Proving Independence

Example

Given a fair coin $(H/T)$ Given a biased coin $(H/H)$ If we pick one and flip twice, are the flip results independent?

Soln

With events $H_1$ and $H_2$ being flipping heads on the first flip and heads on the second flip. The events are independent if: $P(H_1\cap H_2)=P(H_1)P(H_2)$

• Note that $P(H_1\cap H_2|F)=P(H_1|F)\times P(H_2|F)=\frac{1}{4}$ given a fair coin is chosen
• Note that $P(H_1\cap H_2|B)=P(H_1|B)\times P(H_2|B)=1$ given a biased coin is chosen Then, we have shown that the two events have Conditional Independence Then, consider $P(H_1\cap H_2)=P(H_1\cap H_2\cap B)+P(H_1\cap H_2\cap F)$
• $⟹P(H_1\cap H_2)=P(H_1\cap H_2|B)\ast P(B)+P(H_1\cap H_2|F)\ast P(F)$
• $⟹P(H_1\cap H_2)=\frac{5}{8}$ Then, consider $P(H_1)=P(H_1|B)P(B)+P(H_1|F)P(F)$
• $⟹P(H_1)=\frac{3}{4}$ Similarly, we compute $P(H_2)=\frac{3}{4}$

Then, since $\frac{5}{8}\ne \frac{3}{4}\ast \frac{3}{4}$ the two are not independent