Dan Wigodsky
May 2, 2018
Imagine an insurance case where we have 8 claims from an exponential distribution. We know that 2 have a deductible of $10,000 applied to them. The remaining claims are: $4000, $5500, $6200, $7100, $8500 and $9000. How can we fit a parametric distribution from this experience? The good news is that a Gamma distribution, which the exponential is a special case of, can be fit for a likelihood distribution by its moments. In this case, however, we have censored data so we will go ahead with its likelihood.
\(\large\frac{4000}{\theta}e^{\frac{-4000}{\theta}}\frac{5500}{\theta}e^{\frac{-5500}{\theta}}\frac{6200}{\theta}e^{\frac{-6200}{\theta}}*\)
\(\large\frac{7100}{\theta}e^{\frac{-7100}{\theta}}\frac{8500}{\theta}e^{\frac{-8500}{\theta}}\frac{9000}{\theta}e^{\frac{-9000}{\theta}}(e^{\frac{-10000}{\theta}})^{2}\)
To find the appropriate mean, we want to maximize this function. We can find a maximum or minimum of a function by finding where its derivative is 0. Our function turns out to be a difficult function to maximize, though. Luckily, the natural log (or any log) is a continuously increasing function:
We can take the natural log of our likelihood function and the resulting transformation will reach a maximum at the same place.
\(\large\sum_1^6(ln(x_{i}))-6(ln(\theta))-{\frac{\sum_1^6(x_{i})}{\theta}}-\frac{20000}{\theta}\)
Now, to maximize the function, We set its derivative equal to zero. Note that numeric coefficients disappear when we take the derivative.
\(\large0=\frac{-6}{\theta}+\frac{\sum_1^6(x_{i})}{\theta^{2}}+\frac{20000}{\theta^{2}}\)
\(6\theta=4000+5500+6200+7100+8500+9000+20000\)
\(\theta=10,050\)
Our estimate of the parameter, \(\theta\), which is our mean, is $10,050. The censored points skewed our distribution upwards because we don’t know what the true value of the losses for those claim were.