The random variable $Y_i$ is i.i.d. and normally distributed $N(0,\sigma^2)$ for all $i$. How will I prove that
$$E\left(\frac{Y^2}{\sigma^2}\right) = 1$$ and $$W = \frac{1}{\sigma^2}\sum_{i=1}^n Y_i^2$$ is distributed $\chi_n^2$, and
$$E(W) = n$$
Here are my steps in proving:
Since we are dealing with standard normal r.v., the equation $Z = \frac{Y - \mu}{\sigma}$ can be applied. So that $E(\frac{Y^2}{\sigma^2})$ is equal to $E(\frac{Y}{\sigma})^2$, which in turn is equal to $E(Z)^2$ given that $\mu = 0$. Using theorem that states that if $X \thicksim \chi^2(r)$, then $E(Y)=r$, therefore $E(Y)=1$ since $r = 1$.
As for $W \thicksim \chi^2(n)$, if $Z_1,Z_2,...,Z_n$ are independent normal random variables with different means and variances, that is: $Z_i \thicksim N(\mu_i,\sigma_i^2)$ for $i = 1,2,...,n.$ Given that $W = \sum_{i=1}^{n} \frac{Y_i^2}{\sigma^2} = \sum_{i=1}^{n} Z_i^2$. Therefore, $W \thicksim X^2(n)$.
For the last problem, we know that $W = \frac{1}{\sigma^2}\sum_{i=1}^n Y_i^2$. So that $E[\frac{1}{\sigma^2}\sum_{i=1}^n Y_i^2]$. This is also equal to $\frac{1}{\sigma^2}\sum_{i=1}^n E[Y_i^2]$ by linearity of expectation. Since $E[Y_i^2] = Var[Y] + (E[Y])^2$, $\frac{1}{\sigma^2}\sum_{i=1}^n [Var[Y] + (E[Y])^2]$. From the given above, $E[Y] = \mu = 0$, what is left from the equation is $\frac{1}{\sigma^2}\sum_{i=1}^n Var[Y]$. The term $\sum_{i=1}^n Var[Y]$ is the same as $[\sigma^2+\sigma^2+...+\sigma^2]$ because $Y_i$ is identically distributed, which means they have the same variance $\sigma^2$. The summation is also equal to $n\sigma^2$ since there are $n$ variances of $Y$. Therefore we have $\frac{1}{\sigma^2}[n\sigma^2]$, which is equal to $n$. I hope they are correct.^_^