# Proof that Polya's Urn is a Martingale

In these Wisconsin Lecture Notes on probability, they sketch that, in Polya's Urn, the fraction of green balls at a step $$n$$ defines a martingale. I would like to justify these steps and make them rigorous. I will include the relevant section below: My main question is how to justify the first equality that gets rid of the conditional expectation. It makes sense intuitively I guess. I suspect that it either has to do with breaking $$M_n$$ into indicators, or using some events that partition the sample space to decompose the condtional expectation. Any thoughts?

An urn contains 1 red ball and 1 green ball. At each time, we pick one ball and put it back with an extra ball of the same color. Let $$R_n$$ (resp. $$G_n$$) be the number of red balls (resp. green balls) after the nth draw. Let $$F_n = σ(R_0, G_0, R_1, G_1, . . . , R_n, G_n)$$. Define $$M_n$$ to be the fraction of green balls. Then $$E[M_n | F_{n−1}] = \frac{R_{n−1}} {G_{n−1} + R_{n−1}} \frac{G_{n−1}} {G_{n−1} + R_{n−1} + 1} + \frac{G_{n−1}} {G_{n−1} + R_{n−1}} \frac{G_{n−1 + 1}} {G_{n−1} + R_{n−1} + 1} = \frac{G_{n−1}} {G_{n−1} + R_{n−1}} = M_{n−1}.$$

• It is essentially by definition. The sentence "at each time, we pick one ball [independently of previous picks] and put it back with an extra ball of the same color" means that the conditional distribution of $G_n$ given $G_{n-1}$ and $R_{n-1}$ has a mass of $\frac{G_{n-1}}{G_{n-1}+R_{n-1}}$ on $G_{n-1}+1$ and one minus that on $G_{n-1}$. The implied [independence] clause I added in means that conditioning on $F_{n-1}$ gives the same distribution. – Mike Earnest Feb 13 at 2:57