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Propability function Question

  1. Dec 11, 2005 #1
    Hi Guys
    I have Propability function that has caused me some trouble.
    X is a stochastic variable which is Poisson distributed with the parameter
    [tex] \lambda > 0[/tex]
    The Propability function is therefore:
    [tex]
    P(X=x) = \left\{ \begin{array}{ll}
    \frac{{e^{- \lambda}{\lambda ^{x}}}}{{x!}} & \textrm{where} \ x \in (0,1,2,\ldots)&\\
    0 & \textrm{other.}&\\
    \end{array} \right.
    [/tex]
    I'm suppose to show
    [tex]P(X \geq 1) = 1 - e^{- \lambda}[/tex]
    (step1) I get by inserting into the top formula
    [tex]P(X=1) = \lambda e ^ {- \lambda}[/tex]
    My question is how do go from P(X=1) to [tex] P(X \geq 1) [/tex] ???
    Sincerley
    Fred
     
    Last edited: Dec 11, 2005
  2. jcsd
  3. Dec 11, 2005 #2
    G'day, Fred.

    [tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
     
  4. Dec 11, 2005 #3
    Thanks Mate,

    I have second question

    If Lambda =1 then [tex]P(X \leq 2) = \frac{c e^{-1}}{2}[/tex]

    where c = 5

    Can I show that in a simular way like the first?

    Best Regards
    /Fred

     
    Last edited: Dec 11, 2005
  5. Dec 11, 2005 #4
    It's similar in that

    [tex]P(X\leq2) = P(X=0) + P(X=1) + P(X=2)[/tex]

    Apply the formula for each term and add the fractions (as e-1= 1/e).
     
  6. Dec 11, 2005 #5
    Okay thank You again

    This function which now is a to-dimension discrete stochastic vector has the probability function [tex]p_{X,Y}[/tex]

    [tex]P(X=x,Y=y) = \left\{ \begin{array}{ll}\frac{{c e^{- \lambda}{\lambda ^{x}}}}{{x!}} & \textrm{where} \ x \in (-2,-1,0,1) \ \textrm{and} \ \ y \in (0,1,\ldots)&\\
    0 & \textrm{other.}&\\\end{array} \right.[/tex]

    My question is that support [tex]supp \ p_{X,Y} = (-2,-1,0,1)[/tex]???

    Best regards
    Fred
     
    Last edited: Dec 11, 2005
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