MLE

BERRY · #1 $Old$ November 15th, 2009, 05:48 PM

Find the MLE of the unknown parameter $\theta$ when $X_{1},...X_{n}$ is a sample from the following distribution:

$f_{X}(x) = \frac{1}{2}e^{-|x-\theta|}, -\infty < x < \infty$

I have the following, not sure if it's correct.

$f(x_{1}...x_{n}|\theta) = \prod_{i=1}^{n} \frac{1}{2}e^{-|x_{i}-\theta|}$

$=\frac{1}{2^{n}}e^{-[\sum_{i=1}^{n}x_{i}] - n\theta}$

$\log f(x_{1}...x_{n}|\theta) = \frac{1}{2^{n}} [-\sum_{i=1}^{n}x_{i} - n\theta]$

$\frac{d}{d\theta}\log f(x_{1}...x_{n}|\theta) = \frac{-n}{2^{n}}$

Is this correct?

theodds · #2 $Old$ November 15th, 2009, 06:53 PM

Seems to me that maximizing the likelihood function of this guy is equivalent to minimizing $\sum _ {i = 1} ^ n |x_i - \theta|$ . The value that accomplishes this is a well known summary statistic, albeit not the sample mean.

matheagle · #3 $Old$ November 15th, 2009, 06:54 PM

No the absolute value is around the Xi-theta
Use logic not calc here

You want to maximize $2^{-n}e^{-\sum_{i=1}^n|X_i-\theta|}$ wrt theta

that the same as minimizing $\sum_{i=1}^n|X_i-\theta|$ wrt theta

Moo · #4 $Old$ November 15th, 2009, 11:35 PM

Hello,

Note that you can simplify the writing in latex : when there's a single element in the exponent or in the indice, you don't have to put { }
X_1, X_2, ... will produce the same result

BERRY · #5 $Old$ November 23rd, 2009, 09:09 AM

Quote:

Originally Posted by theodds $View Post$

Seems to me that maximizing the likelihood function of this guy is equivalent to minimizing $\sum _ {i = 1} ^ n |x_i - \theta|$ . The value that accomplishes this is a well known summary statistic, albeit not the sample mean.

Which statistic is that?

Also, if I were to try to estimate theta through method of moments...how would I do it?

matheagle · #6 $Old$ November 23rd, 2009, 09:22 AM

the median

kman320 · #7 $Old$ November 23rd, 2009, 10:26 AM

^ what he said.

BERRY · #8 $Old$ November 23rd, 2009, 10:33 AM

What about using method of moments?

matheagle · #9 $Old$ November 23rd, 2009, 12:57 PM

$E(X)= .5\int_{-\infty}^{\theta}xe^{x-\theta}dx +.5\int_{\theta}^{\infty}xe^{\theta -x}dx$

and set that equal to the sample mean.

I rushed but I got theta as I expected

Let $u=\theta -x$ in the first integral and $u=x-\theta$
in the second

$=.5(\theta\Gamma(1)-\Gamma(2))+.5(\Gamma(2)+\theta\Gamma(1))=\theta$