We know the sum of independent observations is a sufficient statistic for intensity parameter of the Poisson distribution.
My question: does there exist a sufficient statistic for a finite Poisson mixture?
For instance for the following Poisson mixture model with parameter $\lambda$ $(\Lambda=c \cdot \lambda$ and $c \ge 1$),
$$f(x)=p\frac{e^{-\lambda } \lambda ^x}{x!}+(1-p)\frac{e^{-\Lambda } \Lambda ^x}{x!}$$
taking two independent samples $x$ and $y$ we have, $$f(x,y)=\frac{e^{-2 (\lambda +\Lambda)} \left(e^\lambda (p-1) \Lambda ^x-e^\Lambda p \lambda ^x\right) \left(e^\lambda (p-1) \Lambda ^y-e^\Lambda p \lambda ^y\right)}{x! y!}$$
Taking Log we have, $$\log \left(\frac{e^{-2 (c+1) \lambda } \left(p e^{c \lambda } \lambda ^x-e^{\lambda} (p-1) (c \lambda)^x\right) \left(p e^{c \lambda} \lambda^y-e^{\lambda} (p-1) (c \lambda)^y\right)}{x! y!}\right) \\ = -\log (x! y!)+\log \left(e^{-2 (c \lambda +\lambda)}\right)+\log \left((e^{\lambda} (p-1) (c \lambda)^x - p e^{c \lambda} \lambda ^x\right)\left(e^\lambda (p-1) (c \lambda)^y -p e^{c \lambda} \lambda^y)\right) \\ =\boxed{-\log (x! y!)+\log \left(e^{-2 (c \lambda +\lambda)}\right)+\log \left( e^{2c\lambda} p^2\lambda^{(x+y)} + e^{-2c\lambda+\lambda}(1-p)p c^y \lambda^{(x+y)}+{}\\ e^{\lambda-c\lambda}(1-p)c^xp\lambda^{(x+y)}+e^{2\lambda}(1-p)^{2}c^{(x+y)}\lambda^{(x+y)} \right)}$$
In above equation it looks like parameter $\lambda$ is interacting with the data only through $(x+y)$, that means sum of data is sufficient statistic for finite Poisson mixture too. Is this correct?