# Marginal probability mass function

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 $$f(v)$$?

• Do you have $y_1,y_2 \in \mathbb{N}_0$? – Dieter Kadelka Aug 6 at 10:18
• 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 $Po(\lambda \dot c)$. – Dieter Kadelka Aug 6 at 10:24
• Yes the distributions are independent with $Y_1 \sim Po(\lambda)$ and $Y_2 \sim Po(c\lambda)$. – Orongo Aug 6 at 12:19

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