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Insurance company has to pay payments at the rate of $d$ per year. They are payable continuously as long as the person remains sick.

The length of the payment period in years is a random variable with the gamma distribution with mean $m$ and variance $d$.

Find the actuarial present value of these payments (force of interest $σ$ is constant).

Solution:

I need to calculate:

$$\bar a = \int_{0}^{\infty} \frac{1-e^{-σt}}{σ} f(t) dt,$$

here $f(t) = \frac{θe^{-θt} (θt)^{α-1}}{ {\Gamma (\alpha )}}$

First of all I want to find these $\alpha$ and $\theta$. From gamma distribution (mean and variance):

$\alpha \theta = m$ and $\alpha \theta ^2=d,$ I get $\alpha=\frac{m^2}{d}$ and $\theta=\frac{d}{m}.$

So now the integral looks like:

$$\bar a = \int_{0}^{\infty} \frac{1-e^{-σt}}{σ} \frac{\frac{d}{m} e^{-\frac{d}{m}t} (\frac{d}{m}t)^{\frac{m^2}{d}-1}}{ {\Gamma (\frac{m^2}{d} )}} dt$$

Next,

$\Gamma (\frac{m^2}{d} ) = \int_{0}^{\infty} e^{-y} y^{\frac{m^2}{d}-1} dy.$

Can someone help me to calculate $\Gamma (\frac{m^2}{d} )$ and $\bar a$?

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Let $t=\frac m dx$ to make $$\bar a = \frac{1}{\sigma \Gamma \left(\frac{m^2}{d}\right)} \int_{0}^{\infty}\Big( x^{\frac{m^2}{d}-1}\,e^{-x}-x^{\frac{m^2}{d}-1} e^{-\frac{ (d+m \sigma )}{d}x} \Big)\,dx$$ Now, using $$\int_{0}^{\infty} x^a e^{-bx}\,dx=b^{-(a+1)} \Gamma (a+1) \qquad \text{if} \qquad\Re(b)>0\land \Re(a)>-1$$ you should obtain (after some minor simplifications) $$\bar a=\frac{1}{\sigma }\left(1-\left(\frac{d}{d+m \sigma }\right)^{\frac{m ^2}d}\right)$$

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  • $\begingroup$ Can you explain where did the fraction $\frac{d}{m}$ disappear? $\endgroup$ – Begri Apr 26 at 12:10

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