# 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.

The issue is that you aren't considering the full support of cdf ofmax$$\{y_1,_2,_y\}$$. 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! – Kitsune Cavalry May 13 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 :) – Lucas Roberts May 13 at 2:04