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Suppose an urn contains one blue and one red ball and that we perform the following random experiment:

  • In each round $n\in\mathbb{N}$ we randomly draw a ball
  • If the drawn ball is blue, we replace it into the urn together with another blue ball

Let $X_n$ be the fraction of red balls after round $n\in\mathbb{N}_0$ and $\mathcal{F}_n:=\sigma(X_0,\ldots,X_n)$. I want to show that $(X_n)_{n\in\mathbb{N}_0}$ is a $(\mathcal{F}_n)_{n\in\mathbb{N}_0}$-martingale.


Obviously, there are only two possible realisations of $X_n$: $$X_n=\frac{1}{n+1}\;\;\;\text{or}\;\;\;X_n=0$$ Let $Y_n$ be the count of red balls after round $n\in\mathbb{N}_0$. We've got $$\Pr[Y_0=1]=1\;\;\;\text{and}\;\;\;\Pr[Y_n=1]=\frac n{n+1}\;\;\;\text{for all }n\in\mathbb{N}$$ So, $$\Pr\left[X_n=\frac 1{n+1}\right]=\Pr[Y_n=1]=\frac n{n+1}$$ and therefore $$\operatorname{E}[X_n]=\frac n{(n+1)^2}$$

How do we need to proceed?

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It boils down to showing that $\mathbb{E} [ X_{n} | \mathcal{F}_{n-1}] = X_{n-1}$.

Now, $\mathbb{E} [ X_{n} | \mathcal{F}_{n-1}]$ = $P(\mbox{drawing a red ball}|\mathcal{F}_{n-1})$(new fraction in this case) + $P(\mbox{drawing a blue ball}|\mathcal{F}_{n-1})$(new fraction in this case)

$ = X_{n-1}\frac{l.X_{n-1} + 1}{l+1} + (1 - X_{n-1})\frac{l.X_{n-1}}{l+1}$ , where $l$ is the total number of balls at stage $n-1$.

Now, this is equal to $\frac{(l+1)X_{n-1}}{l+1} = X_{n-1}$.

Therefore, we are done.

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