- #1
Thaizasaskand
- 1
- 0
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!
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!
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