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January 29th, 2018, 06:16 PM   #1
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Exclamation Sampling Distribution of Normal Random Variables

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 I have absolutely no justification so it's probably wrong. So what is the sampling distribution of $R$, and how do you justify it?

Last edited by John Travolski; January 29th, 2018 at 06:43 PM.
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January 30th, 2018, 11:59 AM   #2
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what formula for sample variance are you using?

If you are using the unbiased version, i.e.

$S_X^2 = \dfrac{1}{K-1}\sum \limits_{k=1}^K~(X_m-\mu)^2$

Then

$\dfrac{m-1}{\sigma^2}S_X^2 = (m-1)\left[\dfrac{1}{m-1}\sum \limits_{k=1}^m ~\left(\dfrac{(X_k-\mu_1)}{\sigma}\right)^2\right] =

\sum \limits_{k=1}^m ~\left(\dfrac{(X_k-\mu_1)}{\sigma}\right)^2
$

This is seen to be the sum of the squares of $m$ standard normals.

Similarly the $Y$'s produce the sum of squares of $n$ standard normals.

Thus $R$ is Chi-Square with $(m+n)$ degrees of freedom.

If you use the biased version of the sample variance you get these $\dfrac{m-1}{m}$ and $\dfrac{n-1}{n}$ factors which make things a mess.
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January 31st, 2018, 04:05 PM   #3
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I was using the unbiased version, that helps out a lot. Thank you.
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