I can across the following lemma in my class lecture notes:
The ith highest draw from a uniform distribution on [0,x] has expectation $\frac{n + 1 - i}{n + 1}$.
I'm now attempting to prove this to myself. In order for a number to be the ith highest we have to have $i - 1$ larger numbers and $n - i + 1$ numbers that are less than or equal to our ith highest.
Let $A_1, A_2, ..., A_n$ be the draws from the uniform distribution. Let $Y_i$ denote the ith largest number. The CDF on $Y_i$ is going to be: $$P(n - i + 1\ of\ the\ draws \leq y\ \cap\ i - 1\ of\ the\ draws \geq y) $$
Since each of our draws are iid(given), $$P(n - i + 1\ of\ the\ draws \leq y\ \cap\ i - 1\ of\ the\ draws \geq y) = P(n - i + 1\ of\ the\ draws \leq y)P(i - 1\ of\ the\ draws \geq y)$$ $$P(n - i + 1\ of\ the\ draws \leq y)P(i - 1\ of\ the\ draws \geq y) = P(A_1 \leq y)P(A_2 \leq y) ... P(A_{n - i + 1}\leq y) \times (1- P(A_{n - i + 2} \leq y)(1- P(A_{n - i + 3} \leq y)(1- P(A_{n} \leq y)$$ $$ = [F(y)]^{n - i + 1} * [1-F(y)]^{i - 1}$$
Then I would take the derivative and then take an integral to find the expectation of the ith highest draw. However, taking a derivative of the above function is rather tedious. Is there a better way to do this? And is this even the right way?
You said you wanted to do this on $[0,x]$ but then instead quoted the result for $[0,1]$. I’ll go with $[0,1]$.
This doesn’t require any calculus.
Close the interval to a circle, choose $n+1$ points uniformly and independently, then uniformly choose one of them as the point at which you break the circle into a line segment. The resulting $n$ points on the line segment are uniformly distributed. By the symmetry of the circle, all $n+1$ intervals have the same expected length, which is thus $\frac1{n+1}$. The value of the $i$-th largest variable is $1$ minus the length of $i$ intervals, and taking the expectation and using the linearity of expectation shows that the expected value of the $i$-th largest variable is $\frac{n+1-i}{n+1}$.