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This is distribution of the sample mean when $\sigma$ is known

Theorem 6.1: Let $X_1, ..., X_n$ be a random sample from a distribution with mean $\mu$ and standard deviation $\sigma$

Then Var($\bar{X})$ = $\cfrac {\sigma^2}{n}$

Proof in the book: By independence, Var($\bar{X})$ = Var($\sum_{i=1}^n \cfrac{X_i}{n}$) = $\cfrac{1}{n^2}$Var($\sum_{i=1}^nX_i$) = $\cfrac {1}{n^2}(n\sigma^2) = \cfrac{\sigma^2}{n}$

Now I know sample mean is just ($\sum_{i=1}^n \cfrac{X_i}{n}$)

But where does the $\cfrac {1}{n^2}$ and $n\sigma^2$come from?

2 Answers2

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Note that $Var(kY)=k^2Var(Y)$, hence when we move $\frac1n$ out from the variance term, we get $\frac1{n^2}$.

Also for independent random variable, the sum of the variance is equal to the variance of the sum. $$Var(\sum_{i=1}^n X_i) = \sum_{i=1}^nVar( X_i)=n\sigma^2$$

Siong Thye Goh
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If you look at the basic formulas: $E(Y^k)=\int y^kdF(y)$ where $F(y)$ is the distribution function of a random variable $Y$. The variance of $Y$ is $E(Y^2)-E(Y)^2$. Now let $Y=\frac{X_i}{n}$ so $Var(\frac{X_i}{n})=\frac{E(X_i)^2}{n^2}-\frac{E(X_i)^2}{n^2}=\frac{Var(X_i)}{n^2}=\frac{\sigma^2}{n^2}$.
Since the $X_i$ are independent, the variance of the sum is $n\frac{\sigma^2}{n^2}$

herb steinberg
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  • Thank you, I just have one question, is it dF(y) or F(y) in the basic formula? Because I have a feeling it should be the latter and want to make sure –  Oct 10 '20 at 03:43
  • Stieltjes integral representation (using $dF(y)$) is the most general way of expressing the integrals involving an arbitrary distribution function (CDF) of a random variable. If you have a PDF $f(y)$ the integral would be over $f(y)dy$. – herb steinberg Oct 10 '20 at 03:53
  • I see, Thank You –  Oct 10 '20 at 04:11