The sum $Y$ of independent Bernoulli variables with Poissonian upper limit $N$ is independent of $N-Y$

The random variables are $$N,X_1,X_2,..$$ are independent,$$N \in po(\lambda)$$ and $$X_k \in Be(1/2)$$, $$k \geq 1$$.

Set $$Y_1 = \sum\limits_{k=1}^{N}X_k$$ and $$Y_2 = N - Y_1$$. Here $$Y_1 = 0$$ for $$N = 0$$. Show that $$Y_1$$ and $$Y_2$$ are independent, and determine their distributions.

My idea, as for the independency,

Pr($$Y_1 = z , N-Y_1 = k$$ ) = Pr($$Y_1 = z, N = z + k$$ ) = Pr($$\sum\limits_{k=1}^{z+ k}X_k = z , N = z + k$$) = Pr($$Y_1 = z)Pr(N - Y_1 =k)$$ , the last equality because $$N$$ and $$X_1,X_2,...$$ are independent so are $$N$$ and $$g(X_1,X_2,...)$$ for any real function $$g$$ , the last statement i can't prove maybe someone could show me?. anyhow Iam a bit unsure about my reasoning since $$N$$ is poissonian $$Y_2 = N-Y_1$$ could take on values that $$Y_1$$ cannot ( e.g any fraction ) , but then this: $$Pr(Y_1 = z , N-Y_1 = k ) = Pr(Y_1 = z)Pr(N - Y_1 =k)$$ only holds for integer values $$z$$. Is there anyone how could clarify these things further?

• for distibution of $Y_1$ have a look at math.stackexchange.com/a/877094/75923 Aug 2, 2014 at 18:39
• First mistake in your proof: N=z−k should read N=z+k.
– Did
Aug 2, 2014 at 18:53
• let me edit.... Aug 2, 2014 at 19:01
• Try to format your posts in a more readable way; this is in your own interest (more people will be willing to read them). Long formulas can be made displayed $$..$$ and there are ways to nicely split them between lines, described here. Proper use of uppercase letters would help too.
– user147263
Aug 2, 2014 at 20:01

For $Y_1$, $Y_2$ to be independent it must hold $\mathbb{P}(Y_1=k_1,Y_2=k_2) = \mathbb{P}(Y_1=k_1) \mathbb{P}(Y_2=k_2)$, i.e., the joint distribution is equal to the product of marginal distributions.

We know the following: $$\mathbb{P}(N=n)=e^{-\lambda}\frac{\lambda^n}{n!},\; n=0,1,\ldots,$$

and $$\mathbb{P}(Y_1=k_1|N=n) = {n \choose k_1}p^{k_1}(1-p)^{n-k_1},\; k_1 = 0,1,\ldots,n,$$ with $p=1/2$.

Now, \begin{align} \mathbb{P}(Y_1=k_1,Y_2=k_2)&=\mathbb{P}(Y_1=k_1,N-Y_1=k_2)\\ &=\mathbb{P}(Y_1=k_1,N=k_1+k_2)\\ &=\mathbb{P}(Y_1=k_1|N=k_1+k_2)\mathbb{P}(N=k_1+k_2) \end{align}

From this point on you should be able to find (after some algebra) that $Y_1$, $Y_2$ are indeed independent as well as their marginal distributions.

Remark: The above result is a special case of a more general result concerning homogeneous Poisson point processes (HPPPs) which states that when the points of a HPPP of intensity $\lambda$ are independently ''marked'' (selected) with probability $p$, the result is two independent HPPPs of intensities $\lambda p$ and $\lambda (1-p)$, respectively.

Note that $Y_{2}=\sum_{k=1}^{N}Z_{k}$ for $Z_{k}=1-X_{k}$.

It is evident that the $Z_{k}$ are independent and Bernouilli$(\frac{1}{2})$-distributed so the distributions of $Y_{1}$ and $Y_{2}$ are the same. In fact their distribution is Poisson$\left(\frac{\lambda}{2}\right)$ and a proof for that can be found here.

$$P\left\{ Y_{1}=r\wedge Y_{2}=s\right\} =P\left\{ Y_{1}=r\wedge N=r+s\right\} =P\left\{ Y_{1}=r\mid N=r+s\right\} P\left\{ N=r+s\right\} =\dbinom{r+s}{r}2^{-r-s}e^{-\lambda}\frac{\lambda^{r+s}}{\left(r+s\right)!}$$

This equals: $$P\left\{ Y_{1}=r\right\} P\left\{ Y_{2}=s\right\} =e^{-\frac{\lambda}{2}}\frac{\left(\frac{\lambda}{2}\right)^{r}}{r!}\times e^{-\frac{\lambda}{2}}\frac{\left(\frac{\lambda}{2}\right)^{s}}{s!}$$ Proved is now that $Y_{1}$ and $Y_{2}$ are independent.