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Consider two normally distributed populations with unknown means $\mu_1$ and $\mu_2$ respectively and unknown variances $\sigma_1^2$ and $\sigma_2^2$ respectively. Let $X_1,X_2,\ldots,X_{n_1}$ and $Y_1,Y_2,\ldots,Y_{n_2}$ be i.i.d. random samples drawn independently from these respective populations. Show that a likelihood ratio test of hypotheses $\mathrm{H}_0: \sigma_1^2=\sigma_2^2$ vs. $\mathrm{H}_1: \sigma_1^2 \ne \sigma_2^2$ is equivalent to the usual $F$-Test, i.e. with critical region $\displaystyle\frac{s_1^2}{s_2^2} \ge f_{1-\alpha/2,n_1-1,n_2-1}$ if $s_1^2 \ge s_2^2$ and $\displaystyle\frac{s_2^2}{s_1^2} \ge f_{1-\alpha/2,n_2-1,n_1-1}$ if $s_1^2 < s_2^2$ (where $s_1^2=\displaystyle\frac{1}{n_1-1}\displaystyle\sum_{i=1}^{n_1} (x_i - \bar{x})^2$, $s_2^2=\displaystyle\frac{1}{n_2-1}\displaystyle\sum_{i=1}^{n_2} (y_i - \bar{y})^2$, and $f_{1-\alpha/2,n_1-1,n_2-1}$ is the upper $1-\frac{\alpha}{2}$ quantile of the $F$ distribution with $n_1-1$ and $n_2-1$ degrees of freedom).

For brevity, let $SS_x=\sum_{i=1}^{n_1} (x_i-\bar{x})^2$ and $SS_y=\sum_{j=1}^{n_2} (y_j-\bar{y})^2$.

I proceeded as follows: Under $\mathrm{H}_0$, the parameter space is $\{-\infty<\mu_1<\infty,-\infty<\mu_2<\infty,\sigma_1=\sigma_2>0\}$. Thus \begin{align*} L_0 &= (2\pi \sigma_1^2)^{-n_1/2} e^{-\frac{1}{2\sigma_1^2}\sum_{i=1}^{n_1}(x_i-\mu_1)^2}(2\pi \sigma_1^2)^{-n_2/2} e^{-\frac{1}{2\sigma_1^2}\sum_{j=1}^{n_2}(y_j-\mu_2)^2}\\ &=(2\pi\sigma_1^2)^{-(n_1+n_2)/2} e^{-\frac{1}{2\sigma_1^2}\left[\sum_{i=1}^{n_1}(x_i-\mu_1)^2 + \sum_{j=1}^{n_2} (y_j-\mu_2)^2\right]}\\ \log L_0 &= -\displaystyle\frac{n_1+n_2}{2} \log(2\pi \sigma_1^2) - \displaystyle\frac{1}{2\sigma_1^2}\left[\sum_{i=1}^{n_1}(x_i-\mu_1)^2 + \sum_{j=1}^{n_2} (y_j-\mu_2)^2\right] \\ \displaystyle\frac{\partial L_0}{\partial \mu_1}&=\displaystyle\frac{1}{\sigma_1^2}\sum_{i=1}^{n_1}(x_i-\mu_1)=0 \implies \hat{\mu}_1=\displaystyle\frac{1}{n}\sum_{i=1}^{n_1} x_i = \bar{x} \text{; similarly } \hat{\mu}_2=\bar{y}\\ \displaystyle\frac{\partial L_0}{\partial \sigma_1^2}&=-\frac{n_1+n_2}{2} \left(\frac{2\pi}{2\pi \sigma_1^2}\right)+\frac{1}{2}\left[\sigma_1^2\right]^{-2}\left[\sum_{i=1}^{n_1}(x_i-\mu_1)^2 + \sum_{j=1}^{n_2} (y_j-\mu_2)^2\right]=0\\ &\left[\sigma_1^2\right]^{-2}\left[\sum_{i=1}^{n_1}(x_i-\mu_1)^2 + \sum_{j=1}^{n_2} (y_j-\mu_2)^2\right]=(n_1+n_2)\left[\sigma_1^2\right]^{-1}\\ \hat{\sigma}_1^2&=\frac{1}{n_1+n_2}\left[\sum_{i=1}^{n_1}(x_i-\hat{\mu}_1)^2 + \sum_{j=1}^{n_2} (y_j-\hat{\mu}_2)^2\right]=\frac{1}{n_1+n_2}\left[SS_x+SS_y\right]\\ \therefore \max L_0 &= (2\pi\hat{\sigma}_1^2)^{-(n_1+n_2)/2} e^{-\frac{1}{2\hat{\sigma}_1^2}\left[SS_x+SS_y\right]}\\ &=(2\pi e)^{-(n_1+n_2)/2} \left(\frac{1}{n_1+n_2}\left[SS_x+SS_y\right]\right)^{-(n_1+n_2)/2}\\ L &= (2\pi \sigma_1^2)^{-n_1/2} e^{-\frac{1}{2\sigma_1^2}\left[\sum_{i=1}^{n_1}(x_i-\mu_1)^2\right]} (2\pi \sigma_2^2)^{-n_2/2} e^{-\frac{1}{2\sigma_2^2}\left[\sum_{j=1}^{n_2}(y_j-\mu_2)^2\right]}\\ &\text{M.L.E.'s are $\hat{\mu}_1=\bar{x},\hat{\mu}_2=\bar{y},\hat{\sigma}_1^2=\frac{1}{n_1}SS_x,\hat{\sigma}_2^2=\frac{1}{n_2}SS_y$}\\ \max L &= (2\pi \hat{\sigma}_1^2)^{-n_1/2} e^{-\frac{1}{2\hat{\sigma}_1^2}SS_x} (2\pi \hat{\sigma}_2^2)^{-n_2/2} e^{-\frac{1}{2\hat{\sigma}_2^2}SS_y}\\ &=(2\pi e)^{-(n_1+n_2)/2} \left(\frac{1}{n_1}SS_x\right)^{-n_1/2} \left(\frac{1}{n_2}SS_y\right)^{-n_2/2} \end{align*} Now, the critical region is of the form $\lambda =\frac{\max L_0}{\max L} \le k$ where $k$ is some constant. \begin{align*} \lambda =\frac{\max L_0}{\max L}=\frac{(2\pi e)^{-(n_1+n_2)/2} \left(\frac{1}{n_1+n_2}\left[SS_x+SS_y\right]\right)^{-(n_1+n_2)/2}}{(2\pi e)^{-(n_1+n_2)/2} \left(\frac{1}{n_1}SS_x\right)^{-n_1/2} \left(\frac{1}{n_2}SS_y\right)^{-n_2/2}} &\le k\\ \text{Taking both sides to power of -2, }\frac{\left(\frac{1}{n_1+n_2}\left[SS_x+SS_y\right]\right)^{n_1}\left(\frac{1}{n_1+n_2}\left[SS_x+SS_y\right]\right)^{n_2}}{\left(\frac{1}{n_1}SS_x\right)^{n_1} \left(\frac{1}{n_2}SS_y\right)^{n_2}} &\ge k^\prime\\ \left(\frac{n_1}{n_1+n_2}\left[1+\frac{SS_y}{SS_x}\right]\right)^{n_1}\left(\frac{n_2}{n_1+n_2}\left[1+\frac{SS_x}{SS_y}\right]\right)^{n_2} \ge k^\prime\\ \left(1+\frac{SS_y}{SS_x}\right)^{n_1}\left(1+\frac{SS_x}{SS_y}\right)^{n_2} \ge k^{\prime\prime}\\ \end{align*} This is as far as I could simplify it; I am close to expressing the critical region in terms of a ratio of sample variances but cannot see how to get rid of the powers of $n_1$ and $n_2$.

