# Poisson process and exponential distribution arrival times

• lotsofmoxie
In summary: U/2, say), or a 2 that arises purely from algebra, through squaring or something.No, I was doing the correct problem (d) but I lost track of how I got the 1-1/e^20 result. It was because I used the wrong intial conditions on the integral which I am aware of now.I'm not sure I agree with your last statement because I didn't use numbers till the last step. I used symbols for everything else and I'm just plugging in the values at
lotsofmoxie

## Homework Statement

Customers arrive in single server queue to be serviced according to Poisson process with intensity 5 customers an hour.
(a) If the customers begin to arrive at 8am, find the probability that at least 4 customers arrived between 9am and 10am.
(b) Find the probability that the 6th customer will arrive after 12 noon.
(c) Find the variance of the time of the arrival of the 3rd customer.

If the service times of the customers are independent uniform r.v.’s on [0, 4] and also are independent of the interarrival times:
(d) Find the probability that the second customer to arrive will have to wait before being served.
(e) Find the expected value of the time that the second customer will have to wait before being served.

## Homework Equations

P(x=x) = (λt)^(x)*e^(-λt)/x!

where lambda = poisson parameter and t=ime interval

[/B]

## The Attempt at a Solution

a) P(at least 4 arrivals between 9-10am)=1-P(0,1,2 or 3 arrivals between 9-10am)
=1-e^(-5*1)(5*1)^(0)/0!+(5*1)^(1)/1!+(5*1)^(2)/2!+(5*1)^(3)/3!)
=0.7349

b)
P(T6>4) = P(number of arrivals between 8-12 is exactly 0, 1, 2, 3, 4, or 5)
= e^(-5*4)*(5*4)^(0)/0!+(5*4)^(1)/1!+(5*4)^(2)/2!+(5*4)^(3)/3!+...)
=0.0000719

c) The 3rd arrival time is the sum of the first three individual arrival times. These are all distributed exponentially with Var=1/λ^2=1/25. Since the three random variables are independent, we can sum their variances to get Var(T3)=1/25+1/25+1/25.[/B]

d) Let U = time taken to serve 1st customer
T = time between arrival of 1st and second customer
f(U)=1/(4-0)=1/4 when 0<=u<=4
f(T) = 5e^-5t when t>0, 0 elsewhere
Because of independence, the joint probability density function f(U,T)=1/4*5e^(-5t)
customer has to wait ------> we want P(U>T)=1-P(T>U) because it is easier to integrate

∫ ∫1/4*5e^(-5t)dudt where inner integral is over (4, infinity) and outer is over (0,4)
e) same as d but with the integrand as multiplied by u-t since that is the waiting time

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especially on a, b, c I'm not sure if I'm doing it right (my answer for b seems intuitively too small).

lotsofmoxie said:
especially on a, b, c I'm not sure if I'm doing it right (my answer for b seems intuitively too small).

Your methods and answers to (a)-(c) are OK. You did not write down the complete, final solution to (d), so I cannot tell if it is OK or not. (The method you used is fine, but I cannot tell if you made any errors when doing the integrations.)

As for (b): it is small, but that is not too hard to fathom: the mean number of arrivals from 8 to noon is 5*4 = 20, with standard deviation = √20 ≈ 4.472. Your upper limit of 5 arrivals is a bit more than 3.5 standard deviations below the mean, so, the answer should be small. However, it is surprising to see exactly how small it is. Anyway, the answer you give is correct.

Thank you for responding. It makes sense now.

I'm getting 1-1/e^20 for d). Is this correct?

lotsofmoxie said:
Thank you for responding. It makes sense now.

I'm getting 1-1/e^20 for d). Is this correct?

That is not what I get. If you write out the details, step-by-step, I would be in a better position to comment.

Ray Vickson said:
That is not what I get. If you write out the details, step-by-step, I would be in a better position to comment.

Gah

Ok.

∫∫1/4*5e^(-5t)dtdu where inner integral is over (4, infinity) and outer is over (0,4)

Is this the integral you are starting with?

So starting with the inner integral ∫1/4*5e^(-5t)dt
pull out the 1/4 so we have 1/4 ∫5e^(-5t)dt over (4,oo)
use a substition w=-5t, dw=-5dt--->dt=dw/-5

so the integral becomes 1/4 ∫5e^(w)dw/-5 over (some interval that doesn't matter because I'll convert it back to t before using them)
=-1/4 ∫e^w dw since the 5 and 1/5 cancel and I pulled out the negative
=-1/4(e^(-5*infinity)-e^(-5*4))
=-1/4(0-e^(-20))

=e^(-20)/4

integrating this with respect to u over (0, 4) I get 1/e^20.

