# Actuarial Discussion Forum - Professional Discussions for Professional Actuaries

 D.W. Simpson & Co, Inc. - Worldwide Actuarial Jobs Life Jobs Health Jobs Pension Jobs Casualty Jobs Salary Apply Pauline Reimer, ASA, MAAA - Pryor Associates Nat'l/Int'l Actuarial Openings: Life P&C Health Pensions Finance ACTEX Publications and MadRiver Books Serving students worldwide for over 40 years Advertise Here - Reach Actuarial Professionals Advertising Information Actuarial Careers, Inc.® - Actuarial Jobs Worldwide Search positions by geographic region, specialization, or salary Ezra Penland Actuarial Recruiters - Top Actuarial Jobs Salary Surveys  Apply Online   Bios   Casualty   Health   Life   Pension

# Thread: mgf (insurance)Question - Australia

1. ## mgf (insurance)Question - Australia

Hi,

im studying actuarial studies in Australia, and i have a problem on moment generating functions (im not sure where this fits in with your curriculum)....can anyone help me?

Claim amounts for a certain insurance portfolio, x1,x2..., follow a distribution with pdf:

f(x;y) = 1/(2y)*e^(-x/y) + 1/y*e^(-2x/y)

calculate the moment generating function of X and show that E(X) = 3/4*y and var(x) = 11/16*y^2

2. Originally Posted by foofighters26
Hi,

im studying actuarial studies in Australia, and i have a problem on moment generating functions (im not sure where this fits in with your curriculum)....can anyone help me?

Claim amounts for a certain insurance portfolio, x1,x2..., follow a distribution with pdf:

f(x;y) = 1/(2y)*e^(-x/y) + 1/y*e^(-2x/y)

calculate the moment generating function of X and show that E(X) = 3/4*y and var(x) = 11/16*y^2
I guess the moment generating function turns out to be

M(t) = 1/(2y*(t-1/y)) + 1/(y*(t-2/y)) [ it is the integral of e^(tx)f(x,y) from 0 < x < infinity; note that t < 1/y otherwise the integral diverges.

<X> = - (3/4)*y [ TAKE the derivative with respect to 't', keeping y fixed and put t = 0] [Note, E[X] is negative, (please check the calculation again)]

<X^2> = (5/4)*y^2 [Take the 2nd Derivative with respect to 't', keeping y fixed and put t = 0]

So, Var(X) = <X^2> - <X>^2 = (11/16)*Y^2

Does that help ?

If you need to see all the steps, let me know ( but it is kinda hard to write down those mathematical symbols though ).

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts