Question

2. SUPPLEMENTAL QUESTION 1 (a) Toss a fair coin so that with probability pheads occurs and with probability p tails occurs. Let X be the number of heads and Y be the number of tails. Prove X and Y are dependent (b) Now, toss the same coin n times, where n is a random integer with Poisson distribution: n~Poisson(A) Let X be the random variable counting the number of heads, Y the random variable counting the number of tails. Prove X, Y are independent

Answer 1

$(a)~ P(X=1,Y=1)=P(\{HT,TH\})=\frac{2}{4}=\frac{1}{2}\neq P(X=1)P(Y=1)\\ where,~P(X=1)=P(\{HT,TH\})=1/2,\\ ~P(Y=1)=P(\{HT,TH\})=1/2\\ Hence~X~and~Y~are~not~independent.\\$

$(b)~p=probability~of~getting~head\\ The~joint~distribution~of~X~and~Y~is\\ P(X=x,Y=n-x)=P(X=x,Y=n-x|N=n)P(N=n)\\ =\binom{n}{x}p^x(1-p)^{n-x}\frac{e^{-\lambda}\lambda^n}{n!}=\left( \frac{e^{-\lambda p}(\lambda p)^x}{x!}\right )\times \\ \left( \frac{e^{-\lambda (1-p)}(\lambda (1-p))^{(n-x)}}{(n-x)!}\right )~x=0,1,2,3,...,n\\ =0~otherwise$

$The~marginal~distribution~of~X~is\\ P(X=x)=\sum_{n=x}^\infty P(X=x,Y=n-x|N=n)P(N=n)\\ =\sum_{n=x}^\infty \binom{n}{x}p^x(1-p)^{n-x}\times \frac{e^{-\lambda}\lambda^n}{n!}=\frac{1}{x!}\times e^{-\lambda}(\lambda p)^x\sum_{n=x}^\infty \frac{\left(\lambda (1-p) \right )^{n-x}}{(n-x)!}\\ =\frac{e^{-\lambda p}\left(\lambda p \right )^x}{x!},~x=0,1,2,... =0~otherwise\\ Similarly~marginal~distribution~of~Y~is\\ P(Y=n-x)=\frac{e^{-\lambda (1-p)}\left(\lambda (1-p) \right )^{n-x}}{(n-x)!},~x=0,1,2,...\\ =0~otherwise\\$

$Hence~P(X=x,Y=y)=P(X=x)P(Y=y)~\forall ~p\in (0,1)\\ Hence~X~and~Y~are~independent.$