# Integration of gamma function

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 $$\bar a$$?

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) $$\bar a=\frac{1}{\sigma }\left(1-\left(\frac{d}{d+m \sigma }\right)^{\frac{m ^2}d}\right)$$
• Can you explain where did the fraction $\frac{d}{m}$ disappear? – Begri Apr 26 at 12:10