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Let $X_1,X_2,...,X_m$ be i.i.d. from a $N(\mu_1,\sigma_1^2)$ distribution, and let $Y_1,Y_2,...,Y_n$ be i.i.d. from a $N(\mu_2,\sigma_2^2)$ distribution, and let the $X_i$'s be independent from the $Y_j$'s. Determine the sampling distribution of the following quantity:

$$R=\frac{(m-1)S_X^2+(n-1)S_Y^2}{\sigma^2}$$ under the condition that $\sigma_1=\sigma_2=\sigma$, where $S_X^2$ and $S_Y^2$ are the respective sample variances of $X$ and $Y$.

My intuition tells me that the sampling distribution is that of a chi squared with $m+n-2$ degrees of freedom, but that's completely unjustified and probably wrong. So what is the sampling distribution of $R$, and how do you justify it?

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From the standard result on the sample variance, $\frac{m-1}{\sigma^2}S_X^2$ is $\chi^2(m-1)$ and $\frac{n-1}{\sigma^2}S_Y^2$ is $\chi^2(n-1).$ And the fact that the sum of independent $\chi^2(d)$ and $\chi^2(d')$ is $\chi^2(d+d')$ is clear from the definition of $\chi^2(d)$ as the sum of $d$ independent square normals. So you're exactly right. $R$ is $\chi^2(n+m-2).$

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