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Moment generating function?
Suppose Y has a probability density function f(y) = e^(2-y) for y >2 and 0 otherwise.
Find m(t) and then use m(t) to find E(Y^2)
1 Answer
- kbLv 710 years agoFavorite Answer
Since Y has a continuous pdf,
m(t) = ∫(-∞ to ∞) e^(ty) f(y) dy
.......= 0 + ∫(2 to ∞) e^(ty) e^(2 - y) dy
.......= e^2 * ∫(2 to ∞) e^((t-1)y) dy
.......= e^2 * e^((t-1)y)/(t-1) {for 2 to ∞}
.......= e^2 * -e^(2(t-1))/(t-1) if t < 1
.......= -e^(2t)/(t - 1) if t < 1.
--------------------
E(Y^2) = (d^2/dt^2) -e^(2t)/(t - 1) {at t = 0}
...........= -2e^(2t) (2t^2 - 6t + 5)/(t - 1)^3 {at t = 0}
...........= 10.
I hope this helps!
