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Assume $\xi_i \sim \mathbb{F}_{\lambda_i}(x)$ are random variables from Poisson distribution.

Consider random variables $\eta_i \sim \tilde{F}_{\lambda_i,t}(x)$, where $\tilde{F}_{\lambda_i,t}(x) = \frac{\int \limits^{x}_{-\infty} e^{ts}dF_{\lambda_i}(s)}{\int \limits^{\infty}_{-\infty} e^{ts}dF_{\lambda_i}(s)}$.

Obtain $\tilde{F}_{\lambda_i,t}(x)$ is a conjugate distribution function family to $F_{\lambda_i}(x)$. (Maybe here i am wrong, it's term called 'exponentiation of mesuare', please, clear my knowledge).

However, let's compute conjugate distribution function.

$\int\limits_{-\infty}^{x} e^{ts}dF(s) = \sum\limits_{k=0}^{n} e^{tk}\frac{\lambda^k}{k!}e^{-\lambda e^t} e^{(e^t - 1)\lambda} \cdot \mathbb{I}_{\lfloor x\rfloor=n}$

$\int\limits_{-\infty}^{\infty} e^{ts}dF(s) = e^{(e^t - 1)\lambda}$

$\tilde{F}_{\lambda,t}(x) = \frac{\int\limits_{-\infty}^{x} e^{ts}dF(s)}{\int\limits_{-\infty}^{\infty} e^{ts}dF(s)} = \sum\limits_{k=0}^{n} e^{tk}\frac{\lambda^k}{k!}e^{-\lambda e^t} \cdot \mathbb{I}_{\lfloor x\rfloor=n}$

is it right?

After that, i have trouble.
Convolution of distribution functions for sum of random variables $\eta_i$, $i=\overline{1,n}$ .

Simplification: sum of 2 random variables $\eta_i + \eta_j$.
Let assume $\gamma = \eta_i + \eta_j$.
Obtain $*\tilde{F}_{\lambda_i,\lambda_j,t_i,t_j} = \int \limits_{-\infty}^{\infty}\tilde{F}_{\lambda_i,t_i}(x-y)d\tilde{F_{\lambda_j,t_j}}(y)$ - $\gamma$-distribution function.

Here i stuck. How to compute (if it is possible) this convolution?

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