38

What is the expected value of the maximum of 500 IID random variables with uniform distribution between 0 and 1?

I'm not quite sure of the technique to go about solving something like this. Could anyone point me the right direction?

Thanks

Galen
  • 1,946
Matt
  • 3,119

5 Answers5

51

Suppose the maximum is $X_{500}$, then $$P(X_{500}\le x)=P(X_i \le x ,i=1,2,...,500)$$ Note that this is so because if the maximum is less than $x$ , then every other order statistic is less than $x$. Now since the $X_i's$ are IID, it follows that; $$P(X_{500}\le x)=\prod_{i=1}^{500} P(X_i\le x)=x^{500}$$ which is the CDF and so the PDF is $500x^{499}$ (which is obtained by differentiation). Now the expected value of the maximum is found as follows; $$E[X]=\int _0^1 x (500x^{499})dx=\int _0^1 500x^{500}dx=\frac {500}{501}$$

The Phenotype
  • 5,300
smanoos
  • 3,191
  • X500 is max implies every order statistic is less than X500 , X500 < x => X(1, 2 .. 499) <= X500 and X500 <= x , so i am confused why is this equivalent to what you have written as first equation ? – user179156 Feb 05 '18 at 07:18
31

There is one trick useful for uniform distribution.

The trick: If you have 500 independent uniform random variables on $[0;1]$ then you may think that you have 501 independent uniform random variables on a circumference with unit length. You just consider the first random variable as a cut-point, which transforms the circumference into a unit segment $[0;1]$.

The average distance between points on the circumference is $1/501$ and ...

the average length of 500 intervals is equal to $500/501$.

Community
  • 1
Roah
  • 2,162
  • 2
    Could you provide more elucidation-- I don't understand the intuition for this way of thinking about the problem, and it seems interesting – Tina Jun 21 '18 at 20:55
  • 3
    @Tinatim You just replace the experiment by an equivalent one. Choose 501 points on a circumference instead of 500 points on a segment. The average distance between points is 1/501 by symmetry. One point is considered as a cut, that transforms circumference into the segment. The distance between the maximum point and the end of the segment is the distance between cut-point and the point to the right (or to the left, does not matter). And this distance is 1/501 on average. So the remaining distance is 500/501 on average. – Roah Jun 23 '18 at 07:24
  • @Tinatim: the 500 points will (on "average") divide the $[0,1]$ segment into 501 segments of length $1/501$. The maximum will sit at the end of the 500th segment, at position $500\times1/501$. – roygbiv Oct 18 '22 at 10:11
  • @Roah I do not see how we get n+1 arcs. Is this because we consider arc lengths from, say, the 12 o'clock position? So with two points on a circle, Arc1 = 12o'c to p1, A2 = p1 to p2, A3 = p2 to 12o'c? Given linearity of expectation isn't concerned with independce, we can say E[A1 + A2 + A3] = 1 (even though A3 and A1 depend on each other i.e length from p3 to p1) giving us the average size of 1/4? With the distance to 12 o'clock, why not just use a line? – user2330624 Jul 30 '25 at 13:54
  • @user2330624 We have $n$ original points plus one special "cut"-point that transforms the circumference into a segment. The advantage of considering circumference is the symmetry, we can conclude that all expected values of all segments are equal. This fact is not so obvious on the line, as on the line we have the leftmost and the rightmost segments. – Roah Jul 31 '25 at 14:21
  • So we are pretty much placing the points at the distance from, say, the 12 o'clock position? I don't see why it's not obvious on a line or really the same exact thing as doing it on a line. So for 1 point, we have the arc length from 12 o'clock to the point and the point to 12 o'clock? @Roah thank you – user2330624 Aug 05 '25 at 14:32
  • @user2330624 Yes, you may rotate the circle and locate the special cut-point at 12 o'clock position. The goal of introducing additional cut-point is to clarify symmetry and identical distribution (but not independence) of the lengths of all the resulting pieces. – Roah Aug 16 '25 at 08:22
4

George Pólya says "Is there a simpler problem of the same type that you do know how to solve?"

The way to go about solving something like this is to look at simpler cases first. What's the expected value of the maximum of one IID uniform random variable? What is the expected value of the maximum of two?

Mouse over if you want the actual answer.

I believe the answer is $\frac{500}{501}$.

MJD
  • 68,126
  • 44
  • 311
  • 628
1

Let $Y$ be the maximum. Then $P(Y\le y)=y^{500}$ (if $0\le y\le 1$). So now we know the cumulative distribution function of $Y$, and therefore the density, and therefore the mean.

André Nicolas
  • 516,285
0

This can be solved by taking repeated expectations. Fix $n$ (500 in your example), and let $Y_k = \mathrm{max}\,\left\{X_1, \ldots, X_k\right\}$ for $1 \leq k \leq n$. Set $R_k = \mathbb E\left(Y_k^{n + 1 - k}\right)$.

Then $R_1 = \int_0^1 x^n\,dx = \frac{1}{n + 1}$, and for $k > 1$, $$R_k = \mathbb E\left(\mathbb E Y_k^{n+1-k}\right|Y_{k-1}).$$

Now, $Y_k$ is $X_k$ if $X_k \geq Y_{k-1}$, and $Y_{k-1}$ otherwise. Thus

$$ R_k = \mathbb E \left(\int_0^{Y_{k1}}Y_{k-1}^{n+1-k}\,dx + \int_{Y_{k-1}}^1x^{n+1-k}\,dx\right) = \mathbb E\left(Y_{k-1}^{n+2-k} + \frac{1 - Y_{k-1}^{n+2-k}}{n+2-k}\right),$$ which in turn gives $$R_k = \frac{1}{n+2-k}+\frac{n+1-k}{n+2-k}R_{k-1}.$$

Using this expression for $R_k$ in terms of $R_{k-1}$, it is a matter of basic algebra to prove via induction that

$$R_k = \frac{1}{n+2 - k} + (n+1 -k)\sum_{i=n+2-k}^n \frac{1}{i(i+1)}.$$

Then $$\mathbb E Y_n = R_n = \frac{1}{2} + \sum_{i=2}^n \frac{1}{i(i+1)} = \sum_{i=1}^n \frac{1}{i(i+1)},$$ and it is again easy to prove by induction that the sum on the right equals $\frac{n}{n+1}$.

Brennan Vincent
  • 249