Originally Posted by

**woody198403** So the distribution of $\displaystyle T_1 $ is geometric.

But the distribution of $\displaystyle T_n $ is not geometric, right? Mr F says: Correct. The distribution of $\displaystyle {\color{red}T_n }$ is not geometric. It has a negative binomial distribution.

I am having trouble determining the distribution of $\displaystyle T_n $. My original thought was that it was perhaps the exponential distribution because the exponential distribution can be considered the continuous version of the geometric distribution, and unless I have made an error in my calculations, the Laplace Transform of the exponential dist. is not the same as the solution given in the question. Mr F says: $\displaystyle {\color{red}T_n }$ is clearly a discrete random variable and therefore cannot have a continuous distribution.

Just to double check though, the Laplace transform of a continuous random variable X is

$\displaystyle L(s) = \mathbb{E} e^{-sX} = \int_0^{\infty} e^{-sx} f(x) \ dx $ where $\displaystyle f(x) $ is the pdf of $\displaystyle X $ ??