For a random sample from a Bivariate Normal distribution with $\rho=\frac{1}{2}$ and equal variances, i.e. $\sigma^2_x=\sigma^2_y=\sigma^2$, I would like to derive the Likelihood Ratio Test for the hypothesis $\mu_x=\mu_y=0$, against all alternatives.
The maximum likelihood estimates for $\mu_x$ and $\mu_y$ are $\bar{X}$ and $\bar{Y}$ respectively, thus the LRT calls to reject the null hypothesis if
$$\frac{
\sum_{i=1}^{n} \left(X_i-\bar{X} \right)^2+\sum_{i=1}^{n} \left(Y_i-\bar{Y} \right)^2-\sum_{i=1}^{n} \left(Y_i-\bar{Y} \right) \left(X_i-\bar{X} \right) }{\sum_{i=1}^{n}X_i^2+\sum_{i=1}^{n}Y_i^2-\sum_{i=1}^n X_iY_i } \leq c$$
Is it possible to simplify this further? Many LRTs reduce to well-known statistics so I wonder if that can be done here as well. Because of the two restrictions imposed on the means and the equality of variances, an F- statistic comes to mind as a possible candidate but it's not obvious to me how to get there. Any hints maybe?
Thank you in advance.
Best Answer
As has been made clear in the comments, the OP is interested in the Likelihood ratio when the common variance is also estimated, and not known.
The joint density of one pair of $\{X_i, Y_i$}, given also the maintained assumptions on the parameter values is
$$ f(x_i,y_i) = \frac{1}{2 \pi \sigma^2\sqrt{3/4}} \ \exp\left\{ -\frac{2}{3}\left[ \frac{(x_i-\mu_x)^2}{\sigma^2} + \frac{(y_i-\mu_y)^2}{\sigma^2} - \frac{(x_i-\mu_x)(y_i-\mu_y)}{\sigma^2} \right] \right\}$$
So the joint Likelihood of the sample (not log likelihood) is
$$ L(\mu_x, \mu_y, \sigma^2 \mid, \mathbf x, \mathbf y, \rho=1/2) = \left(\frac{1}{2 \pi \sigma^2\sqrt{3/4}}\right)^n \\ \times \exp\left\{ -\frac{2}{3\sigma^2}\left[ \sum_{i=1}^n(x_i -\mu_x)^2 +\sum_{i=1}^n(y_i -\mu_y)^2 - \sum_{i=1}^n(x_i-\mu_x)(y_i-\mu_y) \right] \right\}$$
Denote $L_1$ the maximized likelihood with the sample means (MLEs for the true means), and $L_0$ the likelihood with the means set equal to zero. Then the Likelihood Ratio (not the log such) is $$ LR \equiv \frac {L_0}{L_1} = \frac {\hat \sigma^{2n}_1\cdot \exp\left\{ -(2/3\hat \sigma^2_0)\cdot\left[ \sum_{i=1}^nx_i^2 +\sum_{i=1}^ny_i^2 - \sum_{i=1}^nx_iy_i \right] \right\}}{\hat \sigma^{2n}_0 \cdot\exp\left\{ -(2/3\hat \sigma^2_1)\cdot\left[ \sum_{i=1}^nx_i^2 -n\bar x^2 +\sum_{i=1}^ny_i^2 -n\bar y^2 - \sum_{i=1}^nx_iy_i+n\bar x\bar y \right] \right\}}$$
where $\hat \sigma^2_1$ is the estimate with unconstrained means and $\hat \sigma^2_0$ is the estimate with the means constrained to zero.
The OP has (correctly) calculated the MLEs for the common variance as
$$\hat \sigma^2_1 = \frac{2}{3n} \sum_{i=1}^n \left[ \left(x_i-\bar{x}\right)^2+\left(y_i-\bar{y} \right)^2-\left(x_i-\bar{x}\right)\left(y_i-\bar{y} \right) \right]$$
$$\hat \sigma^2_0 = \frac{2}{3n} \sum_{i=1}^n \left( x_i^2+y_i^2-x_iy_i \right)$$
If we plug these into the LR, inside the exponential, both in the numerator and the denominator, things cancel out and we are left simply with
$$ LR = \frac {\hat \sigma^{2n}_1}{\hat \sigma^{2n}_0 } $$
Our goal is not to derive the LR per se -it is to find a statistic to run the test we are interested in. So let's consider the quantity (which is the reciprocal of quantity presented in the question)
$$\left(LR\right)^{-1/n} = \frac {\hat \sigma^{2}_0}{\hat \sigma^{2}_1}$$
$$ = \frac{2}{3n} \frac{\sum_{i=1}^{n}x_i^2+\sum_{i=1}^{n}y_i^2-\sum_{i=1}^n x_iy_i }{\hat \sigma^{2}_1}$$
$$= \frac {1}{3n}\cdot\left[\sum_{i=1}^n\left(\frac {x_i}{\hat \sigma_1}\right)^2 + \sum_{i=1}^n\left(\frac {y_i}{\hat \sigma_1}\right)^2 + \sum_{i=1}^n\left(\frac {x_i-y_i}{\hat \sigma_1}\right)^2\right]$$
Note that $\hat \sigma^{2}_1$ is a consistent estimator of the true variance, irrespective of whether the true means are zero or not. Also (given equal variances and $\rho =1/2$),
$$Z_i = X_i - Y_i \sim N(\mu_x-\mu_y, \sigma^2)$$
Under the null of zero means, then, all $(x_i/\hat \sigma_1)^2$, $(y_i/\hat \sigma_1)^2$ and $(z_i/\hat \sigma_1)^2$ are chi-squares with one degree of freedom (and i.i.d., per sum). Each sum (denote the three sums for compactness $S_x, S_y, S_z$) has expected value $n$ and standard deviation $\sqrt {2n}$ (under the null). So subtract $n$ 3 times and add $n$ 3 times, and also divide and multiply by $\sqrt {2n}$ and re-arrange to get
$$\sqrt {n}\left(LR\right)^{-1/n} = \frac {\sqrt 2}{3}\cdot\left[\frac {S_x - E(S_x)}{SD(S_x)} + \frac {S_y - E(S_x)}{SD(S_x)} + \frac {S_z - E(S_z)}{SD(S_z)}\right] + 1$$
The three terms inside the bracket, are the subject matter of the Central Limit Theorem, and so each element converges to a standard normal. Therefore we have arrived (due to initial bi-variate normality) at
$$\frac {3}{\sqrt 2} \left[\sqrt{n}\left(LR\right)^{-1/n} -1\right] \xrightarrow{d} N(0, AV)$$
Of course in order to actually use the left-hand side as a statistic in a test, we need to derive the asymptotic variance -but for the moment, I do not feel up to the task. I just note that one should determine whether the three $S$'s are asymptotically independent or not.