Suppose we have a Bernoulli random variable $X\sim\mathrm{Binomial}(1,p)$ and $(X_i : i \in \mathbb{N})$ are independent copies of $X$. Let $$Y_n := \sum_{i=1}^{n \wedge \tau} X_i$$ where $\tau$ is a stopping time. Is it true that $\mathbb{E}[Y_n/n] = p$?
I can't see how to make a martingale out of the $X_i$s in a useful way, but the result feels intuitively true.
My friend posed this question to me in terms of parents having children, a boy with probability $p$, and wondering about the distribution of boys and girls given parents "stopping preferences".
This was the natural setup that I came up with, but unfortunately we have not yet covered martingales and stopping times in my probability course, so I don't really know how to proceed.
If $X\sim \mathrm{Binomial}(1,p)$ then $\mathsf E[X - p] = 0$. So that you know that $$ S_n = \sum_{k=1}^n X_k - np $$ is a martingale. From that you have $\mathsf E[S_\eta] = 0$ for any stopping time $\eta$ with a finite expectation. So you have $$ Y_n = S_{n\wedge\tau} + p(n\wedge\tau) $$ where $\eta = \tau\wedge n$, whence $$ \mathsf E[Y_n] = p\mathsf E[n\wedge\tau] $$ In your model, if parents are eager to stop in a finite time (quite natural assumption), you shall not divide by $n$.