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.
 A: 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:
$$
200 + 300 + 0.
$$
A: 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
$\ p(m)=P(max(X1,X2,X3))=
\\P(X1)⋅P(X2≤X1)⋅P(X3≤X1)+P(X2)⋅P(X1≤X2)⋅P(X3≤X2)+P(X3)⋅P(X1≤X3)⋅P(X2≤X3)=$
\begin{equation}
\\ \sum^{n=3}_{i=1} [1/(b-a)].[(m-a)/(b-a)]^{n-1}=[n/(b-a)].[(m-a)/(b-a)]^{n-1}
\end{equation}
