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I have the joint PMF

$\exp(y_1 \ln(\lambda)+y_2 \ln(c)+y_2\ln(\lambda)-\ln(y_1!y_2!)-\lambda(1+c))$

for a constant $c>0$. In canonical representation and mixed parameterization I have $\mathbf{\theta}=(\ln(c),\ln(\lambda))$ and $\mathbf{t}=(v,u)^T$ with $v=Y_2$ and $u=Y_1+Y_2$.

What is the marginal PMF $f(v)$?

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    $\begingroup$ Do you have $y_1,y_2 \in \mathbb{N}_0$? $\endgroup$
    – Dieter Kadelka
    Commented Aug 6, 2019 at 10:18
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    $\begingroup$ Do you have $y_1,y_2 \in \mathbb{N}_0$? If this is true, then $Y_1$ and $Y_2$ are independent, $Y_1$ has Poisson distribution $Po(\lambda)$ and $Y_2$ has distribution $Po(\lambda \dot c)$. $\endgroup$
    – Dieter Kadelka
    Commented Aug 6, 2019 at 10:24
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    $\begingroup$ Yes the distributions are independent with $Y_1 \sim Po(\lambda)$ and $Y_2 \sim Po(c\lambda)$. $\endgroup$
    – Orongo
    Commented Aug 6, 2019 at 12:19

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The pdf of $Po(c \dot \lambda)$ is just $v \to e^{-c\lambda} \frac{(c\lambda)^v}{v!}$, $v \in \mathbb{N}_0$.

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