# Math Help - Poisson process, I'm stuck

1. ## Poisson process, I'm stuck

------------------------------------------------------------------------
Landings at an airport are exponentially distributed with a mean of 1 plane per hour.

a) What is the probability that more than 3 airplanes land within an hour?
b) If 15 separate 1 hour intervals are chosen, what is the probability that no interval contains more than 3 landings.
c) What is the probability that exactly 2 planes land in 2 hours.
d) What is the probability that the waiting period for 3 arrivals will be more than 3 hours?
f) What is the mean and variance for the waiting time for 3 arrivals?
-------------------------------------------------------------------------

So I tried to solve this and failed. I started by getting the exponential distribution function:

e^(-x) which didn't give me the answer to a) so I turned towards the general expression:

e^(-1) / x!

This gave me an answer 0.019 that does not make sense...can someone be kind enough to walk me through this exercise? I would offer \$ via paypal but I think the forum doesn't allow it...lmk

Ty!

2. You may want to prove this statement first:
Let X_t be a process such that at time t, X_t waits at an integer i for an exponential (with parameter lambda) amount of time after which it moves to i+1. Then $\mathbb{P}(X_t=i)=e^{-\lambda t}\frac{(\lambda t)^k}{k!}$.

Then X_t would represent the amount of planes that arrive by time t (a Poisson process).

3. Originally Posted by Focus
You may want to prove this statement first:
Let X_t be a process such that at time t, X_t waits at an integer i for an exponential (with parameter lambda) amount of time after which it moves to i+1. Then $\mathbb{P}(X_t=i)=e^{-\lambda t}\frac{(\lambda t)^k}{k!}$.

Then X_t would represent the amount of planes that arrive by time t (a Poisson process).
I'm not asked anything like that, the answer is straight forward, I can just write the numerical answer and get the points. I'm just an engineer doing his math homework man, keep it simple

ty tho

4. I still can't find...how do you do this lol

5. Originally Posted by Provoke
I'm not asked anything like that, the answer is straight forward, I can just write the numerical answer and get the points. I'm just an engineer doing his math homework man, keep it simple

ty tho
In that case just use the formula $\mathbb{P}(X_t=n)=e^{-\lambda t}\frac{(t \lambda)^n}{n!}$.

a) You want $\mathbb{P}(X_1 \geq 3)=1-\mathbb{P}(X_1 \leq 2)$

b) If the intervals do not overlap, then they are independent, so first find $\mathbb{P}(X_1 \leq 3)$, then use independence (and stationary increments) to find the result. By that I mean if s_1<t_1<s_2<t_2, then $\mathbb{P}(X_{t_1}-X_{s_1},X_{t_2}-X_{s_2})=\mathbb{P}(X_{t_1-s_1})\mathbb{P}(X_{t_2-s_1})$

c) $\mathbb{P}(X_2=2)$

d),e) You need to look at some stopping times for this, let $\tau_n=\inf\{t \geq 0: X_t=n\}$. Now what is $\mathbb{P}(\tau_n \geq t)$? Well, that means by time t, you have had less than n arrivals.

I have no idea to which level you have done things, a little indications of which bits you don't understand would be nice.

6. Originally Posted by Focus
In that case just use the formula $\mathbb{P}(X_t=n)=e^{-\lambda t}\frac{(t \lambda)^n}{n!}$.

a) You want $\mathbb{P}(X_1 \geq 3)=1-\mathbb{P}(X_1 \leq 2)$

b) If the intervals do not overlap, then they are independent, so first find $\mathbb{P}(X_1 \leq 3)$, then use independence (and stationary increments) to find the result. By that I mean if s_1<t_1<s_2<t_2, then $\mathbb{P}(X_{t_1}-X_{s_1},X_{t_2}-X_{s_2})=\mathbb{P}(X_{t_1-s_1})\mathbb{P}(X_{t_2-s_1})$

c) $\mathbb{P}(X_2=2)$

d),e) You need to look at some stopping times for this, let $\tau_n=\inf\{t \geq 0: X_t=n\}$. Now what is $\mathbb{P}(\tau_n \geq t)$? Well, that means by time t, you have had less than n arrivals.

I have no idea to which level you have done things, a little indications of which bits you don't understand would be nice.
My statistics level: Knew nothing of statistics until 1 month ago. Shouldn't matter tho, I know the calculation is straight forward and takes 1 line, I just can't figure it out.

Thx a lot. I don't understand in the a) why its more or equal to 3, I think the problem sais it's greater than 3 so shouldn't the 3 not be included?
So according to the formula I get e(-1) / 4! which gives 0.01532831 probability, is that ok? Or should I do it e(-1) / 5!, e(-1) / 6!.....e(-1) / n! and add the results?

In other words, is this the probability that exactly 3 airplanes land in 1 hour: e(-1) / 4! which gives 0.01532831 ?
And If I wanted to know in 2 hours should I just change the t? That's what I try to work out atm.

thx

Edit: That doesn't add up to one so It can't be right...ok I give up...What's going on? I'm an engineer, I don't think in theory I think in numbers.

7. Originally Posted by Provoke
Thx a lot. I don't understand in the a) why its more or equal to 3, I think the problem sais it's greater than 3 so shouldn't the 3 not be included?
Yes you are probably right. I do have a habit of assuming less than or equal to when people say more.
So according to the formula I get e(-1) / 4! which gives 0.01532831 probability, is that ok? Or should I do it e(-1) / 5!, e(-1) / 6!.....e(-1) / n! and add the results?

In other words, is this the probability that exactly 3 airplanes land in 1 hour: e(-1) / 4! which gives 0.01532831 ?
And If I wanted to know in 2 hours should I just change the t? That's what I try to work out atm.
e(-1) / 4! is the probability that exactly 4 planes land in 1 hour. You change the t to t=2 if you want to work it out for 2 hours.

If you want to know the probability of n or less planes arriving within the hour then you sum $e^{-1}\sum_{i=0}^n \frac{1}{i!}$.

For a) what you want is the infinite sum
$e^{-1}\left (\frac{1}{4!}+\frac{1}{5!}+...\right)$ (*)
which is not really nice to compute, but you know (if you didn't, you do now) that
$e^{-1}\left(1+\frac{1}{2!}+\frac{1}{3!}+...\right)=1$ (**)

What is (**)-(*)? You should be able to compute that quantity, then obtain (*).

8. Shlt I got it lol thx. I'll pay you a beer if you ever come to Canada...which you wouldn't want to do right now unless you want the fuel to freeze inside your tank. Prolly in the summer haha

9. No worries. Canada is a little too far for me.