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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 $$S_{p_1}^2=\dfrac {(n_1-1)S_1^2+(n_2-1)S^2_2} {n_1+n_2-2}\quad\text{and}\quad S^2_{p_2}=\dfrac {1} {2}\left( S_1^2+S_2^2 \right).$$

questions

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    $\begingroup$ If you have finished part (b), part (c) should not be very difficult. $\endgroup$ – afedder May 20 '14 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 '14 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 '14 at 18:29
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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 $S_{p1}^{2} = S_{p2}^{2}$.

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  • $\begingroup$ Thanks so much, ive solved the rest of the question no problems. Thankyou!! $\endgroup$ – James May 22 '14 at 5:08
  • $\begingroup$ No problem, happy to help!! @James $\endgroup$ – afedder May 22 '14 at 6:53

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