Apologies for the format of this question - I am new to this website. I am having trouble with part (c) of the question below, if anyone could assist that would be great.

Thanks in advance

Suppose that two independent random samples of size $n_1$ and $n_2$ observations are selected from normal populations. Let $X_1,\ldots,X_{n_1}$ and $Y_1,\ldots,Y_{n_2}$ be the two random samples and suppose that $X_i\overset d=N(\mu_1,\sigma^2)$ and $Y_i\overset d= N(\mu_2,\sigma^2).$ Note that we are assuming that the populations have a common variance $\sigma^2$. Define the sample variance from each sample as follows $$S_1^2=\dfrac {\sum_{i=1}^{n_1}(X_i-\overline X)^2} {n_1-1}\quad\text{and}\quad S_2^2=\dfrac {\sum_{i=1}^{n_2}(Y_i-\overline Y)^2} {n_2-1}. $$ Also define two pooled variance estimators $$S_{p_1}^2=\dfrac {(n_1-1)S_1^2+(n_2-1)S^2_2} {n_1+n_2-2}\quad\text{and}\quad S^2_{p_2}=\dfrac {1} {2}\left( S_1^2+S_2^2 \right).$$


  • 1
    $\begingroup$ If you have finished part (b), part (c) should not be very difficult. $\endgroup$
    – afedder
    May 20, 2014 at 18:22
  • $\begingroup$ Sorry part b) is where im having the problem. Part a is quite easy but I just have no idea where to start for part b) $\endgroup$
    – James
    May 20, 2014 at 18:26
  • $\begingroup$ Well, it boils down to finding the variances of $\sum_{i=1}^{n_1} (X_i - \bar{X})^2$ and $\sum_{i=1}^{n_2} (Y_i - \bar{Y})^2$ $\endgroup$
    – afedder
    May 20, 2014 at 18:29

1 Answer 1


We have $$\begin{align} \text{Var}(S_{p1}^2) &=\text{Var}\left(\frac{(n_1-1)S_1^2 +(n_2-1)S_2^2}{n_1+n_2-2}\right) \\&= \left(\frac{n_1-1}{n_1+n_2-2}\right)^2\text{Var}(S_1^2)+\left(\frac{n_2-1}{n_1+n_2-2}\right)^2\text{Var}(S_2^2) \,\, \end{align}$$ and $$\begin{align} \text{Var}(S_{p2}^2) &=\text{Var}\left(\frac{1}{2}(S_1^2+S_2^2)\right) \\&= \frac{1}{4}\text{Var}(S_1^2+S_2^2) \\=& \frac{1}{4}\left[\text{Var}(S_1^2) \, + \,\text{Var}(S_2^2)\right]\,\,. \end{align}$$ Thus, we need to compute $\text{Var}(S_i^2)$ for $i=1,2$ . To this end, we have $$\begin{align} \text{Var}(S_1^2) &= \text{Var}\left(\frac{\sum_{i=1}^{n_1} (X_i - \bar{X})^2}{n_1 - 1}\right) \\&= \frac{1}{(n_1 - 1)^2} \text{Var}\left(\sum_{i=1}^{n_1} (X_i - \bar{X})^2\right)\end{align}$$ and $$\begin{align} \text{Var}(S_2^2) &= \text{Var}\left(\frac{\sum_{i=1}^{n_2} (Y_i - \bar{Y})^2}{n_2 - 1}\right) \\&= \frac{1}{(n_2 - 1)^2} \text{Var}\left(\sum_{i=1}^{n_2} (Y_i - \bar{Y})^2\right) \,\,.\end{align}$$ It remains to compute $\text{Var}\left(\sum_{i=1}^{n_1} (X_i - \bar{X})^2\right)$ and $\text{Var}\left(\sum_{i=1}^{n_2} (Y_i - \bar{Y})^2\right)$, which I will leave to you. Let me know if you need help with this as well. Useful identity: $\text{Var}(X) = \text{Cov}(X,X)$ for any random variable $X$. Also, notice that if $n_1=n_2$, then $S_{p1}^{2} = S_{p2}^{2}$.

  • $\begingroup$ Thanks so much, ive solved the rest of the question no problems. Thankyou!! $\endgroup$
    – James
    May 22, 2014 at 5:08
  • $\begingroup$ No problem, happy to help!! @James $\endgroup$
    – afedder
    May 22, 2014 at 6:53
  • $\begingroup$ @afedder . Hello ! I am looking for the notion of pooled variance" In the questions, (c) Derive the efficiency of the estimator $S_{p 2}^{2}$ relative to the $S_{p 1}^{2} .$ When is the relative efficiency equal to $1 ?$ (d) Show that both estimators are consistent estimators of $\sigma^{2}$. (e) Which estimator would you prefer? 1) What does efficiency of an estimator relative to another ? 2) why does "both estimators are consistent" mean ? and finally 3) which is the best estimator to take ? . I ask these questions since I am looking for applying pooled variance on Fisher matrices. $\endgroup$
    – youpilat13
    Feb 2, 2021 at 12:03
  • $\begingroup$ Any precision is welcome. $\endgroup$
    – youpilat13
    Feb 2, 2021 at 12:04

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.