Determine distribution, Prove Laplace Transform

**woody198403** · August 14th 2009, 04:36 AM

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 $\mathbb{E} e^{-sT_n} = (2e^s - 1)^{-n}, s \geq 0$

**woody198403** · August 15th 2009, 06:07 PM

Could someone tell me if the distribution of $T_1$ is a geometric distribution?

**mr fantastic** · August 15th 2009, 06:43 PM

Originally Posted by woody198403

Could someone tell me if the distribution of $T_1$ is a geometric distribution?

Yes, it is.

**woody198403** · August 15th 2009, 10:27 PM

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 $X$ ??

**mr fantastic** · August 15th 2009, 10:46 PM

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 $X$ ??

..

**woody198403** · August 15th 2009, 10:52 PM

Thank-you Mr F. I dont know why I was looking at continuous distributions

. Your help is greatly appreciated.

**wishfulpenguin** · April 1st 2013, 08:53 PM

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

Thread: Determine distribution, Prove Laplace Transform

LinkBack

Thread Tools

Display

Determine distribution, Prove Laplace Transform

Re: Determine distribution, Prove Laplace Transform

Similar Math Help Forum Discussions

Help me determine the Laplace Transform and the ROC?

Laplace transform and Fourier transform what is the different?

Fourier Transform and Laplace distribution

Laplace transform

Laplace Transform - Determine response of the system

Search Tags