Answer to Question #187044 in Statistics and Probability for sphoorthi

Question

Suppose that X

X has pdf given by

f(x)={x=1 if 1<x<2

3-x if 2<x<3

0 elsewhere

the MGF of X is

Answer 1

The given pdf is-

f(x)=x, if 1<x<2

3-x, if 2<x<3

0, elsewhere

The moment generating function is-

$M=E(e^{tx})$

$=\int_0^{\infty}f(x) e^{tx}dx$

$=\int_1^2xe^{tx}dx+\int_2^3(3-x)e^{tx}dx$

$=x\dfrac{e^{tx}}{t}|_1^2-\dfrac{e^{tx}}{t^2}|_1^2+3\dfrac{e^{tx}}{t}|_2^3-(x\dfrac{e^{tx}}{t}|_2^3-\dfrac{e^{tx}}{t^2}|_2^3)$

$=2\dfrac{e^{2t}}{t}-\dfrac{e^{t}}{1}-(\dfrac{e^2t-e^t}{t^2})+3(\dfrac{e^{3t}}{t}-\dfrac{e^{2t}}{2})-[3\dfrac{e^{3t}}{t}-2\dfrac{e^{2t}}{t}-\dfrac{e^{3t}}{t^2}+\dfrac{e^{2t}}{t^2}]$

$=-\dfrac{3e^{2t}}{2}+\dfrac{4e^{2t}}{t}-e^{-t}+\dfrac{e^t}{t^2}-\dfrac{2e^{2t}}{t^2}+\dfrac{e^{3t}}{t^2}$