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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?

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  • $\begingroup$ You forgot to do the sum. $\endgroup$ – GeoMatt22 Dec 7 '16 at 4:51
  • $\begingroup$ Yes! When you sum 1/n n times you do get 1! $\endgroup$ – Michael Chernick Dec 7 '16 at 6:04
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As pointed out in the comments by @Michael Chrenick and @GeoMatt22 you need to sum the last one, you've written:

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

but in fact for the last equality you need:

$$ Var(Y) = Var(\sqrt{n}\bar{X_n}) = Var(\sqrt{n}\frac{1}{n}\sum_{i=1}^n{X_i}) = \frac{1}{n}Var(\sum_{i=1}^n{X_i}) = \frac{1}{n}\sum_{i=1}^nVar(X_i) = \frac{1}{n}(Var(X_1) + Var(X_2) + \dots + Var(X_n)) = \frac{1}{n}(Var(X_1)n) )=Var(X_1) $$

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 $n\times Var(X_1)$.

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