I'm currently hearing a probability theory course, we have the following exercise:
An urn contains a red and a blue ball, at each (integer) time-step a ball is drawn from the urn and placed back into the urn along with an additional ball of the same colour. Let $R_n, B_n$ denote the random variables that count how many red and blue balls have been drawn at step $n$. Let $\mathcal F_n$ be the smalles $\sigma$-Algebra so that $R_n, B_n$ are measurable and let $M_n$ be the fraction of blue balls in the urn.
Show that $(M_n,\mathcal F_n)$ is a martingale.
I think this exercise has two errors, the first is that $(\mathcal F_n)$ as defined here do not actually give a filtration (ie $\mathcal F_n \subset \mathcal F_{n+1}$ does not hold), the second is that even if $\mathcal F_n$ is modified in the obvious way, that $M_n$ is not a martingale.
The first can be corrected by asking that $\mathcal F_n$ be the $\sigma$-Algebra generated by $\{R_1,..,R_n, B_1,..,B_n\}$, so I don't see this as a big deal, I'd just be interested in whether I am correct in thinking that the above definition does not make $\mathcal F_n$ a filtration.
For the second I note that $M_n = \frac{B_n +1}{n+2}$ and $\mathbb E[B_n\mid \mathcal F_{n-1}]= B_{n-1}+\frac12$, so $$\mathbb E[M_n\mid \mathcal F_{n-1}]= \frac{B_{n-1}+\frac32}{n+2}=\frac{(n+1)(M_{n-1}-\frac1{n+1})}{n+2}+\frac{3}2\frac1{n+2},$$ which is not the same as $M_{n-1}$.
Am I correct in thinking that the exercise has an error? I have not included actual justifications, since I think they would harm readability of the question. However I can provide these if my conclusions are not correct and somebody would like to identify my errors.
It is not the case that $\mathbb{E}[B_n \mid \mathcal{F}_{n-1}] = B_{n-1} + \frac12$.
I imagine your reasoning here was that this conditional expectation should be $B_{n-1} + \mathbb{P}( \text{pick a blue ball at step $n$} \mid \mathcal{F}_{n-1})$. However, since it is not the case that half the balls at any given step are blue, $\mathbb{P}( \text{pick a blue ball at step $n$} \mid \mathcal{F}_{n-1}) \neq \frac12$ in general.
Instead, the portion of balls that are blue at time ${n-1}$ is $M_{n-1}$ so that $\mathbb{E}[B_n \mid \mathcal{F}_{n-1}] = B_{n-1} + M_{n-1} = B_{n-1} + \frac{B_{n-1} + 1}{n+1}$. Plugging this in should show that $M_n$ is an $\mathcal{F_n}$ martingale (for the filtration you suggest, not the one in the question).