Questionares

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 . Hint: Conditioning on the value of N, the number of heads is a binomial random variable. Use the total probability theorem to conclude.

ANSWER

Here by the problem,

$N \sim Poisson (\lambda)$ with pmf,

$f_N(n)=P(N=n)=e^{-\lambda}.\frac{\lambda^n}{n!} \ ,n=0,1,2,...$

Now we flip a coin a random number N where each time a head will come with probability p ( and a tail will come with probability (1-p) , independently of N.

So here we assume X be the number of heads and Y =N-X be the number of tail.

Clearly we note that, X|N=n~Binomial(n,p) with pmf,

$f_X(x|N=n)=\binom{n}{x}p^x(1-p)^{n-x} \ \ ,x=0,1,...,n$

and similarly, Y|N=n~Binomial(n,1-p)

So the distribution of X (unconditional) is obtained as follows where we note that,

$P(X=x)=\sum_n P(\left \{X=x \right \}\cap \left \{ N=n \right \})$ $=\sum_n P(X=x| N=n )P(N=n)$

Now clearly as $0\leq x\leq n$ then the lower limit of n be x , not 0.

$=\sum _{n=x}^\infty \binom{n}{x}p^x(1-p)^{n-x} .e^{-\lambda}.\frac{\lambda^n}{n!}=e^{-\lambda}\lambda^x p^x\sum _{n=x}^\infty \frac{n!}{x!(n-x)!} .\frac{[\lambda(1-p)]^{n-x}}{n!}$