# Find variance of a function of a sequence of iid N(0,1) r.v.s

If $X_1, X_2, ..., X_n$ are iid $N(0,1)$ r.v.s, then $Y = \sqrt{n}\bar{X_n} \tilde{} N(0,1)$ (we can show this by calculating the MGF of $Y$). However, when I try to find the variance of $Y$ directly, I get a wrong answer:

$$Var(Y) = Var(\sqrt{n}\bar{X_n}) = Var(\sqrt{n}\frac{1}{n}\sum_{i=1}^n{X_i}) = \frac{1}{n}Var(\sum_{i=1}^n{X_i}) = \frac{1}{n}\sum_{i=1}^nVar(X_i) = \frac{1}{n}$$

I expect the variance to be $1$, not $1/n$. What did I do wrong?

• You forgot to do the sum. Dec 7, 2016 at 4:51
• Yes! When you sum 1/n n times you do get 1! Dec 7, 2016 at 6:04

$$Var(Y) = Var(\sqrt{n}\bar{X_n}) = Var(\sqrt{n}\frac{1}{n}\sum_{i=1}^n{X_i}) = \frac{1}{n}Var(\sum_{i=1}^n{X_i}) = \frac{1}{n}\sum_{i=1}^nVar(X_i) = \frac{1}{n}$$
$$Var(Y) = Var(\sqrt{n}\bar{X_n}) = Var(\sqrt{n}\frac{1}{n}\sum_{i=1}^n{X_i}) = \frac{1}{n}Var(\sum_{i=1}^n{X_i}) = \frac{1}{n}\sum_{i=1}^nVar(X_i) = \frac{1}{n}(Var(X_1) + Var(X_2) + \dots + Var(X_n)) = \frac{1}{n}(Var(X_1)n) )=Var(X_1)$$
where I've added 2 additional equalities for emphasis, the first writes out the summation of the variances and the second uses the i.i.d. assumption which tells you all the variances are the same constant value, so the sum of $n$ numbers (variances) is $n\times Var(X_1)$.