Proving the poisson distribution is normalized

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SUMMARY

The discussion focuses on proving that the Poisson distribution is normalized, specifically addressing the equation P(n) = (μ^n)/n! * e^(-μ). The user correctly identifies that the sum of probabilities from n=0 to ∞ must equal 1, leading to the expression Ʃ P(n) = e^(-μ) * Ʃ (μ^n)/n!. The key step involves recognizing that the series Ʃ (μ^n)/n! converges to e^μ, thereby confirming the normalization of the Poisson distribution.

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Homework Statement



I am trying to prove that the poisson distribution is normalized, I think I've got an ok start but just having trouble with the next step.


Homework Equations



A counting experiment where the probability of observing n events (0≤n<∞) is given by:


P(n) = (μ^n)/n! * e^(-μ)

Where μ is a real number.


The Attempt at a Solution



Background (possibly incorrect)

So it's discrete, as n will take integer values; I need a sum not an integral.

Ʃ P(n) from n=0 to ∞ is just given by:

Ʃ (μ^n)/n! * e^(-μ)

And e^(-μ) does not vary with n, so:

Ʃ P(n) = e^(-μ) * Ʃ (μ^n)/n!

Important bit

Now as I am trying to prove it is normalized, i need to get Ʃ P(n) = 1, so I assume my problem is getting from:

(μ^n) / n!

to

e^μ


Any tips or help would be much appreciated, thanks in advance.
 
Last edited:
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You are doing probability applications but have never seen this material before? Oh, well: see, eg., http://www.mcs.sdsmt.edu/tkowalsk/notes/Common-Taylor-series.pdf .

RGV
 
Last edited by a moderator:
Yeah it's come up in my 2nd year Quantum module. Thanks for the link, I think I'd just forgotten that, curse of the double gap year.

Makes sense now, thanks again.
 

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