Subtracting this from 1 I get 1-1/e^20

where did I go wrong?

lotsofmoxie said:
Gah

Ok.

∫∫1/4*5e^(-5t)dtdu where inner integral is over (4, infinity) and outer is over (0,4)

Is this the integral you are starting with?

No. For ##f_T(t) = \lambda e^{-\lambda t} 1\{t > 0 \}## and ##f_U(u) = (1/U) 1\{ 0 \leq u \leq U \}## we have
$$EW = \int_0^U f_U(u) \, E(W|u) \, du, \; \text{where} \; E(W|u) = \int_0^u (u-t) f_T(t) \, dt$$
You can do the integral to get
$$E(W|u) = \frac{\lambda \, u - 1 + e^{- \lambda u}}{\lambda}$$
The integral ##\int_0^U E(W|u) \, du## is do-able. For ##\lambda = 5, U = 4## I finally get
$$EW = A - B e^{-20}$$
for two constants ##A,B## that I will let you work out for yourself. Numerically, I get ##EW \doteq 1.81## (actually 1.810000000 to 9 decimal places).

Wait, the problem you just explained was part e, right? I was trying to do d).

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lotsofmoxie said:
Wait, the problem you just explained was part e, right? I was trying to do d).

OK:
$$P(W=0) = \frac{1}{U} \int_0^U P(T > u) \, du = \frac{1}{U} \int_0^U e^{- \lambda u} \, du =\frac{1}{\lambda U} \left( 1 - e^{-\lambda U} \right).$$
Plug in ##U = 4, \lambda = 5## and see what you get.

I really think that part of your problem stems from using numerical values too soon. It would be better to carry out the computations symbolically as far as you can, then plug in numbers at the last minute. Doing that makes error-tracing much easier. Part of the problem with using numbers throughout is that you may, for example have a '2' somewhere in the calculation, and it is not always easy to tell whether that is something like a 4/2 from the probability distribution parameters (as U/2, say), or a 2 that arises purely from algebra, through squaring or something.

Ray Vickson said:
OK:
$$P(W=0) = \frac{1}{U} \int_0^U P(T > u) \, du = \frac{1}{U} \int_0^U e^{- \lambda u} \, du =\frac{1}{\lambda U} \left( 1 - e^{-\lambda U} \right).$$
Plug in ##U = 4, \lambda = 5## and see what you get.

I really think that part of your problem stems from using numerical values too soon. It would be better to carry out the computations symbolically as far as you can, then plug in numbers at the last minute. Doing that makes error-tracing much easier. Part of the problem with using numbers throughout is that you may, for example have a '2' somewhere in the calculation, and it is not always easy to tell whether that is something like a 4/2 from the probability distribution parameters (as U/2, say), or a 2 that arises purely from algebra, through squaring or something.

I got 0.04999999989. I don't see where I went wrong when I did the integral - it didn't seem too messy when I did it putting the numerical values in first, but for some reason it is wrong. I tried plugging it into wolfram alpha, and the answer it gave was the same as the one I calculated, so I'm thinking my integral ∫∫1/4*5e^(-5t)dtdu where inner integral is over (4, infinity) and outer is over (0,4) is set up wrong and does not actually give P(U<T) but I'm really not seeing why.

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****. I wrote my integral down messily and I mistook a u for a 4 in the bounds and never caught my mistake later. It should be ∫∫1/4*5e^(-5t)dtdu where inner integral is over (u, infinity) and outer is over (0,4), which gives the same answer as using your method above.

I really hate myself.

Thanks for all of your help! I really get it now.

## What is a Poisson process?

A Poisson process is a mathematical model used to describe the arrival of events over a continuous period of time. It is based on the assumption that the events occur independently and at a constant rate.

## What is the relationship between Poisson process and exponential distribution?

The arrival times in a Poisson process follow an exponential distribution. This means that the time between each event follows an exponential distribution with a certain rate parameter. The exponential distribution is often used to model the time between independent events.

## Can the arrival times in a Poisson process be predicted?

No, the arrival times in a Poisson process are random and cannot be predicted. However, the probability of a certain number of events occurring within a given time interval can be calculated using the Poisson distribution.

## What is the mean and variance of the arrival times in a Poisson process?

The mean and variance of the arrival times in a Poisson process are both equal to the rate parameter, denoted by λ. This means that the average time between events and the spread of the arrival times are both determined by the rate at which events occur.

## How is a Poisson process different from a normal distribution?

A Poisson process is different from a normal distribution in that the arrival times are discrete rather than continuous. Additionally, the arrival times in a Poisson process are not independent, unlike the observations in a normal distribution. Finally, the mean and variance of a Poisson process are equal, whereas in a normal distribution they can be different.

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