# Thread: Moment generating function and distribution function?

1. ## Moment generating function and distribution function?

I got part a out easily enough. I'm having big problems on how to do part b and c. I'm like this . Thanks for any help

2. Originally Posted by DCU

I got part a out easily enough. I'm having big problems on how to do part b and c. I'm like this . Thanks for any help
(b) I suppose they want you to calculate the cdf of X: $F(x) = \int_0^x \lambda e^{-\lambda u} \, du$.

This should be a simple integral for you to calculate.

(c) Apply the definition: $M_X (t) = E\left(e^{tX}\right) = \int_0^{+ \infty} e^{tx} \lambda e^{-\lambda x} \, dx = \lambda \int_0^{+ \infty}e^{-(\lambda - t) x} \, dx$.

Again, this should be a simple integral for you to calculate. What happens if $\lambda - t < 0$ ....?

3. Originally Posted by mr fantastic
(b) I suppose they want you to calculate the cdf of X: $F(x) = \int_0^x \lambda e^{-\lambda u} \, du$.

This should be a simple integral for you to calculate.

(c) Apply the definition: $M_X (t) = E\left(e^{tX}\right) = \int_0^{+ \infty} e^{tx} \lambda e^{-\lambda x} \, dx = \lambda \int_0^{+ \infty}e^{-(\lambda - t) x} \, dx$.

Again, this should be a simple integral for you to calculate. What happens if $\lambda - t < 0$ ....?
if $\lambda - t < 0$, the value of the integral will go to infinity?

4. Originally Posted by DCU
if $\lambda - t < 0$, the value of the integral will go to infinity?
So that tells you the answer to the last part of (c).

5. Originally Posted by mr fantastic
So that tells you the answer to the last part of (c).
so when lambda is greater than t?