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.^_^


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. $Var(X) = \mathbb{E} \left[ \left(X-\mathbb{E}[X] \right)^2 \right]$.

  • $\begingroup$ Thanks for the 1st problem. Can I assume that the 2nd problem is correct? $\endgroup$ – Zander Assand Mar 15 '17 at 7:42
  • $\begingroup$ @ZanderAssand Yes. $\endgroup$ – JohnK Mar 15 '17 at 7:42
  • $\begingroup$ As a follow-up, how will i get the expected value of W? $$E[\frac{1}{\sigma^2} \sum_{i=1}^n Y_i^2]$$ $\endgroup$ – Zander Assand Mar 15 '17 at 8:18
  • $\begingroup$ @ZanderAssand Use again the linearity of the expectation and examine the expectation of each term in the sum separately. $\endgroup$ – JohnK Mar 15 '17 at 8:29
  • 1
    $\begingroup$ That would be:$\frac{1}{\sigma^2}\sum E[Y^2] = \frac{1}{\sigma^2}\sum (Var[Y]+(E[Y])^2)$. The right-hand side of the equation becomes $\frac{1}{\sigma^2}\sum(\sigma^2+0)$, which is equal to $\sum 1$? $\endgroup$ – Zander Assand Mar 15 '17 at 9:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.