Probability through Poisson Distributon

[Thaizasaskand]

Hi!

I am aware of the steps used to show that (e^-λ*λ^r)/r! is P(X=r) for X~Po(λ), where λ = E(X) = Var(X). I have two questions regarding this:

- I'm aware that all of the probabilities add up to 1, but how do we know that they're all probabilities and not just a set of values that add to 1?
- Why e specifically? I do actually remember hearing this before, but I've since forgotten.

As I'm sure that it's probably helpful to post how to reach P(X=r) = (e^-λ*λ^r)/r! anyway, I'll post it:


e^λ =
∞
∑ (λ^r)/r! = λ^0 + (λ^1)/1! + (λ^2)/2! + (λ^3)/3! + ... + (λ^r)/r! + ...
r=0

Dividing both sides by e^λ gives:

1 = e^-λ * λ^0 + (e^-λ * λ^1)/1! + (e^-λ * λ^2)/2! + (e^-λ * λ^3)/3! + ... + (e^-λ * λ^r)/r! + ...

As such, the probability function is given through the previously stated formula.

If my communication is unclear, I apologise - it has been hotter than usual in SE England this week and I'm not so sharp as a result.

Thanks for your time!

 

[LCKurtz]

Hi!

I am aware of the steps used to show that (e^-λ*λ^r)/r! is P(X=r) for X~Po(λ), where λ = E(X) = Var(X). I have two questions regarding this:

- I'm aware that all of the probabilities add up to 1, but how do we know that they're all probabilities and not just a set of values that add to 1?

They are the same thing in the sense that a set of non-negative numbers adding to 1 can be used to define a discrete probability function.

- Why e specifically? I do actually remember hearing this before, but I've since forgotten.

As I'm sure that it's probably helpful to post how to reach P(X=r) = (e^-λ*λ^r)/r! anyway, I'll post it:


e^λ =
∞
∑ (λ^r)/r! = λ^0 + (λ^1)/1! + (λ^2)/2! + (λ^3)/3! + ... + (λ^r)/r! + ...
r=0

Dividing both sides by e^λ gives:

1 = e^-λ * λ^0 + (e^-λ * λ^1)/1! + (e^-λ * λ^2)/2! + (e^-λ * λ^3)/3! + ... + (e^-λ * λ^r)/r! + ...

As such, the probability function is given through the previously stated formula.

If my communication is unclear, I apologise - it has been hotter than usual in SE England this week and I'm not so sharp as a result.

Thanks for your time!

The Poisson distribution is used to model memoryless waiting times. If you make that assumption, the exponential function enters automatically. The discussion at this link might be helpful:

https://www.google.com/url?sa=t&rct...8oKgCQ&usg=AFQjCNFexUgy7SW72RieBJiFqQRH32Tr9Q