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I have two sequences of exponentially distributed independent variables, $X_1, \ldots, X_n$ and $Y_1, \ldots, Y_n$ $$X_{i}\sim \exp(a)\\ Y_i \sim \exp(b)$$ However, I only have the difference of the two sequences $X_1-Y_1, \ldots, X_n-Y_n$ recorded. I was wondering if there is a way to create MLE estimators for the parameters $a$ and $b$ using only the differences between the variables?

Michael Hardy
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zet5000
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1 Answers1

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Sometimes "exponentially distributed with parameter $a$" means the distribution is $e^{-ax}(a\,dx) \text{ for } x>0,$ and sometimes it means $e^{-x/a} (dx/a) \text{ for } x>0.$ For now I will assume the former. I will further assume the sequences are of independent random variables. I would have stated that in the problem.

Robert Israel's answer to a related question tells us that the density of $X-Y$ is $$ \frac{ab}{a+b} \begin{cases} e^{-ax}\,dx & \text{if }x>0, \\ e^{-bx} \,dx & \text{if } x<0. \end{cases} $$ The the likelihood function is \begin{align} L(a,b) & = \left( \frac{ab}{a+b} \right)^n \left( \prod_{i\,:\,x_i \,>\,0} e^{-ax_i} \right) \left( \prod_{i\,:\,x_i\,<\,0} e^{-bx_i} \right) \\[8pt] & = \left( \frac{ab}{a+b} \right)^n \exp\left( -a \sum_{i\,:\,x_i \,>\,0} x_i - b\sum_{i\,:\,x_i \,<\,0} x_i \right). \end{align} Therefore \begin{align} & \ell(a,b) = \log L(a,b) \\[8pt] = {} & n(\log a + \log b - \log(a+b)) -a \sum_{i\,:\,x_i \,>\,0} x_i - b\sum_{i\,:\,x_i \,<\,0} x_i \\[8pt] \text{and so } & \frac{\partial\ell}{\partial a} = \frac n a - \frac n{a+b} - \sum_{i\,:\,x_i \,>\,0} x_i, \\[8pt] \text{and } & \frac{\partial\ell}{\partial b} = \frac n b - \frac n{a+b} - \sum_{i\,:\,x_i \,<\,0} x_i. \\[8pt] \end{align} These are both $0$ when $$\frac b {a(a+b)} = \overline{x}_{>0} \quad\text{and}\quad \frac a{b(a+b)} = \overline{x}_{<0} $$ where $\overline{x}_{>0}$ and $\overline{x}_{<0}$ are respectively the means of the positive and negative $x$-values. Where there are no positive $x$-values and where there are no negative $x$-values, the MLEs for $a,b$ respectively are undefined.

Possibly to be continued$\,\ldots$

Michael Hardy
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