Let $X$ denote the number of tosses required to get the 5th head and $Y$ the number between the 6th and 7th heads. Are $X$ and $Y$ independent?

Question

Let $X$ denote the number of tosses required to get the 5th head and $Y$ the number between the 6th and 7th heads. Are $X$ and $Y$ independent?

Y will always depend on X . NO ? i know geometric distribution has lack of memory property . but in this the underlying distribution is i think negative binomial.

i think 'no, they are not independent ' . help please.

Answer 1

What you want to show is that $$ P(X=x \wedge Y=y) = P(X=x)P(Y=y) $$ for all values $x$ and $y$ of $X$ and $Y$.

Let $Z$ be the number of tosses between the 5th and 6th head. Then \begin{align*} P(X=x \wedge Y=y) &= \sum_{z=1}^\infty \underbrace{\binom{x-1}{4}p^5q^{x-5}}_{X=x} \underbrace{\vphantom{\binom{x-1}{4}}q^{z-1}p}_{Z=z} \underbrace{\vphantom{\binom{x-1}{4}}q^{y-1}p}_{Y=y} \\ &= \binom{x-1}{4}p^5q^{x-5} \cdot q^{y-1}p \cdot \sum_{z=1}^\infty q^{z-1}p \\ &= \binom{x-1}{4}p^5q^{x-5} \cdot q^{y-1}p \cdot \frac{p}{1-q} \\ &= \binom{x-1}{4}p^5q^{x-5} \cdot q^{y-1}p = P(X=x)P(Y=y) \end{align*} Once you separate the events like this it boils down to the tosses being independent of each other.