1. ## Poisson Process Question

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 $\displaystyle \frac{250}{4}=62.5$

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

are these correct?

2. 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 $\displaystyle \frac{250}{4}=62.5$

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

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

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

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

4. 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:

$\displaystyle \lambda e^{-\lambda t} \frac{(\lambda t)^{n-1}}{(n-1)!} = 4 e^{-4(56.4)} \frac{(4(56.4))^{250-1}}{(250-1)!}$
No!

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

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

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

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

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

Note: The Central Limit Theorem is valid here.

5. 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:

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

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

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

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

However, this is not so easy to compute. Like mr fantastic pointed out, a CLT would provide a simpler yet approximate answer.
The text passage contains a series of questions and answers related to a Poisson process, specifically dealing with the mean and variance of the time to observe 250 outcomes, and the probability that the experiment finishes in less than 56.4 minutes.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
1. ## Poisson Process Question

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 $\displaystyle \frac{250}{4}=62.5$

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

are these correct?

2. 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 $\displaystyle \frac{250}{4}=62.5$

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

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

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

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

4. 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:

$\displaystyle \lambda e^{-\lambda t} \frac{(\lambda t)^{n-1}}{(n-1)!} = 4 e^{-4(56.4)} \frac{(4(56.4))^{250-1}}{(250-1)!}$
No!

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

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

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

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

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

Note: The Central Limit Theorem is valid here.

5. 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:

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

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

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

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

However, this is not so easy to compute. Like mr fantastic pointed out, a CLT would provide a simpler yet approximate answer.
user
The text passage contains a series of questions and answers related to a Poisson process, specifically dealing with the mean and variance of the time to observe 250 outcomes, and the probability that the experiment finishes in less than 56.4 minutes.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS