First of all from last question I can obtain some discrete random variable which described below. In this page, we denote this random variable as $Z$.
$P(Z=0) = e^{-\lambda}+(1-e^{-\lambda})(1-\beta)$
$P(Z=n\beta) = \frac{\lambda^{n}}{n!}e^{-\lambda}\beta, (n>0)$
$\beta$ is constant which is lower than 1.
And there is Poisson random variable X with $T\lambda$
$P(n;T\lambda)=\frac{(T\lambda)^{n}}{n!}e^{-T\lambda}$
I made new random variable $K=X+Z$ and i like to know distribution of $T$. Because sum of two random variable`s distribution is formed by convolution of two random variable, i made below equation. $P(K=k)=\sum_{z}P(X=k-z)P(Z=z)$
And using this equation i can obtain pmf like below
$P(K=0)=\sum_{z}P(X=-z)P(Z=z)=P(X=0)P(Z=0)=e^{-T\lambda}(e^{-\lambda}+(1-e^{-\lambda})(1-\beta))$
$P(K=k)=\sum^{k}_{z}P(X=k-z)P(Z=z)=\sum^{k}_{z}\frac{(T\lambda)^{k-z}}{(k-z)!}e^{-T\lambda}\frac{\lambda^{z/\beta}}{(z/\beta)!}e^{-\lambda}\beta$
($z=n\beta$ from above $Z$)
I tried to check $P(K=0)+\sum P(K=k)$ is $1$. But it is not easy to me.
Is there anyone who can help me? If there are some miss approach, please let me know. Thank you!
