Question #187044

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


Expert's answer

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

=∫0∞f(x)etxdx=\int_0^{\infty}f(x) e^{tx}dx


=∫12xetxdx+∫23(3−x)etxdx=\int_1^2xe^{tx}dx+\int_2^3(3-x)e^{tx}dx


=xetxt∣12−etxt2∣12+3etxt∣23−(xetxt∣23−etxt2∣23)=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)


=2e2tt−et1−(e2t−ett2)+3(e3tt−e2t2)−[3e3tt−2e2tt−e3tt2+e2tt2]=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}]


=−3e2t2+4e2tt−e−t+ett2−2e2tt2+e3tt2=-\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}


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