I'm given $Y_1, Y_2, ..., Y_i$ i.i.d. random variables that are $N(1, \sigma ^2)$.
I am asked to obtain the $E \big[\frac{1}{n} \sum_{i=1}^{n} Y_i^2 \big]$
What I did is equate what's in brackets to the mean of $Y_i^2$ so...
MY SOLUTION:
$E\big[\overline {Y_i^2} \big]$
$E\big[\overline {Y_i^2} \big]$ = $V\big[\overline Y_i \big] + E\big[\overline Y_i \big]^2$
Then of course $V\big[\overline Y_i \big]$ = $\sigma ^2 /n$ and $E\big[\overline Y_i \big]^2$ = 1
so $E\big[\overline {Y_i^2} \big]$ = $\frac{\sigma ^2} {n} + 1$
BUT, the solutions say that the answer is simply ${\sigma ^2} + 1$ since they extracted the sum from the intial expected value given so:
THE CORRECT SOLUTION:
$E \big[\frac{1}{n} \sum_{i=1}^{n} Y_i^2 \big]$ = $\frac{1}{n}\sum_{i=1}^{n}E \big[ Y_i^2 \big]$ = $E \big[ Y_i^2 \big]$ = $E\big[ Y_i^2 \big]$ = $V\big[ Y_i \big] + E\big[ Y_i \big]^2$ = $\sigma +1$
Is $E \big[\frac{1}{n} \sum_{i=1}^{n} Y_i^2 \big] $ not equal to $E\big[\overline {Y_i^2} \big]$? If not why not ? My main concern is I do not understand why my solution is wrong
$$ \mathbb{E} \left[\frac 1 n \sum_{i=1}^n Y_i^2\right] = \frac 1 n \sum_{i=1}^n \mathbb{E} \left[Y_i^2\right] = \mathbb{E} Y_1^2 = \sigma^2 + 1. $$ The first equation is due to linearity of expected value operator, second one is due to i.i.d. of $Y_i$s and the third is the formula for variance.
What is wrong with your solution: You write $\overline{Y_i^2}$ but instead it should be $(\overline{Y_i})^2$