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this is my first post so I apologize if the formatting is a little rocky.

I'm currently going through "Probability and Statistics" 4th ed by DeGroot/Schervish, and I was wondering if somebody could help me out on two related problems (7.5.10, 7.6.1).

The first question is as follows: Suppose that $ X_1, \dots, X_n $ form a random sample from a distribution for which the p.d.f. $ f(x|\theta) $ is as follows: $$ f(x|\theta) = \frac{1}{2} e^{-|x-\theta|} \text{ for } -\infty < x < \infty $$ Also, suppose that the value of $ \theta $ is unknown, for $ -\infty < \theta < \infty $. We will find the M.L.E. of $ \theta $.

The likelihood function is given by $$ f_n(\mathbf{x}|\theta) = \frac{1}{2^n} e^{-\sum_{i=1}^n |x_i - \theta|}$$ and will be maximized when $ \sum_{i=1}^n |x_i - \theta| $ is minimized. By choosing $ \theta $ to be a median value of $ x_1, \dots, x_n $, we accomplish the minimization task.

More specifically, note that the likelihood function has log \begin{align*} \log f(\mathbf{x}|\theta) &= -n \log 2 - \sum_{i=1}^n |x_i - \theta| \\ &= n \left ( - \log 2 - \frac{1}{n} \sum_{i=1}^n |x_i - \theta| \right ) \end{align*} Now, we see that the M.L.E. will minimize the sum in the log likelihood shown above. We can also see that $ \frac{1}{n} \sum_{i=1}^n |x_i - \theta| = E(|X-\theta|)$ for $ X $ having a discrete distribution assigning probability $ \frac{1}{n} $ to each of $ x_1, \dots, x_n $, and 0 everywhere else. So now we can see that, by choosing $ \hat{\theta} $ to be a median of $ x_1, \dots, x_n $, then the log likelihood will be minimized. This follows from the fact that the median of a distribution minimizes the mean absolute error.

The second question is to find the MLE of $e^{-\frac{1}{\theta}}$, which by the invariance property of MLEs, should just be $e^{-\frac{1}{\hat{\theta}}}$.

My problem is that the answer in the back of the book (for the second question) is given as $\left ( \prod_{i=1}^n x_i \right)^{\frac{1}{n}}$, and I'm having trouble reconciling that with my answer. You'd think it'd be pretty straightforward, but...

Any help would be appreciated!

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For the second problem what is the mle for theta? It looks like it is not the same theta in the previous problem. Working backwards you would get this answer if the mle for theta is -n/∑lnx$_i$. –  Michael Chernick Sep 8 '12 at 18:51
    
What exactly is your first question? For the second question, I would set $\alpha = e^{1/\theta}$, solve for $\theta$, plug that in $f(x|\theta)$ and find the MLE. –  echoone Sep 8 '12 at 19:03
    
@Michael I also noticed that -- does that mean I calculated the first MLE wrong? –  Chase Uyeda Sep 8 '12 at 19:13
    
@echoone Err, I didn't really have a question about the first part, I was just showing my work to see if maybe I was messing up somewhere there. I'll try your method and see if that works. –  Chase Uyeda Sep 8 '12 at 19:15
    
Here is a hint: In the first problem $X_i \in \mathbb R$. Now take $n = 2$ and suppose $X_1 < 0$ and $X_2 > 0$, each of which can happen with positive probability. Now consider the "estimator" for the second question. Conclusion? :-) –  cardinal Sep 8 '12 at 19:32
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1 Answer

Here is a quick pointer on the optimization step. Your intuition for the problem seems good, it is just a matter of fleshing out the computations.

At this point, you want to maximize the likelihood of the data given $\theta$, \begin{align*} \log f(\mathbf{x}|\theta) &= -n \log 2 - \sum_{i=1}^n |x_i - \theta| \\ &= n \left ( - \log 2 - \frac{1}{n} \sum_{i=1}^n |x_i - \theta| \right ) \end{align*} , which is equivalent to minimizing the term $\frac{1}{n} \sum_{i=1}^n |x_i - \theta|$.

One can write this as an optimization problem:

$$\arg\max_{\theta}{\{-n\log 2 - \sum_{i=1}^n |x_i - \theta|\}}$$

Then let $$ \epsilon_i = |x_i - \theta | \quad \forall i $$

, so we can rewrite the optimization problem as $$\arg\max_{\epsilon,\theta}{\{-n\log 2 - \sum_{i=1}^n \epsilon_i\}}$$ subject to $$ -\epsilon_i \leq x_i - \theta \leq \epsilon_i \quad \forall i$$. The constraints simplify to:

\begin{align*} (x_i - \theta) + \epsilon_i &\geq 0 \\ -(x_i - \theta) + \epsilon_i &\geq 0 \end{align*}

Then, go on to set up the Lagrangian and compute the answer.

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