I perform a series of $N$ coin flips, indexed $i = 1, \ldots, N$. I do not get to see the outcome of the coin flips, but for each one I know the probability of the coin being heads, $p_i(H)$. This changes for each coin flip, $p_i(H) \neq p_j(H)$ for $i \neq j$. Depending on the (hidden) outcome of the coin flip, I win an amount of money given by a random variable $x$ which is sampled from one of two fixed distributions, either $q_H(x)$ or $q_T(x)$, which do not vary with $i$. My task is to determine how much I expect to win if I get heads or tails. So for every $i$, I know the distribution of the coin that was flipped $p_i(H)$, and I know how much I win $x_i$. Can I esimate the expected value of $x$ given the coin was heads or tails, e.g. $E[x \vert T] = \int\limits_x x \, q_T(x) dx$?
1 Answer
A naive estimator of $E[X \mid H]$ might be the weighted average $\dfrac{\sum p_i(H) x_i}{\sum p_i(H)}$
and similarly of $E[X \mid T]$ might be $\dfrac{\sum (1-p_i(H)) x_i}{\sum (1-p_i(H))}$
This has the following properties:
- If the $p_i(H)$ are $0$ or $1$ (some of each), it will give you the obvious estimates for the conditional expectations
- If all the $p_i(H)$ are all equal (not $0$ nor $1$) it will give the same estimates for the two conditional expectations
- The estimate of the unconditional expectation will be the mean of the actual amounts won
- If the two conditional expectations are actually equal then the expected difference between the estimates of the conditional expectations will be $0$
- The difference between the estimates of two conditional expectations may be biased towards $0$ compared with the actual difference between the two conditional expectations