# Thread: Determine distribution, Prove Laplace Transform

1. ## 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 $\displaystyle T_n$ be the number of tosses needed to first enter position n.

By definition $\displaystyle T_0 = 0$

Determine the distribution of $\displaystyle T_1$ and prove that the Laplace Transform of $\displaystyle T_n$ is $\displaystyle \mathbb{E} e^{-sT_n} = (2e^s - 1)^{-n}, s \geq 0$

2. Could someone tell me if the distribution of $\displaystyle T_1$ is a geometric distribution?

3. Originally Posted by woody198403
Could someone tell me if the distribution of $\displaystyle T_1$ is a geometric distribution?
Yes, it is.

4. So the distribution of $\displaystyle T_1$ is geometric.
But the distribution of $\displaystyle T_n$ is not geometric, right?
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.

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 $\displaystyle X$ ??

5. 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 $\displaystyle X$ ??
..

6. Thank-you Mr F. I dont know why I was looking at continuous distributions . Your help is greatly appreciated.

7. ## 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