I have a particular PDF with two parameters, specified as:

$$\alpha \beta e^{-\beta x}(1 - e^{-\beta x})^{\alpha - 1}, \alpha > 0, \beta > 0, x_i > 0$$

Given a random iid sample $(x_1, \dots, x_n)$ from the above, I can then write the likelihood function as:

$$L(\alpha, \beta \mid x_1, \dots, x_n) = (\alpha \beta)^n \prod_{i=1}^n e^{-\beta x}(1-e^{-\beta x})^{\alpha-1}$$

Taking the log thereof to make things slightly easier, we then have:

$$\log(L) = n \log(\alpha) + n \log(\beta) - \beta \sum_{i=1}^n x_i + (\alpha-1) \sum_{i=1}^n \log(1 - e^{-\beta x_i})$$

Taking the derivative with respect to $\alpha$ and setting to zero:

$$\frac{\partial}{\partial \alpha} \log(L) = 0 = \frac{n}{\alpha} + \sum_{i=1}^n \log(1-e^{-\beta x_i})$$ $$\hat{\alpha} = \frac{-n}{\sum_{i=1}^n \log(1 - e^{-\beta x_i})}$$

Then, plugging this back into the likelihood and repeating the process for $\beta$, I get something extremely messy and would have to numerically maximize. Did I screw up some math somewhere? No closed form implies that I will need numerical maximization for $\beta$...


1 Answer 1


The (correct) log-likelihood is

$$\log(L) = n \log(\alpha) + n \log(\beta) - \beta \sum_{i=1}^n x_i + (\alpha-1) \sum_{i=1}^n \log(1 - e^{-\beta x_i})$$

The f.o.c for $\alpha$ is

$$\hat{\alpha} = \frac{n}{-\sum_{i=1}^n \log(1 - e^{-\beta x_i})}$$

Plugging this into the log-likelihood we get

$$\log(L \mid \hat \alpha) = n \log\left(\frac{n}{-\sum_{i=1}^n \log(1 - e^{-\beta x_i})}\right) + n \log(\beta) - \beta \sum_{i=1}^n x_i + \left(\frac{n}{-\sum_{i=1}^n \log(1 - e^{-\beta x_i})}-1\right) \sum_{i=1}^n \log(1 - e^{-\beta x_i})$$

$$ = const. -n\log \left(-\sum_{i=1}^n \log(1 - e^{-\beta x_i})\right) + n \log(\beta) - \beta \sum_{i=1}^n x_i - \sum_{i=1}^n \log(1 - e^{-\beta x_i})$$

I wouldn't call the above,or the consequent derivative, "extremely messy", and yes you will need numerical maximization, which is the rule rather than the exception with ML estimation.

  • $\begingroup$ I suppose 'extremely messy' is just my way of saying that you cannot easily maximize using the derivative (as you might normally expect from a textbook example). Just because I'm curious about your last remark: is this (numerical maximization being the rule) because of the frequency with which one encounters less convenient likelihood functions? $\endgroup$ Mar 6, 2015 at 20:48
  • $\begingroup$ @PatternMatching Yes, that's an observation based on experience. For example, almost any likelihood that includes a cumulative distribution function (say for a censored sample) will need numerical maximization. $\endgroup$ Mar 6, 2015 at 21:23

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