Dependence/independence of functions of Poisson random variables.

Question

The random variables $N,X_{1},X_{2},X_{3}... $ are independent, $N \in Po (\lambda)$, and $X_{k} \in Be (1/2)$, $k\geq1$. Set

$$Y_{1}=\sum_{k=1}^{N} X_{k}$$ and $$Y_{2}=N-Y_{1}$$

($Y_{1}=0$ for $N=0$). Show that $Y_{1}$ and $Y_{2}$ are independent, and determine their distribution.

1) Is it possible to determine if $Y_{1}$ and $Y_{2}$ are independent before determining their distributions? If yes, then how?

2) How can i determine the distribution of $Y_{2}$? I started using the convolution formula, but then realized that i dont know if $N$ and $Y_{1}$ are independent, neither am i able to determine the joint distribution $p_{N,Y_{1}}$.

I feel that im missing something obvious here, but can not get it. What i was able to calculate is that $Y_1$ $\in$ $Po(\lambda/2)$. Then my approach to determine the probability distribution of $Y_{2}$ was the following:

$P (Y_2 = y_2) = P (N-Y_1 = y_2) = P (N=Y_1 + y_2)=$

$$\sum_{y_1} P (N=Y_1 + y_2\mid Y_1 =y_1 )\cdot P (Y_1=y_1)= ...$$ (its in the following im not sure im correct, assuming that i dont know if $N$ and $Y_1$ are independent or not. Is the following correct even if $N$ and $Y_1$ are not independent? ) $$...=\sum_{y_1} P (N=y_1 + y_2 )\cdot P (Y_1=y_1)= \sum_{y_1=0}^{\infty} P (N=y_1 + y_2 )\cdot P (Y_1=y_1)$$

Am i on the right track? All kind of help is appreciated, thanks !

Answer 1

Tip 1: As $Y_1$ is the count of 'successes' in a sequence of $N$ independent Bernoulli events ($N$ itself a Poisson distributed random variable) , then the conditional distribution of $Y_1$ given $N$ is Binomial. You can use this to determine that $Y_1$ has a $\underline{\phantom{\text{Poisson distribution with rate }\lambda}}$, via :$$\mathsf P(Y_1=k)= \sum_{n=k}^\infty \mathsf P(Y_1=k\mid N=n)\mathsf P(N=n)$$

Tip 2: $Y_1=\sum_{k=1}^N X_k\\Y_2~{= N-Y_1 \\ = \sum_{k=1}^N 1 -\sum_{k=1}^N X_k \\ = \sum_{k=1}^N (1-X_k)}$

Tip 3: $\mathsf P(Y_1=k, Y_2=h)~{=\mathsf P(Y_1=k\mid Y_1+Y_2=k+h)\,\mathsf P(Y_1+Y_2=k+h) \\ = \mathsf P(Y_1=k\mid N=(k+h))\,\mathsf P(N=(k+h))}$

Answer 2

$$P(Y_1=k\wedge Y_2=m)=\sum_{n=0}^{\infty}P(Y_1=k\wedge Y_2=m\mid N=n)P(N=n)=$$$$P(Y_1=k\wedge Y_2=m\mid N=k+m)P(N=k+m)=$$$$2^{-(m+k)}\binom{m+k}{k}e^{-\lambda}\frac{\lambda^{k+m}}{(m+k)!}=$$$$e^{-\frac12\lambda}\frac{(\frac12\lambda)^k}{k!}\times e^{-\frac12\lambda}\frac{(\frac12\lambda)^m}{m!}$$

From the fact that we can write $P(Y_1=k\wedge Y_2=m)$ as a product of the form $u(k)v(m)$ it follows that $Y_1,Y_2$ are independent. That is a theorem on its own. Also it is not difficult to avoid the use of this theorem by just finding the PMF of $Y_1,Y_2$ and to check. Here $Y_1$ and $Y_2$ have the same distribution. See tip 2 of @Graham for that.