Probability of Coin toss With Poisson Distribution

Question

Let N ∼ Poisson(λ). You flip a coin a random number of times N. Each time a head will come with probability p, independently of N. Let X be the (random) number of heads outcomes and Y be the (also random) number of tails. Find the distribution of X and Y. Are they independent? Give conditional distribution of N if X = k.

My idea is to condition on value of N and surely the number of heads of a coin toss is a binomial RV. So it's posisble to sum up the conditional probabilities of all values of N, i believe. X and Y are surely independent but I have trouble concluding this (since poisson distributions are uncorrelated) Any hints would be appreciated for second or third one. Thanks in advance.

Answer 1

Let $q:=1-p$ and let $h,t$ denote nonnegative integers.

Then:

$$P(X=h\wedge Y=t)=P(X=h\wedge Y=t\wedge N=t+h)=$$$$P(X=h\wedge Y=t\mid N=t+h)P(N=t+h)=\binom{t+h}{h}p^hq^te^{-\lambda}\frac{\lambda^{t+h}}{(t+h)!}=$$$$e^{-\lambda}\frac{(\lambda p)^h(\lambda q)^t}{h!t!}=e^{-\lambda p}\frac{(\lambda p)^h}{h!}\times e^{-\lambda q}\frac{(\lambda q)^t}{t!}$$

The fact that you can write the probability as function $f(h)g(t)$ is enough already to conclude that $X$ and $Y$ are independent, and makes it easy to find that $X$ and $Y$ both have Poisson-distribution with parameters $\lambda p$ and $\lambda q$ respectively. To check yourself you could also find $P(X=h)$ (and likewise $P(Y=t)$) by working out:

$$P(X=h)=\sum_{n=h}^\infty P(X=h\mid N=n)P(N=n)=\sum_{n=h}^\infty\binom{n}hp^hq^{n-h}e^{-\lambda}\frac{\lambda^n}{n!}=\cdots$$

Further we have the equality:$$P\left(N=n\mid X=k\right)P\left(X=k\right)=P\left(N=n\wedge X=k\right)=P\left(X=k\mid N=n\right)P\left(N=n\right)$$

which enables you to find $P(N=n\mid X=k)$.

Answer 2

The total number $H$ of heads satisfies $$ \begin{aligned} \mathbb{P}(H=x) &=\sum_{n=x}^{\infty} \mathbb{P}(H=x \mid N=n) \mathbb{P}(N=n)=\sum_{n=x}^{\infty}\left(\begin{array}{l} n \\ x \end{array}\right) p^{x}(1-p)^{n-x} \frac{\lambda^{n} e^{-\lambda}}{n !} \\ &=\frac{(\lambda p)^{x} e^{-\lambda p}}{x !} \sum_{n=x}^{\infty} \frac{\{\lambda(1-p)\}^{n-x} e^{-\lambda(1-p)}}{(n-x) !} \end{aligned} $$ The last summation equals 1 , since it is the sum of the values of the Poisson mass function with parameter $\lambda(1-p)$.

This is similar to drhab's solution. (the problem is from the book "Probability and random processes" by Grimmett)