Pooled Estimate of the Variance

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

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 $$\frac{\sum_{i=1}^n (Y_i-\bar Y)^2}{\sigma^2} \sim \chi^2_{m-1}.$$
Apply that to $(n+m-2)S_p^2$ (not to $(n-1)S_p^2$ as your question states).