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#1
11-07-2010, 03:50 PM
 Zraic Join Date: Sep 2010 Posts: 12
Moment Generating Function

Let X be the random variable whose density function is given by f(x)=2(1-x) for 0<x<1 (both are less then or equal to but dont know how to do those symbols). and f(x)= 0 o.w. Find Mx(T). Would it just be the E(e^tx) so it would be the integral of ((e^tx)*2(1-x)) evaluated at 1 and 0? I can't get the integral to match the answer in the back of the book. The answer in the back of the book is (2e^t-2t-2)/t^2

Any help?
#2
11-07-2010, 04:12 PM
 Actuarialsuck Member Join Date: Sep 2007 Posts: 5,324

Then you are not doing your integral right since

$\int_0^1 \, 2e^{tx}(1-x) \, dx \, = \, 2 \frac{e^t \, - \, t \, - \, 1}{t^2}$
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#3
11-10-2010, 05:02 PM
 ActSciMan Member CAS SOA Join Date: Nov 2010 Favorite beer: Beer sucks Posts: 152

So, did that previous post help? You just need to use integration by parts with u = 1 - x and $dv = e^{tx} dx$ (the 2 can just be pulled out in front of the integral). Then, du = -dx and $v = \frac{1}{t}e^{tx}$, right?

$\int_0^1 2e^{tx}(1-x) \,dx = 2[(1-x)\cdot \frac{1}{t}e^{tx}]_0^1 + 2\int_0^1 \frac{1}{t} e^{tx} \,dx = -\frac{2}{t} + 2[\frac{1}{t^2} e^{tx}]_0^1 = -\frac{2}{t} + 2[\frac{e^t}{t^2} - \frac{1}{t^2}] = \frac{-2t + 2e^t - 2}{t^2}$

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