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poisson distribution

Hello there,
Im just looking at a question from a past paper based on the poisson distribution.

"The discrete random variable X has a truncated (0 excluded) Poisson distribution with probability distribution:

P(X = x) = c*(lambda^x)*(e^-lambda) / x ! , x = 1,2,.., infinity (lambda >0) .

Determine :

(a) the constant c "


I know have the answer and it is c = 1 / (1 - e^-lambda) .

But how on earth did they derive that ???

If you could explain I'd be very appreciative. THank you for your time .
Reply 1
You have to show that

x=1\sum_{x=1}^\infty lambda^x e^(-lambda) / x! = 1 - e^(-lambda)

which you can do using

x=0\sum_{x=0}^\infty lambda^x / x! = e^(lambda)

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