# Manipulating a distribution by stopping

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.

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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 $n$.
Two questions then, is there some way to say $E[Y_n/(n \wedge \tau)] = p$? And can we make the expectation something else if we did not assume $E[\tau] < \infty$? – nullUser Jan 31 '13 at 21:41
@nullUser: since you're truncating $\tau$ with $n$, such assumption is not used here. I better re-denote. Let me though think about your first question – Ilya Jan 31 '13 at 21:45
Or what about $E[Y_\tau] = p+E[S_\tau/\tau] = p$? I.e. what is the expected proportion of boys at the time when the parents decide to stop? (Assuming $\tau > 0$) – nullUser Jan 31 '13 at 21:57
Can I say $E[S_\tau/\tau] = E[ E[S_\tau/\tau|\tau]] = E[E[S_\tau|\tau]/\tau] = E[0/\tau] = 0$ or is this logic flawed? – nullUser Jan 31 '13 at 22:01
${\Bbb E}[S_\tau\mid \tau]$ doesn't have to be zero, and neither does ${\Bbb E}[S_\tau/\tau]$. For example, use the rule "stop when you get the first boy". – David Moews Jan 31 '13 at 22:26