2
$\begingroup$

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. $E[x \vert T] = \int\limits_x x \, q_T(x) dx$?

$\endgroup$
1
$\begingroup$

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
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.