# Expected Value of Maximum of Uniform Random Variables

This is a question that is bothering me just because I cannot find a seemingly simple mistake in my work for a question I know the answer to intuitively and through another method.

I was looking at a situation where we had

$$X_1, X_2, X_3 \sim \text{Unif}(200,600)$$ $$Y = \max\{{X_i}\}$$

where these three draws are i.i.d.

It is not hard to see that because of the independence,

$$P(\max(X_1 , X_2 , X_3) \leq y) = P(X_1 \leq y) \cdot P(X_2 \leq y) \cdot P(X_3 \leq y)$$ $$= \left(\frac{y-200}{400}\right)^3$$

Now we know $$E(Y) = \int^{600}_{200} y \cdot (f(y)) \ dy$$

where $$f(y)$$ is the density, easily found with calculus, so we should have:

$$\int^{600}_{200} y \cdot \frac{3(y - 200)^3}{64000000} \ dy$$ $$= \boxed{500}$$

This answer makes complete sense to me. If taking one draw from the uniform distribution, the expected max is just the average, or 1/2 of the way from 200 to 600. If taking two draws, the expected maximum should be 2/3rds of the way from 200 to 600, or 466.666. If taking three draws, the expected maximum should be 3/4ths of the way from 200 to 600, or 500. So on and so forth.

However, I initially tried to solve this problem with a different formula:

$$E(Y) = \int^{600}_{200} (1 - P(Y \leq y)) \ dy$$ $$= \int^{600}_{200} \left[1 - \left(\frac{y-200}{400}\right)^3\right] dy$$

When I plug this into WolframAlpha, I get 300, which clearly makes no sense. I tried it with the case for one draw and two draws as well, and the formula I am using seems to consistently undershoot what I should be seeing. Actually, it consistently undershoots the answer by 200 it seems. Doing the problem by hand also gives me the same curious nonsense. I am baffled at where I have gone wrong in setting up this form of a solution, and am sure I am missing something obvious.

• I find it helpful to work with a simpler way of expressing the numbers: adopt a system of measurement in which the origin is at $200$ and $400$ is one unit. Your variables are uniform on $[0,1]$ in this system, whose CDF is $x$ (for $0\le x\le 1$), whence the CDF of their maximum is $x^3$ (for $0\le x\le1$) and therefore the expectation is $\int_0^1(1-x^3)\mathrm{d}x=3/4.$ This is the formula you tried to use. In these new units it's harder to make a mistake in the computations. In the original units, $3/4$ is equivalent to $200 + 3/4\times 400=500:$ it's now obvious what was missing.
– whuber
Jan 13 at 19:37

The issue is that you aren't considering the full support of cdf of $$Y=\text{max}\{X_1,X_2,X_3\}$$. The full support is $$(0, \infty)$$. Taking a look here: https://en.wikipedia.org/wiki/Uniform_distribution_(continuous) at the definition of $$F(x)$$. Then consider that you'll have 1 minus this value, so for your problem you'd have: $$a=200$$, $$b=600$$ and then $$1-F(y) = 1$$ if $$x < 200$$, $$1-F(y)=0$$ if $$x>600$$ and $$1-\frac{y-200}{400}$$ when $$y \in [200, 600]$$. So the part you are missing in your calculations is:

$$\int_0^{200}dy=200.$$

which is what you're undershooting.

The portion of the integral above $$600$$ is all $$0$$ so can be safely omitted from the calculation. If you wanted to be complete, you'd write:

$$\mathbb{E}(Y_{3:1}) = \int_0^{200}(1-F(y))dy + \int_{200}^{600}(1-F(y))dy + \int_{600}^{\infty}(1-F(y))dy$$

which is: $$\int_0^{200}1dy + \int_{200}^{600}\left(1-\left(\frac{y-200}{400}\right)^3\right)dy + \int_{600}^{\infty}0dy,$$ which simplifies to: $$200 + 300 + 0.$$

• Thank you! This hits the mark! May 13, 2020 at 2:02
• @KitsuneCavalry survival functions are tricky this way when defined on only a partial support. I used to teach this example to Electrical engineers to illustrate exactly this point, I think I chose $a=2$ and $b=6$ to make the calculations simpler though :) May 13, 2020 at 2:04

I think

if $$\ x \sim Uniform(a=200,b=600)$$ ,$$\ n={3}$$

then

$$\ E{[max(X1,X2,X3)=m]}=\int_a^b m.p(m).dm= \\(n/(n+1)).(b-a)+a=(3/4).(400)+200=500$$

where

$$\ p(m)=P(max(X1,X2,X3))= \\P(X1)⋅P(X2≤X1)⋅P(X3≤X1)+P(X2)⋅P(X1≤X2)⋅P(X3≤X2)+P(X3)⋅P(X1≤X3)⋅P(X2≤X3)=$$

$$$$\\ \sum^{n=3}_{i=1} [1/(b-a)].[(m-a)/(b-a)]^{n-1}=[n/(b-a)].[(m-a)/(b-a)]^{n-1}$$$$

• This is the same as your deleted answer, so the comments made there apply here too. Mine was "Although this answer is generally correct, it doesn't address the question itself, which is why the second calculation mysteriously "lost" $200$ from the result."
– whuber
Jan 19 at 16:51