I am struggling to work through the following problem.
Let $\{X_{n}\}_{n\geq 0}$ be i.i.d. random variables, $M$ a Poisson random variable with parameter $\lambda$ that is independent of the random variables $X_{n}$, and $\{\varepsilon_{j}\}_{j\geq 1}$ another sequence of i.i.d. variables that are independent of the variables $X_{n}$ and $M$ and also satisfy
$$\mathbb{P}(\varepsilon_{j}=1)=p\,\,\mathbb{P}(\varepsilon_{j}=0)=1-p$$ Define
$$X:=\sum_{k=1}^{M}\varepsilon_{k}X_{k}\text{ and }Y:=\sum_{k=1}^{M}(1-\varepsilon_{k})X_{k}$$
Show that $X$ and $Y$ are independent.
In calculating an expression for $\mathbb{P}(X\leq x,Y\leq y)$ I ineviatably needed to calculate
$$c_{\sigma}:=\mathbb{P}\left(\bigcap_{i=1}^{k}\varepsilon_{i}=\sigma(i)\right)=\left(\prod_{\sigma(i)=1}p\right)\left(\prod_{\sigma(i)=0}1-p\right)$$
where $\sigma\in\{0,1\}^{k}$ is any function $\sigma:\{1,2,\ldots,k\}\rightarrow\{0,1\}$. I have then calculated the following:
I have also computed expression for $\mathbb{P}(X\leq x)$ and $\mathbb{P}(Y\leq y)$, but should I really be able to factor the monstrosity above? How would you show the independence of $X$ and $Y$?
EDIT: Dean has pointed out a subtle issue in this problem. If the indices of the random variables do not begin at $0$, then the problem doesn't quite make sense. I've edited the problem statement so that the indices start at $0$.
In my interpretation of the question, $X$ and $Y$ are not independent. These random variables are functions of random variables: $$\eqalign{X=X(M,\vec{X},\vec{\epsilon})\\Y=Y(M,\vec{X},\vec{\epsilon})}$$ While the variation from $\vec{X}$ is not shared between $X$ and $Y$, the variations from $M$ and $\vec{\epsilon}$ are shared.
As an example, suppose the probability distribution for $\vec{X}$ is non-zero only for $x>0$. If you consider outcomes $m>\lambda$, the corresponding outcomes $x$ and $y$ will both tend to exceed $E[X]$ and $E[Y]$.
Furthermore, the probability that both $X$ and $Y$ are zero is given by $P(M=0)=e^{-\lambda}$ which differs from the product of the probabilities $P(X=0)P(Y=0)$, and therefore the random variables $X$ and $Y$ are not independent.