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I have $X_1, \ldots, X_n$ and $Y_1, \ldots, Y_n$ as as random samples from two normal distributions with means $0$ and variances $\theta_1$ and $\theta_2$ respectively.The null hypothesis is $\theta_1 = \theta_2$ and the alternative is $\theta_1$ not equal to $\theta_2$ I calculated the likelihood ratio (which is shown below) and now I am trying to figure out what this likelihood ratio is a function of. I believe it is a function of $F$ but I am unsure how to show that it is $F$-distributed with $v_1 = n$, and $v_2 = m$. Thanks for the help. $ \lambda={ { \left\{ {\textstyle 1\over\textstyle 2\pi\bigl[\,(\,\sum x_i^2+\sum y_i^2\,)/(n+m)\, \bigr]} \right\}^{n+m\over2} } \over \biggl[{ {\textstyle1\over\textstyle 2\pi(\sum x_i^2 /n)} }\biggl]^{n/2} \biggl[{ {\textstyle1\over\textstyle 2\pi(\sum y_i^2 /m)} }\biggl]^{m/2} } $

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    And by a function of F, I meant a function of a F-Statistic.2011-12-11

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It is just algebra.

Ok, you have

$\lambda = \frac{\left[\frac{\sum x_i^2}{n}\right]^{n/2} \left[\frac{\sum y_i^2}{m}\right]^{m/2}} {\left[\frac{\sum x_i^2 + \sum y_i^2}{m+n}\right]^{\frac{m+n}{2}}}$

We can easily factor this

$\lambda = \left[\frac{\frac{\sum x_i^2}{n}} {\frac{\sum x_i^2 + \sum y_i^2}{m+n}}\right]^{n/2} \left[\frac{\frac{\sum y_i^2}{m}} {\frac{\sum x_i^2 + \sum y_i^2}{m+n}}\right]^{m/2}$

Now multiply by the appropriate power of $\frac{\sum y_i^2}{\sum y_i^2}$ to get

$\lambda = \left[\frac{(m+n)\frac{\sum x_i^2}{\sum y_i^2}}{n\left(1+\frac{\sum x_i^2}{\sum y_i^2}\right)}\right]^{n/2} \left[\frac{(m+n)}{m\left(1+\frac{\sum x_i^2}{\sum y_i^2}\right)}\right]^{m/2}$

Now we know that $\frac{m}{n}\frac{\sum x_i^2}{\sum y_i^2}$ has a $F_{n,m}$ distribution (it is a ratio of two independent random variables having $\chi^2$ distributions) so we can write $\lambda$ as $\lambda = \left[\frac{\frac{(n+m)n}{m}F_{n,m}}{ \frac{n^2}{m}\left(F_{n,m}+\frac{m}{n}\right) }\right]^{n/2}\left[\frac{m+n}{n\left(F_{n,m}+\frac{m}{n}\right)}\right]^{m/2}$

Now this has $\lambda$ as a function of $F$. Check my algebra.

To be useful it should be a monotone function of $F$. That it is is not immediately clear to me.

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    @MichaelHardy Tha$n$ks. You are of course correct.2011-12-12