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Prob/Stats Exponential and Poisson distributions or processes

  1. Oct 20, 2016 #1
    hi everyone initially I really want to put into words that there is absolutely no source related to following probability in poisson process and distribution $$P(S^1_A<S^1_B<S^1_C)$$ or $$P(S^n_A<S^m_B<S^k_C)$$ where $$S^1_A = \text{first arrival of A event}, S^1_B= \text{first arrival of B event}, S^1_C=\text{ first arrival of C event}$$ I could not carve any book or file out in internet. For instance I want to have some practice on this kind of probability question, but no source or book or pdf files exist. How can I improve myself????? I only have the book of Sheldon Ross, but he only wrote the formula and derivation of $$P(S^1_A<S^1_B)$$. Yes, the formula of $$P(S^n_A<S^m_B)$$ is also written, but that's it. I do not want to memorise the formulas, I am so eager to penetrate into the derivation of the probabilities that I mentioned at the beginning, this is the way how I learn most of things. In a nutshell, I think you have thrived on probability area, and that is why I am asking you some suggestions, books,pdf....
    Could you help me about that?????????? I will be so glad to have your responds, and I am looking forward to your valuable suggestions, and help......
     
  2. jcsd
  3. Oct 20, 2016 #2

    Simon Bridge

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    If you want to derive the general formulas for probabilities, then you should set about deriving them.
    One approach to understanding the probabilities is to write out in English what the notation is saying.
     
  4. Oct 20, 2016 #3
    @Simon Bridge ıs there a document or file or book which can do this stuff??? Because the things you say might be implemented or included in very nice book or document. Also in order to derive well I have to see or observe some examples related to the probabilities I mentioned don't I ??????? Don't you have any source that can satisfy my needs in this realm???????
     
  5. Oct 20, 2016 #4

    Simon Bridge

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    "This stuff"? You want a text on probability theory and your wits.
    If you won't follow advise I cannot help you. Good luck.
     
  6. Oct 20, 2016 #5
    I have sheldon ross books, but not satisfactory for me I am eager to go into deep well, so IS THERE ANY PROBABİLİTY BOOKS YOU MİGHT SUGGEST??? I will appreciate for your returns......
     
  7. Oct 20, 2016 #6

    Ray Vickson

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    Certainly the Ross book has all the material you need. One simple way to proceed is to use a "conditioning" argument, where you condition on the value of ##S_A##, for example; here I will write ##S_A## instead of ##S^1_A##, etc.. Can you figure out how to get ##P(S_C > S_B > t|S_A = t) = P(S_C > S_B > t)##? Alternatively, you can express the probability you want as a three-dimensional integral over a certain region in ##(t_1,t_2,t_3)##-space.
     
  8. Oct 21, 2016 #7

    Ray Vickson

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    I have already made two suggestions for how you can proceed, but so far you have presented no evidence of any effort. Therefore, I will not attempt to help you further.

    By the way: I will not enter into a "conversation" about this topic; if you have requests you should make them right here, in the forum.
     
  9. Oct 22, 2016 #8
    @Ray Vickson , here my work it is I did what you suggested, still no proper result........What is the mistake here???
     

    Attached Files:

  10. Oct 22, 2016 #9

    Ray Vickson

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    Your attachment is hard to read, but your final answer is correct. Why do you think it is wrong?

    In future, please avoid attachment word documents that are photocopies of your work; just attaching jpeg files is probably easier and better. The best solution of all is to type out your work using either plain text or LaTeX. For example, in plain text you can write the LaTeX expression ##\int_a^{\infty} ae^{-ax} \int_x^{\infty} b e^{-by} \, dy \, dx## as int_{0..inf} a e^(-ax) int_{x..inf} b e^(-by) dy dx.
     
  11. Oct 22, 2016 #10
    @Ray Vickson thanks for your return, I thought it was wrong due to fact that I saw different answer on a internet site while I was digging something related to my problem, there was no derivation of it on that site but just wrote down the answer differs from actual. But you know better of course that site actually not as good as here :) By the way what is your suggestion for the case of $$P(S^n_A<S^m_B<S^k_C)$$ What do we have to do here ???????
     
