Computation of the expectation

Question

Assuming points are sampled from a uniform distribution on $ [0,1] $. If we sample $10$ values and compute their mean value $u$, the expectation of $u$ is $1/2$. However, if we sample $10$ values and compute the average $\hat{u}$ of the $9$ smallest of them, what is the expectation of $\hat{u}$?

Answer 1

To make the derivation more clear, let us replace the specific value $10$ by some $n\in \mathbb{N}$. Now let $X_1,\ldots,X_n\sim U(0,1)$ be independent and uniformly distributed in the unit interval $[0,1]$. Then the random variable you are interested in is the average of the $n-1$ smallest $X_i$.

Writing $X^\star := \max_{1\le k\le n} X_k$, we obtain the sum over the smallest $n-1$ of the $X_i$ by subtracting $X^\star$ from the sum over all $X_i$, thus the quantity of interest can be written as $$ M := \frac{1}{n-1} \left( \sum_{i=1}^n X_i - X^\star\right),$$ the expected value of which is $$\mathbb{E} [M] = \frac{1}{n-1} \left( \frac{n}{2} - \mathbb{E}\left[X^\star\right]\right).$$ Hence it remains to find the expected maximum of $n$ independent $U(0,1)$-variables. This can be shown to be $\mathbb{E}\left[X^\star\right] = \frac{n}{n+1}$ (a derivation of which I would be happy to add if you're interested), giving us the final result \begin{align*} \mathbb{E}[M] &= \frac{1}{n-1} \left( \frac n2 - \frac{n}{n+1}\right)\\ &= \frac{1}{n-1} \frac{n(n+1) - 2n}{2(n+1)}\\ &= \frac 12\frac{n}{n+1}. \end{align*} As expected, this is smaller than but gets increasingly close to $\frac 12$ as $n$ increases. In your case, where $n = 10$, this becomes $\mathbb{E}[M] = 10/22 \approx 0.45$.