How to show likelihood ratio test can be based on the random variable F?

Let $X_1,...,X_n$ and $Y_1,...,Y_n$ be independent random samples from the distributions $N(\theta_1,\theta_3)$, $N(\theta_2,\theta_4)$ , respectively.

What I need to show is, in testing $H_0: \theta_3=\theta_4$ against $H_1: \theta_3\neq\theta_4$, that likelihood ratio test statistic $\Lambda=\frac{\mathcal{L}(\hat \omega)}{\mathcal{L}(\hat \Omega)}$ can be based on the random variable $F=\frac{\sum_{i=1}^n(X_i-\bar X)^2/(n-1)}{\sum_{i=1}^m(Y_i-\bar Y)^2/(m-1)}$.

Under the null hypothesis, $\hat \theta_3=\hat \theta_4=\frac{1}{n+m}${$\sum_{i=1}^n(X_i-\bar X)^2+\sum_{i=1}^m(Y_i-\bar Y)^2$}. Let this $\mathcal{u}$. Then, $\mathcal{L}(\hat \omega)=(\frac{1}{2\pi\mathcal{u}})^\frac{n+m}{2}e^{-\frac{1}{2\mathcal{u}}\{\sum_{i=1}^n(X_i-\bar X)^2+\sum_{i=1}^m(Y_i-\bar Y)^2\}}=(\frac{1}{2\pi\mathcal{u}})^\frac{n+m}{2}e^{-\frac{n+m}{2}}$,

And in the general case, $\mathcal{L}(\hat \Omega)=(\frac{1}{2\pi\hat \theta_3})^\frac{n}{2}e^{-\frac{1}{2\hat \theta_3}\sum_{i=1}^n(X_i-\bar X)^2}(\frac{1}{2\pi\hat \theta_4})^\frac{m}{2}e^{-\frac{1}{2\hat \theta_4}\sum_{i=1}^m(Yi-\bar Y)^2}=(\frac{1}{2\pi\hat \theta_3})^\frac{n}{2}(\frac{1}{2\pi\hat \theta_4})^\frac{m}{2}e^{-\frac{n+m}{2}}$.

where $\hat \theta_3=\frac{1}{n}\sum_{i=1}^n(X_i-\bar X)^2$, $\hat \theta_4=\frac{1}{m}\sum_{i=1}^m(Y_i-\bar Y)^2$.

$\mathcal{\Lambda}=\frac{\mathcal{L}(\hat \omega)}{\mathcal{L}(\hat \Omega)}=\frac{\hat \theta_3^\frac{n}{2}\hat \theta_4^\frac{m}{2}}{\mathcal{u}^\frac{n+m}{2}}=\frac{(\frac{1}{n}\sum_{i=1}^n(X_i-\bar X)^2)(\frac{1}{m}\sum_{i=1}^m(Y_i-\bar Y)^2)}{[\frac{1}{n+m}\{\sum_{i=1}^n(X_i-\bar X)^2+\sum_{i=1}^m(Y_i-\bar Y)^2\}]}$.

In order to show that it has something to do with variable $F=\frac{\sum_{i=1}^n(X_i-\bar X)^2/(n-1)}{\sum_{i=1}^m(Y_i-\bar Y)^2/(m-1)}$, I let $\sum_{i=1}^n(X_i-\bar X)^2=\alpha$ and $\sum_{i=1}^m(Y_i-\bar Y)^2=\beta$.($\alpha\ge0,\beta\ge0$).

Then in a big sense, $\Lambda=\frac{(\frac{1}{n}\alpha)^\frac{n}{2}(\frac{1}{m}\beta)^\frac{m}{2}}{(\frac{\alpha+\beta}{n+m})^\frac{n+m}{2}}$ can be seen as $\frac{\alpha^\frac{n}{2}\beta^\frac{m}{2}}{(\alpha+\beta)^\frac{n+m}{2}}$. (As others are all constants.) More simply, it can be $\frac{\alpha^c\beta^d}{(\alpha+\beta)^{(c+d)}}$. Where the $\alpha$ and $\beta$ are the variables of interest. If we let $A=\frac{\alpha}{\alpha+\beta}$, it equals to $A^c(1-A)^d$, and we can graph it and it has maximum.

So the decision rule rejecting $H_0$ when $\Lambda\le c$ is equivalent to rejecting $H_0$ when $\frac{\alpha}{\alpha+\beta}\le k_1$ or $\frac{\alpha}{\alpha+\beta}\ge k_2$. Which can be also equivalent to $\frac{\beta}{\alpha+\beta}=1-\frac{\alpha}{\alpha+\beta}\ge 1-k_1$ or $\frac{\beta}{\alpha+\beta}=1-\frac{\alpha}{\alpha+\beta}\le 1-k_2$.

If we focus on the last part, $\frac{\beta}{\alpha+\beta}=\frac{\sum_{i=1}^m(Y_i-\bar Y)^2}{\sum_{i=1}^n(X_i-\bar X)^2+\sum_{i=1}^m(Y_i-\bar Y)^2} \le 1-k_2$ can be $\frac{\sum_{i=1}^n(X_i-\bar X)^2+\sum_{i=1}^m(Y_i-\bar Y)^2}{\sum_{i=1}^m(Y_i-\bar Y)^2} \ge \frac{1}{1-k_2}$, $\frac{\sum_{i=1}^n(X_i-\bar X)^2}{\sum_{i=1}^m(Y_i-\bar Y)^2} \ge \frac{1}{1-k_2}-1$.

As we know that $\frac{\frac{(n-1)\mathcal{s_1}^2}{\sigma^2}/(n-1)}{\frac{(m-1)\mathcal{s_2}^2}{\sigma^2}/(m-1)}$ ~ $F(n-1,m-1)$ under $H_o$,

We decide whether we reject $H_0$ or not by observing $\frac{\sum_{i=1}^n(X_i-\bar X)^2}{\sum_{i=1}^m(Y_i-\bar Y)^2} \gt \frac{n-1}{m-1}$$F_{0.975}(n-1,m-1)$.

This is the way I think likelihood ratio test can be based on the random variable $F$. Is there any wrong in this reasoning? Actually it is quite analgous to Is it possible to combine the F-test and t-test to test for differences in both the variance and the mean? Please help me out

• What is your statistical question. – Michael Chernick Jan 18 '17 at 18:52
• @MichaelChernick I uploaded correctly. Hope you can teach me out. – HyeonPhil Youn Jan 18 '17 at 19:00