Compute $Var[\frac{1}{n}\sum_{i=1}^n (X_i-E[X])^2]$

Question

Let $X_1,\ldots, X_n$ be positive random variables in the range $[a, b]$ and $\sum_i^n X_i = 1$.

I want to compute

\begin{eqnarray} Var[\frac{1}{n}\sum_{i=1}^n (X_i-E[X])^2] &=& \frac{1}{n^2}Var[\sum_{i=1}^n (X_i-E[X])^2]\\ &=& \frac{1}{n^2}\sum_{i=1}^n Var[X_i^2] - 4Var[X_i E[X]] + Var[E[X]^2]\\ \end{eqnarray}

Here I got stuck. Are there any approximations I can use to go further or to at least estimate an upper bound for the variance?

Answer 1

I simplify the term as far I can think of

$$\frac{1}{n^2}Var[\sum_{i=1}^n (X_i-E[X])^2]$$

Next you multiply out the square brackets and leaving variance opterator still out of the brackets.

$$\frac{1}{n^2}Var[\sum_{i=1}^n (X_i^2-2E[X]X_i+E[X]^2)]$$

$$\frac{1}{n^2}Var[\sum_{i=1}^n X_i^2-2E[X]\sum_{i=1}^n X_i+\sum_{i=1}^nE[X]^2)]$$

$E[X]^2$ is a constant and its variance is 0.

$$\frac{1}{n^2}Var[\sum_{i=1}^n X_i^2-2E[X]\sum_{i=1}^n X_i+n\cdot E[X]^2)]$$

$$\frac{1}{n^2}[Var\left(\sum_{i=1}^n X_i^2\right)-4E[X]^2Var\left(\sum_{i=1}^n X_i\right)]$$

Since $\sum_{i=1}^n X_i=1$, we have a variance of a constant again.

$$\frac{1}{n^2}\cdot Var\left(\sum_{i=1}^n X_i^2\right)$$

To calculate/estimate it explicity further informations are needed.

Answer 2

My thinking was that

• $\frac1n\sum (X_i-E[X_i])^2$ cannot be more than $\frac1n(1-\frac1n)^2+\frac{n-1}n(0-\frac1n)^2 = \frac{n-1}{n^2}$ or less than $0$ if $\sum X_i =1$ and each $X_i\ge 0$

• so its variance cannot exceed $\left(\frac12 \frac{n-1}{n^2}\right)^2 = \dfrac{(n-1)^2}{4n^4}$.

• To achieve this upper bound, have

  • probability $\frac12$ that all the $X_i=\frac1n$ and
  • probability $\frac12$ that one of the $X_i=1$ and the others $=0$.

• The maximum of the upper bound is $\frac1{64}$ when $n=2$.

  • For an example of that, consider $Y\sim \text{Bin}(2,\frac12)$ and $X_1=\frac12Y$ and $X_2=\frac12(2−Y)$.