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Assume now that Ali and Berta independently assign exponentially distributed scores, with rates $a$ and $b$, respectively. But they only record the difference between the two scores (Ali’s score minus Berta’s score, say.) What is the maximum likelihood estimator for $a$ and $b$, using only the score differences?

In a previous part I calculated the MLE of the RV $X$ that was the sum of their scores that was exponentially dist with parameter $\lambda$ as:

$\hat{\lambda} = \frac{n}{\sum_{i=1}^{n} x_i}$

Note that in that part we only observed $X$ and nothing about their scores or score rates individually.

Can I just consider this same result for my new RVs $U$ and $V$ with their parameter's $a$ and $b$? How would you suggest approaching this new RV $Z$ that is their difference?

I see here Find the distribution of $Z=X+Y$ where both $X$ and $Y$ are exponentially distributed. that there might be something to the new variable having the gamma dist. Is this a correct path to take?

I'm just looking for a helpful push or maybe a resource that shows me the right direction.

Thanks for your time.

Edit: $U$, $V$ $\geq 0$ sorry forgot to add this before.

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I don't know if you can find anything interesting.
Based on my calculation, the density of $Z$ would be:
$ f(v) = \frac{ ab}{a+b} e^{ -a|v|}$ if $v \ge 0$
and $f(v)= \frac{ ab}{a+b} e^{ -b|v|}$ if $v \le 0$
So the standard procedure of finding MLE here seems to be not complicated. Though the final result might not be as explicit as when there is only one exponential variable.

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  • $\begingroup$ I forgot to add to the post that u and v are both $\geq 0$. But thank you for your time giving me some answer. Can I ask how you got started making that density calculation? Knowing what we know about $U$ and $V$ what do you then know about $Z$ that makes you able to calculate density? $\endgroup$
    – GoHawkeyes
    Nov 7 '20 at 1:37
  • $\begingroup$ Nothing in particular. I just calculate it. So, I just call out any bounded continuous function $f$ then I use the change of variable to simplify the formula of $\mathbb{E}( f(Z))$. $\endgroup$ Nov 7 '20 at 1:41
  • $\begingroup$ I have to say I need some more information to understand. Sorry I'm still learning about MLE. Thanks for your help $\endgroup$
    – GoHawkeyes
    Nov 7 '20 at 2:24

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