# Determine distribution, Prove Laplace Transform

• Aug 14th 2009, 04:36 AM
woody198403
Determine distribution, Prove Laplace Transform
Consider a game in which a fair coin is tossed indefinitely. Every time heads appears you move 1 metre to the right, and if tails you stay where you are. You start at position 0. Let $T_n$ be the number of tosses needed to first enter position n.

By definition $T_0 = 0$

Determine the distribution of $T_1$ and prove that the Laplace Transform of $T_n$ is $\mathbb{E} e^{-sT_n} = (2e^s - 1)^{-n}, s \geq 0$
• Aug 15th 2009, 06:07 PM
woody198403
Could someone tell me if the distribution of $T_1$ is a geometric distribution?
• Aug 15th 2009, 06:43 PM
mr fantastic
Quote:

Originally Posted by woody198403
Could someone tell me if the distribution of $T_1$ is a geometric distribution?

Yes, it is.
• Aug 15th 2009, 10:27 PM
woody198403
So the distribution of $T_1$ is geometric.
But the distribution of $T_n$ is not geometric, right?
I am having trouble determining the distribution of $T_n$. 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.

Just to double check though, the Laplace transform of a continuous random variable X is
$L(s) = \mathbb{E} e^{-sX} = \int_0^{\infty} e^{-sx} f(x) \ dx$ where $f(x)$ is the pdf of $X$ ??
• Aug 15th 2009, 10:46 PM
mr fantastic
Quote:

Originally Posted by woody198403
So the distribution of $T_1$ is geometric.
But the distribution of $T_n$ is not geometric, right? Mr F says: Correct. The distribution of ${\color{red}T_n }$ is not geometric. It has a negative binomial distribution.

I am having trouble determining the distribution of $T_n$. 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: ${\color{red}T_n }$ 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
$L(s) = \mathbb{E} e^{-sX} = \int_0^{\infty} e^{-sx} f(x) \ dx$ where $f(x)$ is the pdf of $X$ ??

..
• Aug 15th 2009, 10:52 PM
woody198403
Thank-you Mr F. I dont know why I was looking at continuous distributions (Doh). Your help is greatly appreciated.
• Apr 1st 2013, 08:53 PM
wishfulpenguin
Re: Determine distribution, Prove Laplace Transform
This question really intrigues me, I understand to find the Laplace Transform of Tn, you integrate Tn x exp(-st) from the limits of infinity to zero { ∫∞^0 T n × e^(-st) }Please excuse the bad formatting. But based on this question I am confused as to how to find the function of Tn. Could someone please explain?

Thanks