Suppose that $Y_1,Y_2,...,Y_n$ is a random sample of size $n$ where $Y_i$~$N(0,\sigma^2)$. Let $\bar Y$ and $S^2$ denote the usual sample mean and variance.

Find the the variance of $S^2$.

In this problem, I'm assuming that $S^2=\frac{\sum_{i=1}^{n}(Y_i-\bar Y)^2}{n-1}$. I know that the difference of normal random variables is normal (since $\bar Y$ is normal), so I get a chi squared random variable in the numerator and then obtain the following result: $$V(S^2)=\frac{2n(\sigma^2+\frac{\sigma^2}{n})^2}{(n-1)^2}$$ But I have a lingering suspicion that this isn't correct. Did I make an assumption somewhere that I shouldn't have?

  • $\begingroup$ Is there not a separate stack exchange for the stats folks? (I don't have a problem with you posting on here; just genuinely curious.) $\endgroup$ – Tanner Strunk Jan 30 '18 at 5:48

Note the well known result that:

$$ \frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1} $$

(chi-squared distribution with $n-1$ degrees of freedom), and further, the variance of $\chi^2_{n-1}$ is $2(n-1)$.

Now, using this information:

\begin{align*} 2(n-1) = Var \left ( \frac{(n-1)S^2}{\sigma^2} \right) & = \frac{(n-1)^2}{\sigma^4} Var(S^2) \end{align*}

and so:

$$ Var(S^2) = \frac{2(n-1)\sigma^4}{(n-1)^2} = \frac{2\sigma^4}{n-1} $$

So your answer is a bit off, perhaps you could share your steps/assumptions for some advice on where you might have gone wrong

Proof of the result:

note the following useful facts:

  1. For standard normal $Z \sim N(0,1)$, $Z^2 \sim \chi^2_{1}$
  2. $\bar{X} \bot S^2$ (independence)
  3. The moment generating function of a $\chi_p^2$ is $m_{\chi_p^2}(t)= (1-2t)^{-p/2}$
  4. If $X_1, \dots, X_n$ are independent and $X_i = \chi^2_{p_i}$ then:

$$ \sum_{i=1}^n X_i \sim \chi^2_{p1 + p2 + \dots + p_n} $$

The trick for these types of proofs is to use this fact:

\begin{align*} &\sum (X_i - \mu)^2 = (n-1)S^2 + n(\bar{X} - \mu)^2\\ \implies &\sum \left ( \frac{X_i - \mu}{\sigma} \right )^2 = \frac{(n-1)S^2}{\sigma^2} + n \left ( \frac{\bar{X} - \mu}{\sigma} \right ) ^2\\ \implies &\sum \left ( \frac{X_i - \mu}{\sigma} \right )^2 = \frac{(n-1)S^2}{\sigma^2} + \left ( \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \right ) ^2\\ \implies & A = B+C \end{align*}

By fact number 4 $$ A = \sum \left ( \frac{X_i - \mu}{\sigma} \right )^2 \sim \chi_{n}^2 $$

and by fact number 1

$$ C = \left ( \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \right ) ^2 = Z^2 = \chi_1^2 $$

Using facts number 2 and number 3:

\begin{align*} &m_A(t) = m_B(t) m_C(t)\\ \implies &(1-2t)^{-n/2} = M_B(t) (1-2t)^{-1/2}\\ \implies & M_{B}(t) = \frac{(1-2t)^{-n/2}}{(1-2t)^{-1/2}}\\ \implies & M_{B}(t) = (1-2t)^{-(n-1)/2}\\ \end{align*}

which you should note is the moment generating function of a chi square distributed random variable with degrees of freedom $n-1$. The variance follows easily from there.


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