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The probability that a fire will occur is $0.001$. If there is a fire, the amount of damage, $X$, will have a Pareto distribution given by $P(X>x) = \left(\frac{2(10)^6}{2(10)^6+x} \right)^2$. An insurance will pay the excess of the loss over a deductible of $100,000$. For this coverage the one-time insurance premium will be $110 \%$ of the expected payment. Calculate the premium.

So the expected payment is $E[W]$ where $W$ denotes the payment. Then $E[W] = E[W| \text{fire}]P(\text{fire})+E[W| \text{no fire}]P(\text{no fire})$. To calculate $E[W| \text{fire}]$, we could use $\int_{0.1}^{\infty} [1-F(x)] \ dx$? This would be: $\int_{0.1}^{\infty} 1-\left[1-\left(\frac{2(10)^6}{2(10)^6+x} \right)^2\right] \ dx$ which equals $\int_{0.1}^{\infty} \left(\frac{2(10)^6}{2(10)^6+x} \right)^2 \ dx$?

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TO calculate $E(W|fire)$, you first need to calculate the pdf. You can get this by taking derivative of $1-P(X>x)$. Call this function $p(x)$.

Since the insurer pays out any amount greater than the deductible of $100000$, the expected payment is given by

\begin{equation} \int_{100000}^\infty (x - 100000) p(x) dx \end{equation}

The premium should be 110% of the value of the integral.

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  • $\begingroup$ But for a non-negative random variable $X$, we have $E(X) = \int_{0}^{\infty} [1-F(x)] \ dx$. $\endgroup$ – PrimeNumber Nov 5 '10 at 18:26
  • $\begingroup$ @Trevor: Right, but you can also calculate expected value the way Svenkatr does. It's important, too, that you're looking for $E[W|fire]$, not $E[X|fire]$. Svenkatr's answer uses the key fact that $W = X - 100,000$ when $X \geq 100,000$ and $W = 0$ when $X < 100,000$. $\endgroup$ – Mike Spivey Nov 5 '10 at 20:10
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So $E[W| \text{fire}] = \int_{0.1}^{\infty} [1-F(x)] \ dx = 4 \int_{0.1}^{\infty} (2+x)^{-2} \ dx = 4/2.1$. Thus $E[W] = (4/2.1)(0.001)+0 = 0.00190476$. So the premium is $(0.0019047)(10^6)(1.1) =2,095$.

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