Prove mean and variance of $\overline{X}$.

Question

If a random sample of size $n$ is selected without replacement from the finite population that consists of the integers $1, 2, \ldots ,N$, show that

a. $E(\overline{X})=\dfrac{N+1}{2}$.

b. $var(\overline{X})=\dfrac{(N+1)(N-n)}{12n}$.

To answer this question, I construct the p.d.f of discrete random variable $X$ which have uniform distribution, i.e.

\begin{align} p_X(x)=\dfrac{1}{N}, x=1,2,\ldots,N. \end{align} Now I find the mean of $X$, i.e. $$E(X)=\sum\limits_{x=1}^N xp_X(x)=\dfrac{1}{N}(1+2+\ldots+N)=\dfrac{1}{N}\dfrac{N(N+1)}{2}=\dfrac{N+1}{2}.$$ The second moment of $X$ is \begin{align} E(X^2)=\sum\limits_{x=1}^N x^2 p_X(x)=\dfrac{1}{N}(1^2+2^2+\ldots+N^2)=\dfrac{1}{N}\dfrac{N(N+1)(2N+1)}{6}=\dfrac{(N+1)(2N+1)}{6}. \end{align} So, the variance of $X$ is \begin{align} var(X)&=E(X^2)-(E(X))^2\\ &=\dfrac{(N+1)(2N+1)}{6}-\dfrac{(N+1)^2}{4}\\ &=(N+1)\left(\dfrac{2N+1}{6}-\dfrac{N+1}{4}\right)\\ &=(N+1)\left(\dfrac{4N+2}{12}-\dfrac{3N+3}{12}\right)\\ &=\dfrac{(N+1)(N-1)}{12}. \end{align}

Now, I find the expectation and variance of $\overline{X}$ as follows.

\begin{align} E(\overline{X})&=E\left(\dfrac{X_1+X_2+\ldots+X_n}{n}\right)\\ &=\dfrac{1}{n}E(X_1+X_2+\ldots+X_n)\\ &=\dfrac{1}{n}(E(X_1)+E(X_2)+\ldots+E(X_n))\\ &=\dfrac{1}{n}(nE(X))\text{ (random samples are i.i.d.)}\\ &= E(X)\\ &=\dfrac{N+1}{2}. \end{align}

\begin{align} var(\overline{X})&=var\left(\dfrac{X_1+X_2+\ldots+X_n}{n}\right)\\ &=\dfrac{1}{n^2}(var(X_1)+var(X_2)+\ldots+var(X_n))\\ &=\dfrac{1}{n^2}(n\cdot var(X))\\ &=\dfrac{var(X)}{n}\\ &=\dfrac{(N+1)(N-1)}{12n}. \end{align}

I can prove $E(\overline{X})=\dfrac{N+1}{2}$. But I find $$var(\overline{X})=\dfrac{(N+1)(N-1)}{12n},$$ not $$var(\overline{X})=\dfrac{(N+1)(N-n)}{12n}.$$ What my mistake in my answer?

Answer 1

$var\left(\frac{X_1+X_2+\ldots+X_n}{n}\right)\not=\dfrac{1}{n^2}(var(X_1)+var(X_2)+\ldots+var(X_n))$ due to the lack of independence

Instead use $E\left[\left(\frac{X_1+X_2+\ldots+X_n}{n}\right)^2\right] -\left(E\left[\frac{X_1+X_2+\ldots+X_n}{n}\right]\right)^2 = \frac1nE[X_i^2] + \frac{n-1}{n}E[X_iX_j] -(E[X_i])^2$ where $i\not = j$. This is $$\frac1n\frac{(N+1)(2N+1)}{6}+\frac{n-1}{n}\frac{(N+1)(3N+2)}{12}-\frac{(N+1)^2}{4} = \frac{(N+1)(N-n)}{12n}$$