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$p(x) = e^{-x\beta}(1-e^{-\beta})$
Where $x$ is integer takinng values as $x = \{0, 1, 2, ...\}$ and $\beta > 0 $ is a real constant.

Is this a Poissonian distribution? and how could I approximate it in order to sample continuous random variables.

I have tried to google distributions similar to this but I haven't found anything similar so I'm confused on how to proceed.

Thank you.

Magil
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  • Is $x$ integer valued? Your function doesn't integrate to $1$ (for any value of $\beta$) but it sums to $1$ if you assume $x={0,1,2,3,\cdots}$ – lulu Aug 07 '16 at 21:34
  • Now that you know it is a geometric random variable, is your objective of simulation still the same, and what would you like to do ? – Jean Marie Aug 07 '16 at 21:37
  • @lulu It is what I have assumed instinctively. – Jean Marie Aug 07 '16 at 21:38
  • @JeanMarie Sure. It makes sense, and it does give an actual probability distribution. But, really, the OP should specify that. If, to the contrary, the OP simply expects to multiply this by a constant to get it to integrate to $1$ then you'll just get the standard (continuous) exponential distribution. – lulu Aug 07 '16 at 21:40
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    The continuous analogue of the geometric is the exponential(considered as the waiting time for the first arrival) as is the gamma for the negative binomial. – theoGR Aug 07 '16 at 21:42
  • @lulu, as the OP says "discrete variable", an intermediate step would to gain a degree of freedom by introducing a constant $n$ and saying that $X$ can take all values of the form $k/n$... And, later on make $n$ arbitrarily large? But you, the OP, what is your opinion about all this ? – Jean Marie Aug 07 '16 at 21:47
  • @lulu In a dual way, one could set $\beta=\gamma/n$, for a fixed $n$. And then take $n$ arbitrarily large. – Jean Marie Aug 07 '16 at 22:54
  • @JeanMarie I think, whichever way you look at it, the exponential distribution really is the best continuous analog of a geometric distribution. here is a relevant discussion. But, I repeat, absent word from the OP we are really just guessing at what is desired. – lulu Aug 07 '16 at 22:58
  • Yes x is integer taking values as x = {0, 1, 2, ...} The thing is that for the simulation I'm running, for certain values of $\beta$ using the discrete version of the inverse transform sampling, it can be pretty slow because x gets quite large. So I'm thinking that with a continuous approximation I could sample values much faster. – Magil Aug 08 '16 at 06:47
  • @lulu Your assumption is right, sorry I didn't specify that, so x is an integer variable that can take values as you said. I believe that you, JeanMarie and theoGR are correct, it is a geometric distribution, the simulations with an exponential distribution seem to be working as expected. Thank you very much for your input lulu, JeanMarie and theoGR. – Magil Aug 08 '16 at 12:43

1 Answers1

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Written under the form

$$P(X=k)= e^{-k\beta}(1-e^{-\beta})$$

it's clearly a geometrical distribution

$$P(X=k)= p^{k}(1-p)$$

with $p:=e^{-\beta}$, $p < 1$ because $\beta > 0$.

Jean Marie
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