Likelihood function

**starlet** · December 30th 2010, 07:29 AM

Hi guys

..so this is my first time actually posting on this but this problem is really starting to annoy me!! Any help would be much appreciated

(and please excuse the lack of "math text" - i'll figure that out later :P )

The total holding time of successful telephone calls is the sum of two parts, assumed to be indepedent, following negative exponential distributions with different means beta1 and beta2, so that its pdf is:

fT (t; beta1, beta2) = [e^(−t/beta2) − e^(−t/beta1)]/ [beta2 - beta1]

where (t > 0) and the overall mean is beta1 + beta2.

i need to find the likelihood function followed by the log-likelihood function and then derive the equations that would have to be solved in order to find the Maximum Likelihood Estimators for beta1 and beta2.

Every time i try it it just doesn't work and it's sooo frustrating

Thanks

**matheagle** · December 31st 2010, 09:28 PM

the best you can do is

$L(\vec x)=\prod_{i=1}^n{ e^{-x_i/b}- e^{-x_i/a}\over b-a}$

so $\ln L(\vec x)=\sum_{i=1}^n\ln ( e^{-x_i/b}- e^{-x_i/a})-n\ln ( b-a)$

Next you need to differentiate wrt both a and b.

${-1\over b}\sum_{i=1}^n{e^{-x_i/b}\over e^{-x_i/b}- e^{-x_i/a}}-{n\over b-a}=0$

and ${1\over a}\sum_{i=1}^n{e^{-x_i/a}\over e^{-x_i/b}- e^{-x_i/a}}+{n\over b-a}=0$

The next part looks worse, but I'm guessing this is all you were asked to do.
You could muliply the first by -b and the second by a and then subtract to kill that nasty sum.

**starlet** · January 2nd 2011, 04:50 AM

Aaaaah! It was the differentiating wrt both parts that got me!

Thanks matheagle

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