Normally distributed random variables with $N(0,\sigma^2)$

Question

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.^_^

Answer 1

There is nothing wrong with your reasoning, other than mixing the letters in the first problem. Here is a faster way to handle the first problem:

$$\mathbb{E} \left[ \frac{Y^2}{\sigma^2} \right] = \frac{1}{\sigma^2} \mathbb{E} \left[Y^2\right] = \frac{1}{\sigma^2} \left( Var(Y) + \left( \mathbb{E} \left[Y\right] \right)^2 \right) = \frac{1}{\sigma^2} \left( \sigma^2 + 0 \right) = 1$$

where to get the second equality we have used that for a random variable $X$, $Var(X) = \mathbb{E} \left[X^2\right] - \left( \mathbb{E} \left[X\right] \right)^2 $, which may be proven quite easily from the general definition of the variance, i.e. $Var(X) = \mathbb{E} \left[ \left(X-\mathbb{E}[X] \right)^2 \right]$.