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I am given n random variables

$ y_1 = \theta_1 \theta_2 + e_1$

$ y_i = \theta_1 + e_i$ for $ \space i = 2,...,n$

where $\theta = [ \theta_1 \space \theta_2] $ is a vector and $e_i$ are Gaussian independent variables $e_i \sim N(0,1)$

Find the maximum likelihood estimator $\theta^{ML}$ of $\theta$

To solve it I would do this:

$\theta^{ML} = argmax_\theta P[ y = y_i | \theta] = argmin_\theta -log(P[ y = y_i | \theta])$

$P[ y = y_i | \theta] = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}(y_1-\theta_1\theta_2)^2} * \prod_{i=2}^n \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}(y_i-\theta_1)^2}$ where the mean is 0 and the variance 1 so

$-log(P[ y = y_i | \theta]) = \frac{1}{2}log(2\pi)+\frac{1}{2}(y_1-\theta_1\theta_2)^2+\frac{n-1}{2}log(2\pi) + \sum_{i=2}^n\frac{1}{2}(y_i-\theta_1)^2$

now I need to calculate the minimum so I need the first derivative:

$\frac{d(-log(P[ y = y_i | \theta]))}{d\theta_1} = -\theta_2(y_1-\theta_1\theta_2) - \sum_{i=2}^n(y_i-\theta_1) = $

$=\theta_1\theta_2^2-\theta_2 y_1 +(n-1)\theta_1 -\sum_{i=2}^ny_i$

$\frac{d(-log(P[ y = y_i | \theta]))}{d\theta_2} = -\theta_1(y_1-\theta_1\theta_2) = \theta_1^2\theta_2-\theta_1y_1 $

Now making the derivatives equal to 0 I can solve for $\theta_1$ and $\theta_2$

I would like to know if this exercise is solved correctly

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  • $\begingroup$ where is the $n$? $\endgroup$
    – Nikos M.
    Jan 21, 2016 at 13:12
  • $\begingroup$ @NikosM. what do you mean? The sums have n $\endgroup$
    – Davide lu
    Jan 21, 2016 at 13:21
  • $\begingroup$ ok then, seems good to me $\endgroup$
    – Nikos M.
    Jan 21, 2016 at 13:51

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