  12. Oct 22, 2016 #11

    Ray Vickson

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    I will write ##a,b,c## instead of ##\lambda_A, \lambda_B, \lambda_C##. Also, I will write ##A_n, B_m, C_k## instead of ##S^n_A, S^m_B, S^k_C##. Note that ##A_n## has distribution ##E_n(a)## = ##n##-Erlang with rate parameter ##a##. Its density is
    $$f_{A_n}(t) = f_n(t,a) \equiv \frac{a^n t^{n-1}}{(n-1)!} e^{-at} \; 1_{\{ t > 0 \}} $$.
    Similarly, ##B_m## has ##m##-Erlang density ##f_m(t,b)## and ##C_k## has density ##f_k(t,c)##. Thus
    $$P(A_n < B_m < C_k) = \int_{t_1=0}^{\infty} d t_1 \,f_n(t_1,a) \int_{t_2=t_1}^{\infty} dt_2 \, f_m(t_2,b) \int_{t_3=t_2}^{\infty} dt_3 \, f_k(t_3,c)$$.

    This can be evaluated by repeated application of the following result. If ##p_i(\mu) = \mu^i e^{-\mu}/i!## is a Poisson probability, then
    $$\int_t^{\infty} f_L(x,r) \, dx = \sum_{i=0}^{L-1} p_i(rt)$$.
    You can get this in two ways:

    Method (1): Direct--using repeated integration-by-parts, starting with ##u = x^{L-1}, dv = e^{-rx} \, dx##, etc., so that the integral with ##x^{L-1}## in it is expressed in terms of an integral with ##x^{L-2}##, then integrate-by-parts again to get ##x^{L-3}##, etc. Eventually you will arrive at the previous formula.

    Method (2): "Probabilistic" (more clever and a lot easier): The event that the ##L##th arrival is later than ##t## is the same as the event that the number of arrivals before time ##t## is ## \leq L-1##. That is, ##P(S_L > t) = P(N[0,t] \leq L-1)##, and for a Poisson process with rate ##r## we have ##P(N[0,t] \leq L-1) = \sum_{i=0}^{L-1} p_i(rt)##. That gives the same formula.

    So
    $$f_m(t_2,b) \int_{t_3=t_2}^{\infty} f_k(t_3,c) dt_3 = \sum_{i=0}^{k-1} f_m(t_2,b) (c t_2)^i e^{-c t_2}/ i! \\= \sum_{i=0}^{k-1} \frac{b^m c^i}{(m-1)! i!} t_2^{m+i-1} e^{-(b+c) t_2}$$

    By the way, if you were to integrate this with respect to ##t_2## from ##t_2 = 0## to ##t_2 = +\infty## you would have the expression for ##P(B_m < C_k)##, and if you finish the algebra you will get the formula you previously cited for that probability.

    Anyway, we now have
    $$\int_{t_2=t_1}^{\infty}dt_2\, f_m(t_2,b) \int_{t_3=t_2}^{\infty} f_k(t_3,c) dt_3 = \sum_{j=0}^{k-1} \frac{b^m c^i}{i! (m-1)!}
    \int_{t_1}^{\infty} t_2^{m-1+i} e^{-(b+c) t_2} \, dt_2$$
    The integrand in the ##i##th term is ##[(m-1+i)!/(b+c)^{m+i}] f_{m+i}(t_2,b+c)##, so the ##t_2##-integral can be expressed as
    $$\int_{t_1}^{\infty} t_2^{m-1+i} e^{-(b+c) t_2} \, dt_2 = \frac{(m-1+i)!}{(b+c)^{m+i} }\sum_{j=0}^{m+i-1} p_j((b+c)t_1).$$

    So, finally when we multiply this by ##f_n(t_1,a)## and integrate from ##t_1=0## to ##t_1 = +\infty## we will have an answer that is a somewhat complicated double sum of the form ##\sum_{j=0}^{m+i-1} \sum_{i=0}^{k-1} ( \cdots ) ##; you can fill in the details. I don't know offhand if the double sum can be simplified (using some other known formulas) but you can work on that if you wish.

    Note: when we integrate from ##t_1=0## to ##+\infty## we do not obtain a third sum, because when we integrate an Erlang density from 0 to ##\infty## we obtain 1; we would obtain another sum only when the lower limit of integration is ##> 0##.
     
    Last edited: Oct 22, 2016
  13. Oct 24, 2016 #12
    @Ray Vickson Thanks for your nice and remarkable post :D
     
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