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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.

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  • $\begingroup$ @DilipSarwate Thank you for your comment. I have not taken any logs yet though. This is merely the $$\frac{1}{2\pi \sigma^2 \sqrt{1-\left(\frac{1}{2}\right)^2}}$$ part, after canceling a few constants. Given the mles of the means, the mle of the common variance is $$\frac{2}{3n} \sum \left[ \left(X_i-\bar{X}\right)^2+\left(Y_i-\bar{Y} \right)^2-\left(X_i-\bar{X}\right)\left(Y_i-\bar{Y} \right) \right] $$ The constant did not appear in the above, because it cancels in the quotient. The exponentials also cancel after being evaluated at the mles. Have I done something wrong? $\endgroup$ – JohnK Mar 19 '15 at 9:38
  • $\begingroup$ @downvoter May I ask why the downvote? This is a legit question and I have gone to great lengths to explain why two answers do not cover it. $\endgroup$ – JohnK Mar 19 '15 at 16:10
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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.

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  • $\begingroup$ What about the variance? It is assumed equal but unknown. If you maximize w.r.t to $\sigma^2$ as well, you will get to the statistic of my post. $\endgroup$ – JohnK Mar 19 '15 at 16:07
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    $\begingroup$ @JohnK Use a consistent estimator for it and rely on consistency and Slutsky's Theorem to treat it as the true value asymptotically. $\endgroup$ – Alecos Papadopoulos Mar 19 '15 at 16:15
  • $\begingroup$ That might be a good approach but are you certain no exact LRT exists? I could use the asymptotic $\chi^2$ distribution if this is the case. $\endgroup$ – JohnK Mar 19 '15 at 16:17
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    $\begingroup$ Thanks John. In most cases, answers are good when questions are good. $\endgroup$ – Alecos Papadopoulos Mar 19 '15 at 23:27
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    $\begingroup$ @DeepNorth By the assumptions at the beginning of the OP's question, we have that $X,Y$ follow jointly a bivariate normal, so linear combinations of their marginals will also be normal. As regards the variance, we have $$Var(Z) = V(X) + V(Y) - 2Cov(X,Y) = \sigma^2 + \sigma^2 - 2\rho\sigma\cdot \sigma = 2\sigma^2 - 2(1/2)\sigma^2 = \sigma^2$$ $\endgroup$ – Alecos Papadopoulos May 22 '16 at 3:10
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It is my understanding that there are likelihood ratios and log likelihood ratios, the latter being the log of the quotient as opposed to the quotient of logs. I think this is where you have gone wrong. Best of luck!

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  • $\begingroup$ Hi, thanks for your input. I have not used any logs though. This is merely the quotient $$\frac{ \sup_{\mathbf{\theta} \in \omega} L\left(\mathbf{\theta};\mathbf{x,y} \right)}{\sup_{\mathbf{\theta}\in \Omega}L\left(\mathbf{\theta};\mathbf{x,y} \right)} $$ where $\omega$ and $\Omega$ denote the restricted space of the null hypothesis and the full parameter space respectively. $\endgroup$ – JohnK Mar 19 '15 at 9:49
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    $\begingroup$ But the PDF of a gaussian involves an exponential, and the probability of a sequence of independent trials is a product, not a sum. For bi variate normal: mathworld.wolfram.com/BivariateNormalDistribution.html $\endgroup$ – jlimahaverford Mar 19 '15 at 10:44
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    $\begingroup$ I have to disagree John. This is a log-likelihood on the top and the bottom. This is not the likelihood of parameters of a Gaussian, it cannot be. While I have little experience with the likelihood ratio test, I can say definitively, this is not the likelihood on the top and bottom. One thing in my favour is that if we imagine subtracting the denominator from the numerator you can see a ton of cancellation that will take place. This is what will happen in the exponent of the exponentials. $\endgroup$ – jlimahaverford Mar 19 '15 at 12:50
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    $\begingroup$ Let's go with temporary agreement. I can't carry out the calculation now and I'm most interested in the next step. Now, for a single $(x,y)$ plug these values into the equation from the link, completely as is, and if there is any cancellation afterword, demonstrate it. $\endgroup$ – jlimahaverford Mar 19 '15 at 13:27
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    $\begingroup$ Actually, I believe it is $e^{-n}$ but that's not a big detail. Yes, the claim is that the exponentials cancel and eventually you get the quotient of the estimated variances to the power $n$. $\endgroup$ – JohnK Mar 19 '15 at 13:45
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I am fairly confident that it reduces to a statistic with an F distribution. The numerator of the likelihood ratio you have provided is a chi-squared distribution multiplied by a constant (having 2*(n-1)) under HO. Also, under HO, X == Y, therefore the denominator can also be written as a chi-squared variable (having 2n df).

