All density functions must sum to 1, but the integral of 0.75*exp(-y) from 0 to
infinity is 0.75. Therefore, the function 0.75*exp(-y) must be multiplied by the
inverse of 0.75 in order to become a probability density function.
This means that f(y|y>0) is exp(-y).
(0.75*exp(-y) has to be normalized over the domain, y>0. Because 75% of the
distribution function F(y) is contained in the area above y>0, the orginal
density f(y) must be divided by the probability that y is greater than 0, or
0.75. This guarantees that the resultant density will sum to 1 over the domain.)
Sheri De La Boursodiere <Sheri_De_La_Boursodiere@hsbc.ca> on 03/04/99 02:43:16
PM
To: studygroup4b <studygroup4b@lists.casact.org>
cc: (bcc: Christian Coleianne/MCIG/USA/Zurich)
Subject: 4b Intro to Credibility Theory question 2.11
Given f(y)= 0 y<0
.25 y=0
.75e^-y y>0
The solution in the book is E[Y^3|y>0]=Integral from o to infinity of
(y^3)(e^-y)dy = 6
Why isn't the solution E[Y^3|y>0]=Integral from o to infinity of
(y^3)(.75e^-y)dy = .75 x 6 = 4.5 ?
Sheri.
Can someone help me figure out why I am .75 off on this solution?