Moment Generating Function

Question

Let X has a probability density function

f(x) = k/(a-b) ; a<x<b

f(x)= 0 elsewhere

how do we find the moment generating function of Y= 10X+3

Answer 1

First, we notice that X is a continuous uniform random variable from a to b. This means it must have probability density function

f(x) = 1/(b-a)

This corresponds to k = -1 in the equation in the problem statement.

Next, we figure out how Y is distributed. If X is uniformly distribute between a and b then 10X must be uniformly distributed between 10a and 10b, and 10X + 3 must be uniformly distributed between 10a + 3 and 10b + 3.

Finally, we use the formula for the moment generating function of a continuous uniform distribution

M_X(t) = (e^tb -e^ta)/(t(b-a))

Substituting the new lower bound and new upper bound for the Y, we get

M_Y(t) = (e^{t(10b + 3)}-e^{t(10a + 3)})/(t(10b + 3 - 10a - 3)),

which can be simplified to

M_Y(t) = (e^{t(10b + 3)}-e^{t(10a + 3)})/(t(10b - 10a)).