I have a random variable $X_n=\frac{1}{n} \sum^n_{i=1} Z_i$ for $n\in \mathbb{Z}^+$ where $X_0=0$ and the $\{ Z_i \} \sim N(0, 1)$ for all $i$. The $Z_i$ are independent. I need to find the distribution of $X_n$.
I know that a linear combination of normal random variables is also normal. I also know that $\mathbb{E}(X_n)=0$. However, I'm not sure how to calculate the variance of $X_n$. I know $\mathrm{Var}(\sum^n_{i=1} Z_i)=1+1+1...$ n times $= n$. But am unsure what happens when I try to find the $\mathrm{Var}(\frac{1}{n}\sum^n_{i=1} Z_i)$. My friends have gotten answers as $\frac{1}{n+1}$, but I got $\frac{1}{n}$ as my answer from the following working: $$\mathrm{Var}(\frac{1}{n}\sum^n_{i=1} Z_i)=(\frac{1}{n})^2\mathrm{Var}(\sum^n_{i=1} Z_i)$$ $$\mathrm{Var}(\frac{1}{n}\sum^n_{i=1} Z_i)=\frac{1}{n^2}\cdot n$$ $$\mathrm{Var}(\frac{1}{n}\sum^n_{i=1} Z_i)=\frac{1}{n}$$
Can anyone offer some insight?
It should indeed be $\frac1n$. More generally,$$\operatorname{Var}(\sum_ia_iZ_i)=\operatorname{Cov}(\sum_ia_iZ_i,\,\sum_ja_jZ_j)=\sum_{ij}a_ia_j\operatorname{Cov}(Z_i,\,Z_j)=\sum_{ij}a_ia_j\delta_{ij}=\sum_ia_i^2,$$where the penultimate $=$ uses the fact that the $Z_i$ are independent, and each of mean $0$ and variance $1$, to write $\operatorname{Cov}(Z_i,\,Z_j)=\delta_{ij}$ in terms of the Kronecker delta. Now take $a_i=\frac1n,\,1\le i\le n$.