Suppose $X_{1},\dots,X_{m}$ is iid $N(\mu_{1},\sigma^{2})$ and $Y_{1},\dots,Y_{n}$ is iid $N(\mu_{2},\sigma^{2})$. Is it true that the pooled estimate of the variance, $S_{p}^{2}$, has the property $\frac{(n-1)S_{p}^{2}}{\sigma^{2}}\sim\chi^{2}_{n-1}$, as is the case for a single sample variance $S^{2}$? I know that $S_{p}^{2}$ multiplied by a constant has a $\chi^{2}$ distribution, but am not sure if this property holds here.
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No: It's $\chi^2_{m+n-2}$. I'll post an answer below. – Michael Hardy Jul 16 '13 at 22:05
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You omitted to say that the sequence $X_1,\ldots,X_m$ is independent of $Y_1,\ldots, Y_n$. I will take that to be assumed below.
I will take $S_p^2$ to be defined as follows: $$ S_p^2 = \frac{\sum_{i=1}^m (X_i-\bar X)^2 + \sum_{i=1}^n (Y_i-\bar Y)^2}{n+m-2} $$ where $\bar X$ and $\bar Y$ are the sample means.
Then $$ \frac{\sum_{i=1}^m (X_i-\bar X)^2}{\sigma^2} \sim \chi^2_{n-1} $$ and $$ \frac{\sum_{i=1}^n (Y_i-\bar Y)^2}{\sigma^2} \sim \chi^2_{m-1}. $$
The sum of two chi-square random variables that are independent of each other is a chi-square random variable, and the number of its degrees of freedom is just the sum of the numbers of degrees of freedom of the two terms in the sum.
Apply that to $(n+m-2)S_p^2$ (not to $(n-1)S_p^2$ as your question states).
