There is poisson random variable $X$ $$P(X=x)=\frac{\lambda^{x}}{x!}e^{-\lambda}$$ And define random variable $Z=\lceil \beta X \rceil$ ( $\beta$ is rational number which is lower than 1).
How can i find distribution of $Z$.?
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When dealing with ceilings, the proper way to proceed is using probability intervals. $$ P(Z=z)=P(z-1<\beta X <=z)=P(\frac{z-1}{\beta}<X<=\frac{z}{\beta})=F_X(\frac{z}{\beta})-F_X(\frac{z-1}{\beta}) $$ Where $F_X$ is the cumulative distribution of X. Substituting: $$ P(Z=z)=e^{-\lambda}\sum_{i=0}^{\lfloor\frac{z}{\beta}\rfloor}\frac{\lambda^i}{i!}-e^{-\lambda}\sum_{i=0}^{\lfloor\frac{z-1}{\beta}\rfloor}\frac{\lambda^i}{i!}= \begin{cases} e^{-\lambda}\sum_{i=\lfloor\frac{z-1}{\beta}\rfloor}^{\lfloor\frac{z}{\beta}\rfloor}\frac{\lambda^i}{i!} &, z\ge 1 \\ e^{-\lambda} &,z=0 \end{cases} $$
Jsevillamol
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Thank you. Is it possible Z has value between 0 and 1? – Kim Feb 05 '16 at 10:04
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As long as $z>\beta$, $P(Z=z)$ is strictly greater than 0. Thus, Z takes values between $(\beta\ \infty)$. – Jsevillamol Feb 05 '16 at 10:09
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Um... i don`t understand your point. My opinion is that $Z$ have only positive integer due to ceiling function. Having only positive integer is what i want to and reason why i used ceiling function – Kim Feb 05 '16 at 10:17
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@Kim Oops. My blame. I'll edit it. – Jsevillamol Feb 05 '16 at 10:29
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@Kim Note however that Z can be equal to 0. – Jsevillamol Feb 05 '16 at 10:35
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I check this distribution using matlab. But sum of all probability is not one. It is possible that i mistake. Can you check for me? – Kim Feb 05 '16 at 12:21