# I N events occur in one Poisson process before m events

1. Oct 14, 2016

### mertcan

guys, I have a very ımportant question. First let me introduce parameters: $$S^A_1 = \text{first arrival of A event}, and S^B_1= \text{first arrival of B event}, and S^C_1=\text{ first arrival of C event}$$, then probability of $$P(S^A_1<S^B_1) = \frac {\lambda_A} {\{ \lambda_A + \lambda_B \} }$$, and I know how to derive this, also I know how to derive $$P(S^A_2<S^B_5)$$ BUT what is the probability of $$P(S^A_1<S^B_1<S^C_1)$$ ???? AND ALSO what is the probability of $$P(S^A_2<S^B_4<S^C_6)$$ ??? ıf you know could you show me the derivation of that probabilities????
I really endeavour to dig lots of things related to this situation out of internet, but I have not found worthy....

Last edited: Oct 14, 2016
2. Oct 14, 2016

### andrewkirk

You have not stated, but I assume it to be implied, that the processes for A, B and C are independent.

How did you derive $P(S_1^A<S_1^B)$? The way I know involves integration of a bivariate distribution over a region of the number plane, where the pdf of the distribution is the product of the pdfs of two exponential distributions, as arrival times for a Poisson process have an exponential distribution.

Given that, the derivation of $P(S_1^A<S_1^B<S_1^C)$ will involve a triple integral of a pdf that is the product of three independent exponential distributions. The main challenge will be to set the correct upper and lower limits for each integral. But that should not be hard.

3. Oct 15, 2016

### mertcan

Initially, sorry for not saying these are independent random exponential variables. By the way I have found a answer but I do not know if it is true becasue no source exist pertain to this kind of topic.
My answer is $$\frac {\lambda_A*\lambda_B} { \{ \lambda_A + \lambda_B+\lambda_C \} }$$.
If it is not right could you show me the steps which lead to right answer ?????

4. Oct 15, 2016

### andrewkirk

No it's not correct. You can see that by dimensional analysis. The units of each $\lambda$ are events per second. A probability needs to be unitless. Your formula in the OP is unitless as it has units of (events per sec) / (events per sec). But your formula in post 3 has units of
(events per sec) * (events per sec) / (events per sec) = (events per sec), which is not unitless.

As I said in post two, you need to do a triple integral. Start by posting the double integral you used to get the formula in the OP, and see if you can work out how to adapt it to the probability involving three random variables. If you can't work out how to adapt it, hints should be available.

5. Oct 15, 2016

### mertcan

Andrew thanks for your return, but I would like to express that I have so many exams next week and I really got stuck in this question, I can not focus on other exam topics, also I tried to do the same thing I had done for 2 random variables, but obtained false answer as you see. Therefore, would you mind proving this probability result ?????? or if it is available you can upload some pdf files related to this derivation....

6. Oct 18, 2016

### mertcan

Hi, you say that I have a wrong result , and I would like to share my work for the derivation of $$P(S^A_1<S^B_1<S^C_1)$$ And I really ask you to show me the true solution or true derivation of this probability, and show me where I have a mistake in my work in attachment thus I will be so glad to take your help

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7. Oct 18, 2016

### andrewkirk

The proof proceeds in the wrong way. You need to start with the probability density $f_{\langle S_A^1,S_B^1,S_C^1\rangle}$ of the multivariate random variable $\langle S_A^1,S_B^1,S_C^1\rangle$ and then replace that by the individual probability density functions $f_{S_A^1}$ etc, using an assumption that the three random variables are independent - which you have not stated, by the way.

The probability you seek is

$$Prob(S_A^1<S_B^1<S_C^1)=\int_a^b\int_c^d\int_e^f f_{\langle S_A^1,S_B^1,S_C^1\rangle}(x,y,z) dz\,dy\,dx$$

where the integration limits $a,b,c,d,e,f$ are chosen so that the volume of integration is exactly that part of $\mathbb R^3$ for which $x<y<z$. Note that, in this integration, the integration variables $x,y,z$ correspond to the realised values of random variables $S_A^1,S_B^1,S_C^1$ respectively.

What do the values of $a,b,c,d,e,f$ need to be, to get the correct region of integration?

8. Oct 19, 2016

### mertcan

In my opinion a=0,b remains same but c= b, d=e and f= infinite, or may be a=0, b= infinite, c=b, d=infinite, e=d, f=infinite, ublackff actually I really confused...... I think I need more mathematical demonstration about this probability. Could you please do this??? I really get crazy by the way ..., I can not believe myself that I can not solve this probability but can solve $$P(S^1_A<S^1_B)$$ Actually would you mind providing me with total derivation of this probability $$P(S^1_A<S^1_B<S^1_C)$$ ??? I think ıf I see the steps I can totally comprehend the structure of these kind of probabilities...

Last edited: Oct 19, 2016
9. Oct 19, 2016

### haruspex

I don't think any integration is necessary. Just use the fact that Poisson processes have no memory.
What is the probability that the first event is an A? With that event out of the way, what is the probability that the next is a B?

10. Oct 20, 2016

### mertcan

@haruspex @andrewkirk Actually this method also in my mind, but you know that memoryless is proved using condition. I do not want to be intuitive for the result(when I have intuition I can satisfy myself easily but can not write what kind of intuition I have down on paper :| ), I want to see how it is derived, but when I tried to use some condition in probabilites that I asked I can not attain proper result. In short, I cannot solving it using condition as we use in memoryless property. So, I need and really asking you to solve the probability I asked. I am sure your time is valuable, but I will be so glad and appreciate if you can spare your time....

11. Oct 20, 2016

### haruspex

I regard that as backwards. The supposition that a process is memoryless is the very basis for determining that it is a Poisson process. It is a perfectly rigorous approach to the problem.

12. Oct 21, 2016

### mertcan

Hi again :D here, I would like to express $$S^1_n = \text{n^th arrival of event 1}, S^2_m= \text{m^th arrival of event 2}$$ and I really want to know is there a derivation of this equation?????? How do we attain such a result????