simulation suggests that this expected value is $\frac{1}{n+2}$. I'm trying to prove this - brute force calculating doesn't work well, does anyone have any tips on how to proceed?

ringwraith10
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  • What do you mean product of maximum, minimum? Do you mean you have n random variables and you just multiply the largest and smallest together? – ConMan Aug 28 '18 at 23:31
  • yes, precisely, $\mathbb{E}[min*max]$, where min and max are the minimum, maximum of n uniform(0,1) variables respectively – ringwraith10 Aug 28 '18 at 23:32

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Let $U_1, U_2, U_3, \dots , U_n$ be i.i.d. Uniform(0,1) random variables, and let $X = \min_{1 \le i \le n} U_i$ and $Y = \max_{1 \le i \le n} U_i$, so we want to find $E(XY)$. We proceed by finding the joint distribution function of $(X,Y)$, then the density function, and finally integrate to find the expected value of $XY$.

To find the joint distribution function, note that $$\begin{align} F(x,y) &= P(X \le x, Y \le y) \\ &= P(Y \le y) - P(X \ge x, Y \le y) \\ &= y^n - (y-x)^n \end{align}$$ for $0 \le x \le y \le 1$. Then the joint density function is $$f(x,y) = \frac{\partial^2 F(x,y)}{\partial x \; \partial y}=n(n-1)(y-x)^{n-2} $$ so the expected value of $XY$ is $$\begin{align} E(XY) &= \int_0^1 \int_0^y xy \; f(x,y)\; dx\;dy \\ &= n(n-1) \int_0^1 \int _0^y xy(y-x)^{n-2} \;dx \; dy \\ &= \frac{1}{n+2} \end{align}$$

awkward
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