Poisson Process Question

**lllll** · November 16th 2008, 03:58 PM

Assume that you perform and experiment where the outcomes occur according to a Poisson Process with rate 4 per minute, and that the experiment is complete once you observe 250 outcomes. Calculate the mean time and variance time of the experiment.

for the mean would think that it's just $\frac{250}{4}=62.5$

but for the variance I'm not too sure but would think that it's $\frac{250}{4^2}=15.625$

are these correct?

**CaptainBlack** · November 16th 2008, 10:32 PM

Originally Posted by lllll

Assume that you perform and experiment where the outcomes occur according to a Poisson Process with rate 4 per minute, and that the experiment is complete once you observe 250 outcomes. Calculate the mean time and variance time of the experiment.

for the mean would think that it's just $\frac{250}{4}=62.5$

but for the variance I'm not too sure but would think that it's $\frac{250}{4^2}=15.625$

are these correct?

The interval between events has a exponential distribution with a mean of 1/4 min and a variance of (1/4)^2, so a SD of 1/4 min, also the intervals are independent RV's.

Now you are looking at the sum of 250 such intervals so the mean is 250(1/4), and the SD is (1/4)*sqrt(250).

CB

**lllll** · December 10th 2008, 10:51 PM

If I wanted to calculate the probability that the experiment finishes in less then 56.4 minutes? The would it follow:

$\lambda e^{-\lambda t} \frac{(\lambda t)^{n-1}}{(n-1)!} = 4 e^{-4(56.4)} \frac{(4(56.4))^{250-1}}{(250-1)!}$

**mr fantastic** · December 11th 2008, 03:04 AM

Originally Posted by lllll

If I wanted to calculate the probability that the experiment finishes in less then 56.4 minutes? The would it follow:

$\lambda e^{-\lambda t} \frac{(\lambda t)^{n-1}}{(n-1)!} = 4 e^{-4(56.4)} \frac{(4(56.4))^{250-1}}{(250-1)!}$

No!

You need to know the distribution of the sum of the 250 independent exponential random variables each with parameter $\lambda = 4$ (since the mean is 1/4).

The following is a well known result (and not too difficult to prove):

Let $X_1, \, X_2, \, .... \, X_n$ be independent exponential random variables with parameter $\lambda$ . Then $Y = X_1 + X_2 + \, .... \, + X_n$ has a gamma distribution with pdf $f(y) = \frac{y^{n-1} e^{-y/\lambda}}{\Gamma(n) \lambda^n} = \frac{y^{n-1} e^{-y/\lambda}}{(n-1)!\, \lambda^n}$ .

In your case $\lambda = 4$ and $n = 250$ .

Use the pdf to calculate $\Pr(Y < 56.4)$ .

Note: The Central Limit Theorem is valid here.

**Laurent** · December 11th 2008, 09:40 AM

Originally Posted by lllll

If I wanted to calculate the probability that the experiment finishes in less then 56.4 minutes?

If you prefer to use Poisson random variables, you can also say that what you're looking for is:

$P(X_1+\cdots+X_{250}\leq 56.4)=P(N_{56.4}\geq 250),$

where the $X_i$ are those of mr fantastic, and $N_t$ is the number of points in the Poisson process before $t$ . This is just a dual way to express the probability.

For any $t$ , $N_t$ is distributed according to a Poisson distribution of parameter $4t$ .

So, the probability is $P(N_{56.4}\geq 250)=1-P(N_{56.4}<250)=1-\sum_{k=0}^{249} e^{-\lambda} \frac{\lambda^k}{k!}$ where $\lambda=4\times 56.4$ .

However, this is not so easy to compute. Like mr fantastic pointed out, a CLT would provide a simpler yet approximate answer.

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