Moment generating function for geometric

geometric distribution given by

$\displaystyle Pr(X=x) = (1/a)(1-(1/a)^{x-1}) $ where a > 1.

Am I right in saying the distribution above is the same as

$\displaystyle Pr(X=x) = p(1-p)^{x-1} $ which is just the normal geometric pdf or does the change from p to 1/a change something else

Re: Moment generating function for geometric

Hello,

It doesn't change anything, but I just don't understand where the MGF is in there ?

Re: Moment generating function for geometric

Quote:

Originally Posted by

**whiteboard** geometric distribution given by

$\displaystyle Pr(X=x) = (1/a)(1-(1/a)^{x-1}) $ where a > 1.

Am I right in saying the distribution above is the same as

$\displaystyle Pr(X=x) = p(1-p)^{x-1} $ which is just the normal geometric pdf or does the change from p to 1/a change something else

What is raised to the power (x-1) in each of these cases? (and is it a typo?)

CB