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Suppose I have $X_1 \sim Exp(\theta_1)$ and $X_2\sim Exp(\theta_2)$. Then it is not difficult to show that $Y = X_1 - X_2$ will have density:

$f_Y(y) = \frac{1}{\theta_1 + \theta_2}e^{-y/\theta_1}\mathbb{1}_{y > 0} + \frac{1}{\theta_1+\theta_2}e^{y/\theta_2}\mathbb{1}_{y\leq 0}$.

Given $n$ iid observations from $Y$, we can find the sufficient statistics and the MLE of $\theta_1, \theta_2$. I am not sure if there is a shorter way to do this... I seem to arrive at quadratic equations here when I do this.

The likelihood is:

$\frac{1}{(\theta_1 + \theta_2)^n}e^{-\frac{1}{\theta_1}\sum y^+ +\frac{1}{\theta_2}\sum y^-}$; where the $y^+ = \max(0,y)$ and $y^- = \min(0,y)$. By factorization, we have $\left(\sum y^+, \sum y^-\right)$ as the sufficient statistics.

Now for the MLE, we have first the log of the likelihood which is:

$\ell = -n\log(\theta_1 + \theta_2)-\frac{1}{\theta_1}\sum y^+ + \frac{1}{\theta_2}\sum y^-$

Then the first derivatives which are set of 0, are: $\frac{\partial \ell}{\partial \theta_1} = -\frac{n}{\theta_1 + \theta_2} + \frac{\sum y^+}{\theta_1^2}\equiv 0$ and $\frac{\partial \ell}{\partial \theta_2} = -\frac{n}{\theta_1 + \theta_2}-\frac{\sum y^-}{\theta_2^2} \equiv 0$.

So it follows that $\frac{1}{n} \sum y^+ = \left(\frac{\theta_1^2}{\theta_1+\theta_2}\right)_{MLE}$ and $-\frac{1}{n} \sum y^- = \left(\frac{\theta_2^2}{\theta_1+\theta_2}\right)_{MLE}$. This is where I am stuck.

How can we find the $\hat{\theta}_1^{MLE}$ and $\hat{\theta}_2^{MLE}$ individually? I can solve the system of equations on Mathematica, but I am unable to do so by hand.

A direction:

I could take $\frac{1}{n} \sum y^+ +\frac{1}{n} \sum y^-$ and by invariance, I get $\left(\frac{\theta_1^2-\theta_2^2}{\theta_1+\theta_2}\right)_{MLE} = (\theta_1 - \theta_2)_{MLE}$. However, I would still need to find a way to isolate the individual parameters.

Since we now have $(\theta_1 - \theta_2)$ we can also find $(\theta_1 + \theta_2)$. The denominator of $\frac{\theta_1^2}{\theta_1 + \theta_2}$ is telling that we should form $\frac{(\theta_1 + \theta_2)^2}{\theta_1 + \theta_2} \to (\theta_1 + \theta_2)$. Then we can solve an easier systems of equations that is linear on $\theta_1$ and $\theta_2$.

From here on, I use simpler notation, $S_p = \frac{1}{n}\sum y^+$ and $S_n = \frac{1}{n}\sum y^-$. By invariance, $\sqrt{-S_n}$ is the MLE estimate of $\frac{\theta_2}{\sqrt{\theta_1 + \theta_2}}$ and $\sqrt{S_p} = \frac{\theta_1}{\sqrt{\theta_1+\theta_2}}$, thus $2\sqrt{-S_n S_p}$ is the MLE of $\frac{2\theta_1 \theta_2}{\theta_1 + \theta_2}$. So \begin{align*} \frac{\theta_1^2 + 2\theta_1 \theta_2 + \theta_2^2}{\theta_1 + \theta_2} &= \frac{(\theta_1 + \theta_2)^2}{\theta_1 + \theta_2} \\ &= \theta_1 + \theta_2 \end{align*} Now we can estimate the above by $S_p + S_n + 2\sqrt{-S_n S_p}$. Thus we have:

$\theta_1 - \theta_2 = S_p + S_n$ and $\theta_1 + \theta_2 = S_p + S_n +2\sqrt{-S_n S_p}$. This is now easily solvable for $(\theta_1, \theta_2)$.

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    $\begingroup$ A quicker way would be to use the invariance property of the MLE. $\endgroup$
    – Ben
    Jun 29, 2022 at 2:59
  • $\begingroup$ @Ben I updated the post with my ideas of using the invariance of MLE. I was only able to get that far unfortunately. $\endgroup$
    – s l
    Jun 29, 2022 at 3:04
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    $\begingroup$ actually, i'm onto something. will update the answer as I get closer to a solution $\endgroup$
    – s l
    Jun 29, 2022 at 3:09
  • $\begingroup$ @Ben I have posted the final steps of my attempt. I believe I have gotten my minuses and pluses correct, but there is a chance I may have copied my handwriting incorrectly onto stackexchange. But the idea is the same. $\endgroup$
    – s l
    Jun 29, 2022 at 3:25
  • $\begingroup$ good catch indeed it is a typo $\endgroup$
    – s l
    Jul 10, 2022 at 1:25

1 Answer 1

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Following from the direction at the end of the post above, the solution is now \begin{align*} \theta_1 - \theta_2 &= S_p + S_n \\ \theta_1 + \theta_2 &= S_p - S_n + 2\sqrt{-S_n S_p} \\ &\implies 2\theta_1 = 2S_p + 2\sqrt{-S_n S_p} \implies \theta_1^{MLE} = S_p + \sqrt{-S_n S_p} \\ &\implies -2\theta_2 = 2S_n - 2\sqrt{-S_n S_p} \implies \theta_2^{MLE} = -S_n + \sqrt{-S_n S_p} \end{align*}

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