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January 29th, 2018, 06:16 PM  #1 
Senior Member Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0  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: $$R=\frac{(m1)S_X^2+(n1)S_Y^2}{\sigma^2}$$ 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 $m+n2$ 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. 
January 30th, 2018, 11:59 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,931 Thanks: 999 
what formula for sample variance are you using? If you are using the unbiased version, i.e. $S_X^2 = \dfrac{1}{K1}\sum \limits_{k=1}^K~(X_m\mu)^2$ Then $\dfrac{m1}{\sigma^2}S_X^2 = (m1)\left[\dfrac{1}{m1}\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 ChiSquare with $(m+n)$ degrees of freedom. If you use the biased version of the sample variance you get these $\dfrac{m1}{m}$ and $\dfrac{n1}{n}$ factors which make things a mess. 
January 31st, 2018, 04:05 PM  #3 
Senior Member Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 
I was using the unbiased version, that helps out a lot. Thank you.


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