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I have a seemingly innocuous problem that I can't seem to wrap my head around. The following is mentioned in passing on one of my lecture slides, but when I try to arrive at the same conclusion I get stuck.

Consider N independent draws from a uniform distribution over [0, 1]. On average, what is the highest draw? I know that the answer is:

$$ \frac{N}{N+1} $$

but I can't arrive there myself. Honestly, I'm not even sure how to start. Can someone sketch a procedure or show me exactly what needs to be done? Thanks!

Michael Boutros
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    An "intuitive" consideration: on average, the $N$ values must be equally spaced over the interval and from the end points, that is, they must be $$\frac{1}{N+1}, \frac{2}{N+1}, \cdots, \frac{N}{N+1}$$ (Check that the spacings are OK). Thus, on average, the rightmost point is at $\frac{N}{N+1}$. – Dilip Sarwate Apr 20 '15 at 19:05
  • @DilipSarwate I tried to reason the same way but I hit a mental roadblock: if the draws are all independent, why would they be equally spaced over the interval? Shouldn't they all land right in the middle? – Michael Boutros Apr 22 '15 at 19:13
  • I didn't say that the draws are equally spaced over the interval but that one should expect the draws to be equally spaced. Suppose that you are asked to validate the behavior of a new design of a "random number generator" that returns values that_should_ be spread uniformly in $[0,1]$. Would an output that bunches all the values near $\frac 12$ not raise suspicion that something is awry? What would be more persuasive that the generator is reasonable is a sequence of $N$ values whose histogram that looks like a uniform distribution, no? – Dilip Sarwate Apr 22 '15 at 20:33
  • That makes sense! Your post inspired a bit of a eureka moment for me. Every point along the distribution is equally likely to be drawn, so if there's one point, we expect it in the middle of the interval. With two points, we expect them to be uniformly distributed, even if each individual draw would alone be expected at the half-way mark, and so on for N draws. – Michael Boutros Apr 24 '15 at 04:43

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Let $X_i$ be a single draw from the uniform distribution. Then it follows that it has a CDF of $$F(x) = \left\{\begin{array}{lr} x, & 0\le x\le1,\\ 0, & \text{ Otherwise}. \end{array}\right.$$ For $N$ draws, then, what we are looking for is $Y=\max(X_1,X_2,\ldots X_N)$. The CDF of $Y$ is $$G(y)=\Pr(Y\le y)$$ Since $Y$ is the max, this means each $X_i$ can be no greater than $y$, thus, $$=\Pr(X_1\le y, X_2\le y,\ldots,X_N\le y) $$ Now since each draw is independent, we may split these probabilities up, $$=\Pr(X_1\le y)\Pr(X_2\le y)\ldots\Pr(X_N\le y)\\ =F(y)F(y)\ldots F(y)\\ =[F(y)]^N$$ Now, knowing the CDF of $Y$, can you go on to find the PDF and expectation?

Brent
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  • Thanks, Brent, for the headstart. I get that g(y) = N*(F(y)^(N-1))*f(y). Now, f(y) = 1 over the entire uniform distribution. To get E[Y], I integrate y*g(y) from 0 to 1. But doing so get's me (1/2) in the end. Is my expression for g(y) wrong? – Michael Boutros Apr 20 '15 at 05:43
  • $f(y)=1$, yes, but $F(y)=y$, so your integral should be $\int_0^1y[Ny^{N-1}]\cdot1$ – Brent Apr 20 '15 at 05:55
  • Ah, indeed! Thanks for your help! – Michael Boutros Apr 22 '15 at 19:12