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This is the original problem: Suppose that the two-dimensional vectors ($X_1$, $Y_1$), ($X_2$, $Y_2$), ..., ($X_n$, $Y_n$) form a random sample from a bivariate normal distribution for which the means of $X$ and $Y$ are unknown but the variances of $X$ and $Y$ and the correlation between $X$ and $Y$ are known. Find the M.L.E.’s of the means.

(I skipped some steps to get this equation.) Below is the partial derivative equation that I am working on.

$$\frac{\partial L(\mu_1,\mu_2 )}{\partial \mu_1} = \left [ \frac{1}{(1-\rho ^2)\sigma_1^2}\sum_{i=1}^{n}(x_i-\mu_1)\right ]- \left [ \frac{\rho}{(1-\rho ^2)\sigma_1\sigma_2}\sum_{i=1}^{n}(y_i-\mu_2)\right ] = 0$$

In the equation, $\mu_1$, $\mu_2$ are unknown means for two populations, $\sigma_1$, $\sigma_2$ are known variances for two populations, and $\rho$ is the correlation for bivariate variable $(X, Y)$. $L$ is the likelihood function for $(\mu_1, \mu_2)$.

This question is from a statistics textbook. I want to solve for $\mu_1$ in order to get the MLE (Maximum Likelihood Estimate) of $\mu_1$, but I have trouble in getting rid of $\mu_2$.

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  • $\begingroup$ Have you tried working with the loglikelihood $\ell$ rather than $L$? And by the way, are we working with a bivariate normal random variable here? $\endgroup$ – Clarinetist Jan 10 '16 at 2:09
  • $\begingroup$ @Clarinetist Yes, and I have already taken the natural log of the original pdf of the bivariate normal distribution. $\endgroup$ – Ray Jan 10 '16 at 2:10
  • $\begingroup$ $\mu_1$ is a linear function of $\mu_2$, because there is a non zero correlation $\rho$. where is the problem ? $\endgroup$ – reuns Jan 10 '16 at 2:13
  • $\begingroup$ @user1952009 This is the problem: Suppose that the two-dimensional vectors ($X_1$, $Y_1$), ($X_2$, $Y_2$), . . . , ($X_n$, $Y_n$) form a random sample from a bivariate normal distribution for which the means of $X$ and $Y$ are unknown but the variances of $X$ and $Y$ and the correlation between $X$ and $Y$ are known. Find the M.L.E.’s of the means. $\endgroup$ – Ray Jan 10 '16 at 3:57
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Let us move to a matrix representaion. Define ${\bf{R}}=\begin{pmatrix} \sigma_1^2 & \rho \sigma_1 \sigma_2\\ \rho \sigma_1 \sigma_2 & \sigma_2^2 \end{pmatrix}$ to be the covariance matrix, further define $\underline{\mu}=(\mu_1, \mu_2)^T$, $\underline{x}_i=(x_i, y_i)^T$ and $\underline{x}=(\underline{x}_1^T, \underline{x}_2^T, \dots, \underline{x}_n^T)^T$, so we can write

\begin{equation} f\left ( \underline{x} \right ) = \frac{1}{\left ( 2 \pi \right )^n \det\left ( R \right )^n} exp \left ( -\frac{1}{2} \sum_{i=1}^{n} \left ( \underline{x}_i-\underline{\mu} \right )^T {\bf{R}}^{-1}\left ( \underline{x}_i-\underline{\mu} \right ) \right ). \end{equation}

Taking the natural logarithm denoted as $\log$ will yield \begin{equation} \log f\left ( \underline{x} \right ) = -nlog\left ( 2\pi \right ) -n \log \left [ \det \left ( R \right ) \right ] - \frac{1}{2} \sum_{i=1}^{n} \left ( \underline{x}_i-\underline{\mu} \right )^T {\bf{R}}^{-1}\left ( \underline{x}_i-\underline{\mu} \right ). \end{equation}

Deriving with respect to the $2 \times 1$ vector of means, $\underline{\mu}$, will give

\begin{equation} \frac{\partial }{\partial \underline{\mu}}\log f\left ( \underline{x} \right ) = -\frac{1}{2}\sum_{i=1}^{n}\left [ -2 \underline{x}_i^T {\bf{R}}^{-1} + \underline{\mu}^T {\bf{R}}^{-1} \right ] = 0, \end{equation}

which gives the result of the well known sample mean:

\begin{equation} \underline{\hat{\mu}}_{ML} = \frac{1}{n}\sum_{i=1}^{n} \underline{x}_i. \end{equation}

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