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Poisson distribution Question

  1. Nov 28, 2005 #1
    Hello I'm Presented with the following Poisson distribution question

    [tex]P(X = x) = \frac{e^{-\lambda} \cdot \lambda^{x}}{x!}[/tex]

    where [tex]x \in (1,2,3,\ldots)[/tex] and [tex]\lambda > 0[/tex]

    Then I'm suppose to show that the above can be re-written if

    [tex]P(X \leq 1) = 1 - e^{- \lambda}[/tex]

    Any idears on how I do that?

    I'm told [tex]\sum_{x=0} ^ {\infty} p(x) = \frac{e^{- \lambda} \cdot \lambda^{x}}{x!} = e^{- \lambda} \sum _{x=0} ^{\infty} \frac{\lambda ^{x}}{x!}[/tex]

    if [tex] \lambda ^{x} = 1[/tex]

    then [tex] p(x) = e^{- \lambda} [/tex]

    This must give [tex]P(X \leq 1) = 1- e^{- \lambda}[/tex]

    Can anybody please tell me if I'm on the right track here?

    Sincerley Bob
     
    Last edited: Nov 28, 2005
  2. jcsd
  3. Nov 28, 2005 #2

    HallsofIvy

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    No, you are not told that! Obviously
    [tex]\frac{e^{- \lambda} \cdot \lambda^{x}}{x!}[/tex]
    ,for specific x, is not the same as sum for all x:
    [tex]e^{- \lambda} \sum _{x=0} ^{\infty} \frac{\lambda ^{x}}{x!}[/tex]
    What you are told is that the sum for all x is equal to 1: so that this is a valid probability distribution.
    Since x can only take on positive integer values, [itex]P(x \leq 1)[/itex] is exactly the same as [itex]P(x= 1)[/itex] which is
    [tex]\frac{e^{-\lambda}\lambda^1}{1!}= \lambda e^{-\lambda}[/tex]
    Assuming, as you say, that that is [itex]1- e^{-\lambda}[/itex], then you are not told that [itex]e^{\lambda}= 1[/itex]. You are told, rather, that [itex]e^{\lamba}= 1+ \lambda[/itex] so that
    P(X= x) can be written as
    [tex]P(X= x)= \frac{e^{-\lamba}\lamba^x}{x!}= \frac{\lamba^x}{(1+\lambda)x!}[/tex]
     
    Last edited: Nov 28, 2005
  4. Nov 28, 2005 #3
    Hello Hall and Thank You,

    x = 1

    then [tex]\frac{e^{-\lamba}\lamba^1}{1!}= \frac{\lamba^1}{(1+ 1)1!} = e^{- \lambda} [/tex]

    Which is then [tex]P(x \geq 1) = 1- e^{- \lambda}[/tex]

    Am I on the right track now ?

    Sincerely
    Bob
     
    Last edited: Nov 28, 2005
  5. Nov 28, 2005 #4

    HallsofIvy

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    I don't know why the Latex isn't showing properly. What I was trying to say was that IF P(X<= 1)= P(X= 1)= lambda e-lambda is equal to 1- e-lambda, then multiplying both sides of the equation by elamba we have lambda= elambda- 1 or elambda= lambda + 1.

    Now to "rewrite" P(x)= (lambdaxe-lambda/x!, just replace that e-lambda bu 1/(lambda+ 1), getting P(x)= lambdax/(x!(lambda+ 1)).
     
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