Prove that $E(Y_i \bar{Y}) = \frac{\sigma^2}{n}+\mu^2$

Question

Given $Y_1, Y_2,...Y_n$ are i.i.d random variable which follows a distribution of $N(\mu, \sigma^2)$, I'm trying to prove that $$E(Y_i \bar{Y}) = \frac{\sigma^2}{n}+\mu^2$$

Here's what I've tried: $$E(Y_i \bar{Y}) = E(Y_i \frac{1}{n} \sum_{i=1}^{n}Y_i )$$ $$ = \frac{1}{n} \sum_{i=1}^{n}E(Y_i Y_i )$$ Since $E(Y_i^2) = Var(Y_i)+E(Y_i)^2 = \sigma^2 +\mu^2$ $$E(Y_i \bar{Y}) = \frac{1}{n} \sum_{i=1}^{n}E(Y_i^2)$$ $$ = \frac{1}{n} n (\sigma^2 +\mu^2)$$ $$ = (\sigma^2 +\mu^2)$$ which does not prove it right. I am unsure of where i got this wrong. Can anyone help me with this? Thank you.

Answer 1

Here is the error in your argument caused by the choice of the index letter that is already used.

$E(Y_i \bar{Y}) = E(Y_\color{red}{i} \frac{1}{n} \sum_{\color{red}{i=1}}^{n}Y_\color{red}{i} )$

To be correct, \begin{align} E(Y_i \bar{Y}) = E(Y_i \frac{1}{n} \sum_{\color{blue}{j=1}}^{n}Y_\color{blue}{j} ) = E(\sum_{j=1}^{n} \frac{1}{n}Y_iY_j) =\frac{1}{n}\sum_{j=1}^{n} E(Y_iY_j) \end{align}

If $i \neq j$, we have $E(Y_i Y_j)=E(Y_i)E(Y_j)=\mu^2$. If $i=j$, $E(Y_i^2)=Var(Y_i) + E(Y_i)^2=\sigma^2+\mu^2$

Thus \begin{align} E(Y_i \bar{Y}) = \frac{1}{n}\sum_{j=1}^{n} E(Y_iY_j)&=\frac{1}{n} \left( (n-1)\mu^2+\ \sigma^2 + \mu^2\right) \\ &=\frac{1}{n}(n\mu^2 + \sigma^2)=\mu^2 + \frac{1}{n}\sigma^2 \end{align}