I have this probability distribution:
$$P(\{X=x\})=\frac{\lambda^x\exp(-\lambda)}{x!(1-\exp(-\lambda))}$$
where $x\in \mathbb{N}$. If not for this factor $\frac{1}{1-\exp(-\lambda)}$, this would be Poisson distribution. What is its name?
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3$\begingroup$ For what values of $x$? Are you sure the probabilities sum to $1$? $\endgroup$– Juho KokkalaDec 14, 2015 at 16:48
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1$\begingroup$ Hint: what values can $x$ take? $\endgroup$– Scortchi - Reinstate Monica ♦Dec 14, 2015 at 16:48
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$\begingroup$ @JuhoKokkala: Snap! $\endgroup$– Scortchi - Reinstate Monica ♦Dec 14, 2015 at 16:48
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4$\begingroup$ You must specify the range of $x$! Without that, the definition is not precise. We cannot, for instance, decide if the probabilities sum to 1 or not. $\endgroup$– kjetil b halvorsen ♦Dec 14, 2015 at 16:49
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4$\begingroup$ Where does $\mathbb{N}$ start for you? Nought or one? $\endgroup$– Scortchi - Reinstate Monica ♦Dec 14, 2015 at 16:50
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1 Answer
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I have found the name. It's called "Zero-truncated Poisson distribution".
Source: Johnson L., Kemp A., Kotz S., Univariate Discrete Distributions 3rd edition, Wiley, p. 188
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$\begingroup$ (+1) Note that $\exp(-\lambda) = \Pr(X=0)$ when $X$ is a Poisson random variable. So the mass function can be recognized as $\frac{\Pr(X=x)}{\Pr(X >0)}=\Pr(X=x \mid x>0)$, the Poisson probability of a count conditional on its being greater than nought. $\endgroup$ Dec 15, 2015 at 11:41