Thread: moment generating function of exponential dist.

1. moment generating function of exponential dist.

I'm having trouble finding the moment generating function of a exponential distribution.

$\phi = \int e^{ty}f(x) dx = \int_{0}^{\infty} e^{ty}\frac{1}{\beta} e^{-y/\beta} dy$

$\phi = \frac{1}{\beta} \int_{0}^{\infty}e^{y(t-1/\beta)} dy$

$\phi = \frac{1}{\beta}\times \frac{1}{t-\frac{1}{\beta}} e^{y(t-1/\beta)}\bigg{|}^{\infty}_{0}$

$\phi= \left(\frac{1}{\beta} \right) \left( \frac{1}{t-\frac{1}{\beta}} \right)$

$\phi = \frac{1}{\beta t-1}$

but the table in the back of my book gives:

$\phi = \frac{1}{\color{red}{1-\beta t}}$

2. Originally Posted by lllll
I'm having trouble finding the moment generating function of a exponential distribution.

$\phi = \int e^{ty}f(x) dx = \int_{0}^{\infty} e^{ty}\frac{1}{\beta} e^{-y/\beta} dy$

$\phi = \frac{1}{\beta} \int_{0}^{\infty}e^{y(t-1/\beta)} dy$

$\phi = \frac{1}{\beta}\times \frac{1}{t-\frac{1}{\beta}} e^{y(t-1/\beta)}\bigg{|}^{\infty}_{0}$

$\phi= \left(\frac{1}{\beta} \right) \left( \frac{1}{t-\frac{1}{\beta}} \right)$

$\phi = \frac{1}{\beta t-1}$

but the table in the back of my book gives:

$\phi = \frac{1}{\color{red}{1-\beta t}}$
The issue here in integrating this is the the value of $e^{y(t-1/\beta)}$ as $y\to\infty$.

We can fix this by writing $e^{y(t-1/\beta)}=e^{-y(1/\beta-t)}$

In evaluating the integral for the moment generating function, we should have

$\phi = \frac{1}{\beta}\times -\frac{1}{\frac{1}{\beta}-t} e^{-y(1/\beta-t)}\bigg{|}^{\infty}_{0} =-\frac{1}{1-\beta t}\times\left[e^{-\infty}-e^0\right]=-\frac{1}{1-\beta t}\times\left(-1\right)=\color{red}\boxed{\frac{1}{1-\beta t}}$

Does this make sense?

--Chris