Practice Problem: Obtain the moment generating function for an exponential random variable

Let X be an exponential random variable. Recall that the pdf of an exponential random variable is given by

$ \ f_X(x)= \lambda e^{-\lambda x}, \text{ for }x\geq 0 . $

Obtain the moment generating function M_X(s) of X.

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Answer 1

Hint:

$ M_X(s)=E[e^{sX}] = \int e^{sx} f_{X}(x)dx $

get an answer for s < λ

note that the integrating range of x starts from 0

Answer 2

$ M_X(jw) = E[e^{jwx}] $

$ E[e^{jwx}] = \int e^{jwx} f_{X}(x)dx \text{ For } x \geq 0 $

$ M_X(jw) = \int_{0}^{\infty} e^{jwx} \lambda e^{-\lambda x}dx $

$ M_X(jw) = \lambda \int_{0}^{\infty} e^{jwx} e^{-\lambda x}dx $

$ M_X(jw) = \lambda \int_{0}^{\infty} e^{(jw-\lambda)x}dx $

$ M_X(jw) = \lambda \int_{0}^{\infty} e^{-(\lambda-jw)x}dx $

$ M_X(jw) = \lambda \frac {1}{-(\lambda-jw)}(e^{-(\lambda-jw)\infty}-e^{-(\lambda-jw)0}) $

$ M_X(jw) = \lambda \frac {1}{-(\lambda-jw)}(e^{-\infty}-e^{0}) $

$ M_X(jw) = \lambda \frac {1}{-(\lambda-jw)}(0-1) $

$ M_X(jw) = \frac {\lambda}{\lambda-jw} $

$ M_X(s) = \frac {\lambda}{\lambda-s} $

Instructor's comment: Be careful. In general, you can't obtain the Laplace transform by simply replacing $ j\omega $ by s. In fact, the corresponding integral for s diverges for some values of s. -pm

Answer 3

Write it here.

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