I am struggling with coming up with an approach for the following question (pulled from Assignment 4 from MIT's 6.041 Probability course, which I'm trying to self-study as I realized my probability fundamentals are really weak):
- Consider a sequence of independent tosses of a biased coin at times t = 0, 1, 2, . . .. On each toss, the probability of a ’head’ is p, and the probability of a ’tail’ is 1 − p. A reward of one unit is given each time that a ’tail’ follows immediately after a ’head.’ Let R be the total reward paid in times 1, 2, . . . , n. Find E[R] and var(R).
So far, this is what I've been able to observe and my thought process (which seems to be too complicated/incorrect). I'd love some advice/line of questioning/poking holes into my logic (without giving away the solution) so that I may come up with the approach myself:
We get a reward of 1 unit every time we encounter a 'TH' sequence. Then all coin flips outside of the 'TH' occurences are a sequence of H followed by a sequence of T. ex. HHHTHTTT for 1 'TH' occurence.
Given n coin flips, there can be at most floor(n/2) such occurences of this sequence. Then the expected value of the reward, E[R] = $\sum_{x=0}^{n/2} x * p(x)$
Then I considered the case x = 0. If there are no 'TH', then this means the sequence of flips must be a sequence of i heads followed by j tails, where i+j = n. Then p(x = 0) = $\sum_{i=0}^n p^i * (1-p)^{(n-i)}$ which is a geometric progression (right?)
Obviously the value of p(x=0) doesn't matter since x = 0, but that was the easiest case. Then I tried to find what this would look like for x=1 and so on, but then I would have to consider all the partitions of non 'TH' flips created, and this would get massively complicated, so clearly I'm doing something wrong here, but I can't find another direction.
Any advice would be appreciated!