What is the name and application of this probability distribution

Lurco
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Hi.

In my homework I've encountered a discrete probability distribution of this form:

f(k,\lambda)=N \frac{\lambda^k}{k!}

k is the variable, and \lambda is a parameter. I'm curious what is this distribution - what's its name and where can it be applied. I will be grateful for, for example, redirecting me to the proper wikipedia article. Thanks!
 
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Lurco said:
Hi.

In my homework I've encountered a discrete probability distribution of this form:

f(k;\lambda) = \frac{\lambda ^{k} e^{\lambda}}{k!}

k is the variable, and \lambda is a parameter. I'm curious what is this distribution - what's its name and where can it be applied. I will be grateful for, for example, redirecting me to the proper wikipedia article. Thanks!

Are you sure you copied the formula correctly? The Poisson distribution is defined by:

f(k;\lambda) = \frac{\lambda ^{k} e^{-\lambda}}{k!}

where \lambda is the rate parameter (expected number of events per unit time), and k is the number of events observed.

In evaluating Poisson noise the question becomes P(k=N_t) but your formula still doesn't look right since it lacks the exponential term.
 
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I think the number N here is used as a normlization factor.

SW VandeCarr said:
Are you sure you copied the formula correctly? The Poisson distribution is defined by:

f(k;\lambda) = \frac{\lambda ^{k} e^{-k}}{k!}

where \lambda is the rate parameter (expected number of events per unit time), and k is the number of events observed.

In evaluating Poisson noise the question becomes P(k=N_t) but your formula still doesn't look right since it lacks the exponential term.
 
shuxue1985 said:
I think the number N here is used as a normlization factor.

[EDIT]: After reading the wiki page, yes the value depends on lambda not k.

Can't believe I've used this pdf so many times and forgotten it!
 
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Thank you all for responding. Yes, the number N stand for the normalization constant, and in the wikipedia article posted by micromass the exponent is exactly the normalization:
e^{-\lambda},

so i does not vary with k.
 
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