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I would like to suggest this way to simplify the likelihood ratio $\Lambda$. I am not a native English speaker so feel free to correct my grammar errors.

Let me introduce these statistics $U, V$ which are given by linear combination of X and Y as

$\begin{pmatrix} U \\ V \end{pmatrix} =\begin{pmatrix} 1 & -{1 \over 2} \\ 0 & { \sqrt3 \over 2} \end{pmatrix}\begin{pmatrix} X \\ Y \end{pmatrix} $

where $(X, Y)' \sim N_2((\mu_1,\mu_2)',\begin{pmatrix} \sigma^2 & {\sigma^2\over 2} \\ {\sigma^2\over 2} & \sigma^2 \end{pmatrix})$ as given in the problem.

We know that the distribution of $U$ and $V$ is $$(U, V)' \sim N_2((\mu_1-{\mu_2\over2},{\sqrt3 \mu_2\over2})',\begin{pmatrix} {3 \sigma^2\over 4} & 0 \\ 0 &{3 \sigma^2\over 4} \end{pmatrix})$$

We can tell that $U$ and $V$ are independent of each other.

Now we can express $\Lambda^{-1/n}$ with these statistics like this:

$$\frac{\sum_{i=1}^{n}X_i^2+\sum_{i=1}^{n}Y_i^2-\sum_{i=1}^n X_iY_i }{ \sum_{i=1}^{n} \left(X_i-\bar{X} \right)^2+\sum_{i=1}^{n} \left(Y_i-\bar{Y} \right)^2-\sum_{i=1}^{n} \left(Y_i-\bar{Y} \right) \left(X_i-\bar{X} \right) } = \frac{\sum_{i=1}^{n}U_i^2+\sum_{i=1}^{n}V_i^2 } { \sum_{i=1}^{n} \left(U_i-\bar{U} \right)^2+\sum_{i=1}^{n} \left(V_i-\bar{V} \right)^2 } $$

$$=1+\frac{n \bar{U}^2+n \bar{V}^2 } { \sum_{i=1}^{n} \left(U_i-\bar{U} \right)^2+\sum_{i=1}^{n} \left(V_i-\bar{V} \right)^2 }$$

We know that $S_U^2$ the sample variance of U and $S_V^2$ the sample variance of V are independent of $\bar U$ and $\bar V$ respectively. So these four statistics are independent of each other.

And we also know their distributions under null hypothesis : $${n\bar U^2 \over 3\sigma^2/4 } \sim \chi^2 (1) $$ $${n\bar V^2 \over 3\sigma^2/4 } \sim \chi^2 (1) $$ $${{(n-1)S_U^2\over 3\sigma^2/4 } } \sim \chi^2 (n-1) $$ $${ {(n-1)S_V^2\over 3\sigma^2/4 }} \sim \chi^2 (n-1) $$

Let's use these statistics then,

$$\Lambda^{-1/n}= 1+\frac{{n\bar U^2 \over 3 \sigma^2 /4}+{n\bar U^2 \over 3\sigma^2 /4}}{{(n-1)S_U^2\over 3 \sigma^2 /4}+{(n-1)S_V^2\over 3 \sigma^2 /4}}$$

Look at the right term. The numerator has $\chi^2(2)$ distribution and the denominator has $\chi^2(2n-2)$ distribution. We can adopt F distribution here like this:

$$\Lambda^{-1/n}=1+\frac {Q} {n-1}$$ where Q has a F distribution with dof 2 and 2n-2.

Further more, I guess it would have non central F distribution under alternative hypothesis.

Thank you for reading.


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Interestingly $\lim_{n->\infty} \Lambda = e^{-Q}$. Could we interpret this? I am being curious.


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Well, according to this document :http://www.math.wm.edu/~leemis/chart/UDR/PDFs/FChisquare.pdf

I think Q will converge to the distribution $\Gamma(1,1)$ that is $f_{Q_\infty}(q)=e^{-q}$ which means that the limit of the likelihood ratio $\Lambda_{\infty}$ is likely to have uniform distribution, since $f_{\Lambda_{\infty}}=1$ and $0\leq \Lambda_{\infty} \leq 1$ Am I right? I am not sure about this.

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I am fairly confident that it reduces to a statistic with an F distribution too. The numerator of the likelihood ratio you have provided is a chi-squared distribution multiplied by a constant, and the same holds for the denominator. The cross product XY is the sum of two independent chi-squared variables since var (x) = var (y) and XY may be rewritten in the form .25*(X+Y)^2 -.25*(X-Y)^2 which is of the chi-squared form.

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    $\begingroup$ I don't think this is correct. $\endgroup$ – Michael R. Chernick Apr 26 '17 at 23:56

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