Re: Fall 98 Question 19

doris.schirmacher@zurich.com
Fri, 16 Apr 1999 09:33:48 +0200

Hi,
A_bar x is the expected value of v^T, where T is gamma distributed.
So from definition, we have
E[v^T] = Integral (from 0 to infinity) v^t b^a t^(a-1)e^(-delta*t)/Gamma(a)
dt
= b^a/Gamma(a) * Integral v^t t^(a-1) e^(-delta*t) dt
= b^a/Gamma(a) * Integral t^(a-1) e^(-(delta+b)t) dt

Now, the way I did the integral is to recognize that the integrand is the
density function
of a gamma distribution with parameter a and delta+b, with constant factor
(delta+b)^a/Gamma(a)
in front missing. So this integral should be Gamma(a)/ (delta+b)^a. (Remember,
integral
of a density function should give you 1). Therefore,
E[v^T] = b^a/Gamma(a) * Gamma(a)/ (delta+b)^a
= b^a / (delta +b)^a

I hope you understand what I wrote.

Doris

"marc.rothschild" <marc.rothschild@zurichus.com> on 04/15/99 10:13:55 PM

To: studygroup4a <studygroup4a@lists.casact.org>
cc: (bcc: Doris Schirmacher/Re/Germany/Zurich)
Subject: Fall 98 Question 19

If the future lifetime random variable T for (x) is gamma distributed with
parameters a and b, the which of the following is an expression for a-bar x? A
hint is given showing the pdf of a gamma.

The CSM's solution says that a-bar x = (1-A-bar x)/delta, which is OK, then it
goes on to state that A-bar x = (b/(delta + b)^a, which I don't understand.

Could someone explain the latter part of the CSM's solutions?

Thanks, Marc