Question

Let \(\left(X_{1}, Y_{1}\right),\left(X_{2}, Y_{2}\right), \ldots,\left(X_{n}, Y_{n}\right)\) be a random sample from a bivariate normal distribution with \(\mu_{1}, \mu_{2}, \sigma_{1}^{2}=\sigma_{2}^{2}=\sigma^{2}, \rho=\frac{1}{2}\), where \(\mu_{1}, \mu_{2}\), and \(\sigma^{2}>0\) are unknown real numbers. Find the likelihood ratio \(\Lambda\) for testing \(H_{0}: \mu_{1}=\mu_{2}=0, \sigma^{2}\) unknown against all alternatives. The likelihood ratio \(\Lambda\) is a function of what statistic that has a well- known distribution?

Short answer

The likelihood ratio \(\Lambda\) is a function of \(\frac{\sum_{i=1}^{n}Y_{i}^{2}-n\overline{Y^2}}{2\overline{X^2}}\). This statistic has a chi-square distribution with \(n-1\) degrees of freedom.

Step-by-step solution

Establish the joint PDF

First, we must establish that the joint probability density function (PDF) of the bivariate normal distribution for a single observation (Xi, Yi) is given by \(f(X_{i},Y_{i};\theta)=\frac{1}{2\pi\sigma^{2}\sqrt{1-\rho^2}}e^{-\frac{1}{2\sigma^2}\frac{(X_{i}-\mu_{1})^2 + (Y_{i}-\mu_{2})^2-2\rho(X_{i}-\mu_{1})(Y_{i}-\mu_{2})}{\sqrt{1-\rho^2}}}\), where \(\rho, \sigma, \mu_{1}, \mu_{2}\) are the correlation coefficient, the standard deviation, and the means of \(X_{i}\) and \(Y_{i}\) respectively.

Calculate joint PDF of the sample

Next, we calculate the joint PDF of the whole sample by assuming independence, which gives us the product:\(L(\theta)=\prod_{i=1}^{n}f(X_{i},Y_{i};\theta)\)

Establish the hypotheses

Then we establish that for the null hypothesis, \(H_{0}\), \(\mu_{1}=\mu_{2}=0\) and for alternatives \(\mu_{1},\mu_{2}\) can be anything.

Calculate the likelihood ratio

Next, we calculate the likelihood ratio, which is defined as the maximum of the likelihood function under the null hypothesis divided by the maximum of the likelihood function under any hypothesis, which yields \(\Lambda =\frac{maxL(\theta|H_{0})}{maxL(\theta)}\).

Find the statistic

Finally, we find that the statistic \(\Lambda\) is a function of is actually \(\frac{\sum_{i=1}^{n}Y_{i}^{2}-n\overline{Y^2}}{2\overline{X^2}}\), where \(\overline{X^2}\) represents the sample variance of the \(X_{i}\)'s and \(\overline{Y^2}\) represents the sample variance of the \(Y_{i}\)'s. We find that under \(H_{0}\), this statistic has a chi-square distribution with \(n-1\) degrees of freedom as this follows directly from the properties of the bivariate normal distribution when \(\mu_{1}=\mu_{2}=0\).