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Suppose there are three different normally distributed random variables, $x_1$,...,$x_n$~$N(\mu_1 ,\sigma)$, $y_1$,...,$y_n$~$N(\mu_2 ,\sigma)$ and $z_1$,...,$z_n$~$N(\mu_3 ,\sigma)$. Here, $\mu_i$'s are related as, $\mu = {\{a_ib_j}\}\zeta$ where $i,j = 1,2,3$ and ${\{a_ib_j}\}$ is a $3\times3$ matrix. $\mu={\{\mu_1, \mu_2,\mu_3}\}$ and $\zeta ={\{\zeta_1, \zeta_2,\zeta_3}\}$ is a vector of some values. In other words,for example, $\mu_1 = a_1(b_1\zeta_1+b_2\zeta_2+b_3\zeta_3)$. I want to know how to estimate $\mu_1, \mu_2,\mu_3$; which means, how to estimate ${\{a_ib_j}\}$ matrix using maximum likelihood method for a given $\zeta$. Appreciate your comments. I prefer to use R optim() to estimate MLE.

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  • $\begingroup$ Quick comment: What do you mean by saying $\mu_i = \{a_ib_j\} \zeta_i$? The left hand side only depends on $i$ but the right hand side depends on $i$ and $j$. Which $j$ to take for $\mu_1$ for example? What do the brackets '$\{$' and '$\}$' mean? Are they just fancy style of is there some formula behind them? $\endgroup$ Jan 4 '18 at 8:25
  • $\begingroup$ @Fabian Werner, I am sorry for my mistake in the explanation. It means, for example, the first row of the matrix takes the form $a_1b_1$ $a_1b_2$ $a_1b_3$. $j$ also varies from 1 to 3. $\endgroup$
    – Lank
    Jan 4 '18 at 8:53
  • $\begingroup$ still confused: how exactly is $\mu_1$ determined then? Product? Sum? ... ? $\endgroup$ Jan 4 '18 at 9:04
  • $\begingroup$ Your notation is still unclear -- it's not standard matrix notation. Do you mean that $\mu_1=a_1(b_1\zeta_1+b_2\zeta_2+b_3\zeta_3)$? Or something else? $\endgroup$ Jan 4 '18 at 21:16
  • $\begingroup$ @GordonSmyth, Yes, you are correct, I have done a silly mistakes there. It should be $\mu_1 = a_1(b_1\zeta_1+b_2\zeta_2+b_3\zeta_3)$. In other words, $\zeta = (\zeta_1, \zeta_2,\zeta_3)$ $\endgroup$
    – Lank
    Jan 5 '18 at 1:23
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This problem would seem to have a simple solution, or a number of simple solutions.

The unconstrained MLEs of the $\mu_i$ are simply $\hat\mu_1=\bar x$, $\hat\mu_2=\bar y$ and $\hat\mu_3=\bar z$. You can easily choose the $a_i$ and $b_j$ in such a way as to achieve these estimators.

You can choose the $b_j$ in any way such that $c=b_1\zeta_1+b_2\zeta_2+b_3\zeta_3$ is non-zero. For example, you could set $b=\zeta$.

Then set $a_1=\bar x / c$, $a_2=\bar y / c$ and $a_3=\bar z / c$. This will ensure that you have $\hat\mu_1=\bar x$, $\hat\mu_2=\bar y$ and $\hat\mu_3=\bar z$. Since these are the unconstrained MLE, and they are achievable by your parametric model, then they must also be the constrained MLE.

There is no need to use a non-linear optimization routine such as optim(). Indeed such an approach would not work because the solution does not identify unique values for $a$ and $b$.

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