Show rigorously that Pólya urn describes a martingale

Question

We work with the famous Pólya urn problem. At the beginning one has $r$ red balls and $b$ blue ball in the urn. After each draw we add $t$ balls of the same color in the urn.

$(X_n)_{n \in \mathbb N}$ is the share of red balls in the urn after $n$-th draw. I need to show that $(X_n)_{n \in \mathbb N}$ is a martingale rigorously (i.e. using only theorems and without imagination power based on real world experience).

I select a filtration $(\mathcal F_n)_{n \in \mathbb N}=\sigma (X_1,...,X_n)$.

I need to show that $E(X_{n+1}|\sigma (X_1,...,X_n))=X_n$

$E(X_{n+1}|\sigma (X_1,...,X_n))=E(\frac{lX_n+S}{l+t}|\sigma (X_1,...,X_n))=\frac{lX_n}{l+t}+\frac{E(S|\sigma (X_1,...,X_n))}{l+t}$ where $l$ is the number of balls in the urn after $n$-th draw and $S$ is the number of balls added after $n+1$-th draw (random variable), the last step is possible because $X_n$ is $\sigma (X_1,...,X_n)$ measurable.

Further $E(S|\sigma (X_1,...,X_n))=tE(\mathcal I\{$read ball at $n+1$-th draw$\}|\sigma (X_1,...,X_n))$

How do I take it from here?

Please note that I saw The Pólya urn model describes a martingale, and I do not consider it rigorous enough.

Answer 1

Actually, you may want to account for the underlying random process...

Let $\{U_n\}_{n=0}^\infty$ be a sequence of i.i.d. uniform random variables on $[0,1]$. Let $r_n$ denote the number of red balls at time $n$ and $l_n=r+g+t\cdot n$. Then

$$ X_n=\frac{r_n}{l_n}\quad\text{and}\quad r_{n+1}=r_n1\{U_{n+1}>X_n\}+(r_n+t)1\{U_{n+1}\le X_n\}. $$

Letting $\mathcal{F}_n:=\sigma(U_1,\dots,U_n)$, one gets

\begin{align} \mathsf{E}[X_{n+1}\mid\mathcal{F}_n]&=\mathsf{E}\!\left[\frac{r_n}{l_n+t}\times 1\{U_{n+1}>X_n\}+\frac{r_n+t}{l_n+t}\times 1\{U_{n+1}\le X_n\} \mid\mathcal{F}_n\right] \\ &=\frac{r_n}{l_n+t}\mathsf{E}\!\left[1\{U_{n+1}>X_n\}\mid\mathcal{F}_n\right]+ \frac{r_n+t}{l_n+t}\mathsf{E}\left[1\{U_{n+1}\le X_n\}\mid\mathcal{F}_n\right] \\ &=\frac{r_n}{l_n+t}\left(1-\frac{r_n}{l_n}\right)+ \frac{r_n+t}{l_n+t}\left(\frac{r_n}{l_n}\right)=X_n \end{align}

because $r_n$ and $X_n$ are $\mathcal{F}_n$-measurable, and $U_{n+1}$ is independent of $\mathcal{F}_n$.

Answer 2

Firstly if you want to prove it properly you need to show the process is adapdet and integrable. I will skip this.

For the remaining property set the following:

$l_n= b+r+ n t$ the number of balls prior the n+1th extraction; $A_n$ the event: the n+1th extraction is red.

Then $X_n$ can be expressed recursively by $X_{n+1}= \frac{l_n X_n +t}{l_{n+1}} I_{A_n} + \frac{l_n X_n }{l_{n+1}} I_{A_n^c}$

This leads to the following:

$$E[X_{n+1}|F_n] = E[X_{n+1} I_{A_n}|F_n] + E[X_{n+1} I_{A_n^c}|F_n] = E[\frac{l_n X_n +t}{l_{n+1}} I_{A_n}|F_n] + E[\frac{l_n X_n }{l_{n+1}} I_{A_n^c}|F_n]= \frac{l_n X_n +t}{l_{n+1}} X_n + \frac{l_n X_n }{l_{n+1}} (1-X_n)=X_n \frac{l_n+t}{l_{n+1}}=X_n$$

Answer 3

Since I'm new here, I cannot make comments yet. But, answering @Sergey and @BCLC, we can see that $$I_{A_n} = 1, \mbox{with probability} X_n; \quad 0, \mbox{with probability } (1-X_n)$$

So, given $F_n$ (loosely, given $X_n$) we can see that $I_{A_n} \sim Bin(1,X_n)$.

Therefore, $E[I_{A_n}|F_n]=1 \cdot X_n + 0 \cdot (1-X_n) = X_n$.