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Taking the critical region $\left(1+\frac{SS_y}{SS_x}\right)^{n_1}\left(1+\frac{SS_x}{SS_y}\right)^{n_2} \ge k^{\prime \prime}$, let $r=\displaystyle\frac{SS_x}{SS_y}$ and write the left side of the expression as a function of $r$, i.e. $g(r)=(1+r^{-1})^{n_1}(1+r)^{n_2}$. We can show using calculus that this function has a minimum at $r=\frac{n_1}{n_2}$. Thus, $g(r)$ increases as we move left or right from this point, and so the inequality $g(r) \ge k^{\prime \prime}$ will be satisfied for very small $r$ or very large $r$. Thus we can rewrite the critical region as $\displaystyle\frac{SS_x}{SS_y} \le k_1 \cup \displaystyle\frac{SS_x}{SS_y} \ge k_2$ where $k_1,k_2$ are constants. Equivalently, $\frac{SS_x/(n_1-1)}{SS_y/(n_2-1)}=\frac{s_1^2}{s_2^2} \le k_1^\prime \cup \frac{SS_x/(n_1-1)}{SS_y/(n_2-1)}=\frac{s_1^2}{s_2^2} \ge k_2^\prime$

Under $\mathrm{H}_0$, since $\sigma_1^2=\sigma_2^2$, $S_1^2/S_2^2$ has an $F(n_1-1,n_2-1)$ distribution. Thus, a critical region of size $\alpha$ is constructed as follows: \begin{align*} &\Pr\left(\displaystyle\frac{S_1^2}{S_2^2} \le k_1^\prime \cup \displaystyle\frac{S_1^2}{S_2^2} \ge k_2^\prime\right)=\alpha\\ &\Pr\left(\displaystyle\frac{S_1^2}{S_2^2} \le k_1^\prime\right) + \Pr\left(\displaystyle\frac{S_1^2}{S_2^2} \ge k_2^\prime\right)=\alpha\\ &\text{If we divide the type I error prob. equally between the two,}\\ k_1^\prime&= f_{\alpha/2,n_1-1,n_2-1} \text{ and } k_2^\prime=f_{1-\alpha/2,n_1-1,n_2-1} \end{align*} And therefore we can write the critical region as $\displaystyle\frac{s_1^2}{s_2^2} \le f_{\alpha/2,n_1-1,n_2-1} \cup \displaystyle\frac{s_1^2}{s_2^2} \ge f_{1-\alpha/2,n_1-1,n_2-1}$.

Finally, if we prefer, we can rewrite $\displaystyle\frac{s_1^2}{s_2^2} \le f_{\alpha/2,n_1-1,n_2-1}$ as $\displaystyle\frac{s_2^2}{s_1^2} \ge \displaystyle\frac{1}{f_{\alpha/2,n_1-1,n_2-1}}$ and therefore, making use of a property of the $F$ distribution, as $\displaystyle\frac{s_2^2}{s_1^2} \ge f_{1-\alpha/2,n_2-1,n_1-1}$

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