Poisson distribution Question (1 Viewer)

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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:

HallsofIvy

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Bob19 said:
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]\frac{e^{- \lambda} \cdot \lambda^{x}}{x!} = e^{- \lambda} \sum _{x=0} ^{\infty} \frac{\lambda ^{x}}{x!} = e^{- \lambda} e^{\lambda} = 1[/tex]
Then if [tex] \lambda ^{x} = 1[/tex]
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]
 
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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:

HallsofIvy

Science Advisor
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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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