MLE for the mean of exponential fill gaps

Question

A market-making desk records the time in seconds between consecutive fills and models each gap as an independent exponential variable with unknown mean $\theta$ , density

$f(x \mid \theta) = \frac{1}{\theta}\, e^{-x/\theta}, \qquad x > 0.$

Six gaps are logged: 3, 8, 1, 6, 4, 2 seconds.

Derive the maximum likelihood estimator of $\theta$ and compute it for this sample.

Accepted Answer

Write the likelihood of the whole sample as the product of densities, then take logs to turn the product into a sum: \ell(	heta) = \sum{i=1}^n \left( -\ln 	heta - \frac{xi}{	heta} ight) = -n\ln	heta - \frac{1}{	heta}\sum{i=1}^n xi. Differentiate with respect to 	heta and set the derivative to zero: \ell'(	heta) = -\frac{n}{	heta} + \frac{1}{	heta^2}\sumi xi = 0 \;\Longrightarrow\; \sumi xi = n	heta \;\Longrightarrow\; \hat	heta = \frac{1}{n}\sumi xi = \bar x. The second derivative is negative at this point, so it is a maximum. The MLE of the mean of an exponential is just the sam…

MLE for the mean of exponential fill gaps

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