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HELP: Probability Question

  1. Apr 30, 2006 #1
    Hi People,

    I have this Probability Problem which is given me a headach,

    Given two independent Stochastic variables (X, Y) where X is Poisson distributed [tex]Po(\lambda)[/tex] and Y is Poisson distributed [tex]Po(\mu)[/tex].

    Where [tex]\lambda[/tex], [tex]\mu > 0[/tex]. Let [tex]m \geq 0[/tex] and [tex]p = \frac{\lambda}{\lambda+ \mu}[/tex]

    By the above I need to show, that

    [tex]P(X=l| X+Y=m) = \left( \begin{array}{cc} l \\ m \end{array} \right ) p^l (1-p)^{(l-m)}[/tex]

    Proof:

    Its know that

    [tex]\left( \begin{array}{cc} l \\ m \end{array} \right ) = \frac{l!}{m!(1-m)!}[/tex]

    Wherefore the Binormal formula can be written as

    [tex]\frac{l!}{(m!(l-m))!} p^l (1-p)^{(l-m)}[/tex]

    for [tex]m \geq 0[/tex] I get:


    [tex]\frac{-(\frac{\lambda + \mu}{\mu})^{l} \cdot p^{l} \cdot \mu}{l!(l!-1) \cdot (\lambda + \mu)}[/tex]

    Any surgestions on how to processed from here?

    Sincerely Yours
    Hummingbird25
     
    Last edited: Apr 30, 2006
  2. jcsd
  3. Apr 30, 2006 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, no suggestions on how to proceed from there since you seem to be going at it backwards.

    For any probability distribution, P(A and B)= P(A|B)*P(B) so that
    P(A|B)= P(A and B)/P(B).

    You are looking for P(X= l| X+ Y= m) so A is "X= l" and B is "X+ Y= m".
    Then A and B is "X= l and X+ Y= m" which is the same as "X= l and Y= m-l". Now P(A and B)= P(X= l)*P(Y= m) .
     
  4. Apr 30, 2006 #3
    Hi Hall,

    Hello since X and Y is Poisson distributed then

    [tex]P(x=l) * P(Y=m) = \frac{\lambda^l}{l !} e^{-\lambda} * \frac{\lambda^m}{m !} e^{-\mu}[/tex]

    Is this the next?

    Sincerley Yours
    Hummingbird25
     
    Last edited: Apr 30, 2006
  5. Apr 30, 2006 #4
    I think HallsofIvy made a small mistake in his last statement. He probably meant "P(A and B)= P(X= l)*P(Y= m-l)".
     
    Last edited: Apr 30, 2006
  6. Apr 30, 2006 #5
    Hello since X and Y is Poisson distributed then

    [tex]P(X=l) * P(Y=m-l) = \frac{\lambda^l}{l !} e^{-\lambda} *[/tex] (what do I need to add here then)?

    Is this the next?

    Sincerley Yours
    Hummingbird25

     
  7. Apr 30, 2006 #6
    [tex]P(X=l) * P(Y=m-l) = \frac{\lambda^l}{l !} e^{-\lambda} * \frac{\mu^{m-l}}{(m-l)!} e^{-\mu}[/tex]

    And Hummingbird25, can I request that you check the question again? I tried solving the question and my answer was [tex]P(X=l| X+Y=m) = \left( \begin{array}{cc} m \\ l \end{array} \right ) p^l (1-p)^{(m-l)}[/tex]. Perhaps there is something wrong with my working...
     
    Last edited: Apr 30, 2006
  8. Apr 30, 2006 #7
    Hello Pizzasky and thank You,

    I looked at my assigment again I and discovered, that You result was the right one, I had (by mistake swapped l and m.

    The right answer as You formulated:

    [tex]\left( \begin{array}{cc} m \\ l \end{array} \right ) p^l (1-p)^{(m-l)}[/tex]


    Sincerely Hummingbird25

     
  9. Apr 30, 2006 #8
    One last thing Pizzasky,

    Which formula did You use to achive the binormal formula?

    Sincerely Yours
    Hummingbird25

     
  10. Apr 30, 2006 #9
    Reply

    You can use the substitution [tex]p = \frac{\lambda}{\lambda+ \mu}[/tex] to get the Binomial formula. Also, try to figure out what expression "1-p" corresponds to.
     
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