I was going thorough an actuarial exam and came across a problem that I can't figure out. Here is the problem as stated on the practice exam:

You are given:

$\bullet$ An insurance product with a per loss limit of 200 covers losses from an exponential distribution with parameter $\theta$.

$\bullet$ Based on the following table, the maximum likelihood estimate of $\theta$ is 168.

$$\begin{array}{|c | c | c |}\hline \text{Size of loss} & \text{Number of claims} & \text{Sum of losses} \\ \hline \text{Less than } 200 & 1,114 & 142,752\\ \hline \text{At least } 200 & N & 200N \\ \hline \end{array}$$

Calculate $N$ for the table above.

The answer is:

$N$ is less than 250.

My question: How can we know for sure that $N$ is less than 250? It can be thought of as a random variable, couldn't it? It could be that there is a single loss above 200 but very large so that the overall average is still 168, yes?

It seems that we are supposed to assume that the losses above 200 are exponential with mean 168 above 200 by the memoryless property and thus $$\frac{142752+368N}{1114+N}=168$$ gives $N=222$, which aligns well with the official answer.

But with the given information, it could be that the first 1,114 data points are all 128.14 and then we have a single data point of 44568. Of course this is an event of negligible probability, but nonetheless it is theoretically positive.

What am I missing?


The point is that, because of the claim limit, you do not observe the values above $200$, just the numbers of these censored values. What those high values would have been if they had not been censored could affect the likelihood, but you do not know them

Given $\theta$, the probability density for an observation of $x_i$ below $200$ is $\frac1\theta e^{-x_i/\theta}$, while the probability (not density) for an observation $x_j$ above $200$ is $e^{-200/\theta}$

That makes the likelihood for $\theta$ from the observations below $200$ proportional to $\frac1\theta^{1114} e^{-142752/\theta}$ and the likelihood from the observations above $200$ proportional to $e^{-200N/\theta}$. Multiply these together and you get an overall likelihood proportional to $\frac1\theta^{1114} e^{-(142752+200N)/\theta}$. That likelihood is maximised when $\theta = \frac{142752+200N}{1114}$

We are told that $\hat{\theta}= 168$, which suggests $N = \frac{1114 \times 168 - 142752}{200} = 222$

  • $\begingroup$ Ok, I see now. I thought the MLE was using the actual untruncated losses. So that we could have any arbitrary number of observations above 200 and still get 168 as the MLE. That would make the problem ill-posed. Or am I still wrong about that even? $\endgroup$ – jdods Apr 17 at 7:10
  • $\begingroup$ You had $N$ values above $200$ and the maximum likelihood estimator is $\hat\theta = \frac{142752+200N}{1114}$. If $N$ had been $1$ then you would have had $\hat\theta \approx 128.3$; if $N$ had been $1000$ then you would have had $\hat\theta \approx 307.7$. So saying "we could have any arbitrary number of observations above $200$ and still get $168$ as the MLE" is not correct $\endgroup$ – Henry Apr 17 at 7:26
  • $\begingroup$ Yes, I see now that it is the MLE for the loss, and not including any information about what went beyond 200. If the MLE was calculated knowing those numbers above 200 (i.e. standard exponential MLE formula), then N could be anything. Now I see why that shouldn't even be a relevant thought here. We are only estimating $\theta$ given our limited data. $\endgroup$ – jdods Apr 17 at 7:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.