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Consider an urn that initially contains b black balls and w white balls. At every iteration, we draw a random ball is chosen and the chosen ball is replaced by c > 1 balls of the same color. Let $X_i$ denote the fraction of black balls after i-th draw. Prove that $X_0$, $X_1$, . . . is a martingale.

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Let $d=c-1$. Let $k=b+w$ be the initial number of balls. Then $X_{n}(k+dn)$ is the number of black balls at the $n$-th step. Therefore, \begin{align*} \mathbb{E}\left[X_{n+1}\mid X_{n}\right] & =X_{n}\frac{X_{n}\left(k+dn\right)+d}{k+d\left(n+1\right)}+\left(1-X_{n}\right)\frac{X_{n}\left(k+dn\right)}{k+d\left(n+1\right)}\\ & =X_{n}\left[\frac{X_{n}\left(k+dn\right)}{k+d\left(n+1\right)}+\frac{d}{k+d\left(n+1\right)}\right]+\left(1-X_{n}\right)\frac{X_{n}\left(k+dn\right)}{k+d\left(n+1\right)}\\ & =X_{n}\left[\frac{d}{k+d\left(n+1\right)}+\frac{\left(k+dn\right)}{k+d\left(n+1\right)}\right]\\ & =X_{n}. \end{align*}

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  • $\begingroup$ A little doubt, I thought we have to prove that E[$X_{n+1}$|$X_n$,$X_{n-1}$,...,$X_1$] = $X_n$, Why haven't you considered the previous $X_i$'s? Am I missing something? I asked cause we have been taught that way. $\endgroup$ – mskanyal Apr 29 '19 at 3:14
  • $\begingroup$ In our case, $\mathbb{E}[X_{n+1} \mid X_n, X_{n-1}, \ldots, X_0] = \mathbb{E}[X_{n+1} \mid X_n]$ since you don't get any additional information from knowing the previous states. In particular, this is a Markov chain. $\endgroup$ – parsiad Apr 29 '19 at 3:45
  • $\begingroup$ Thanks a lot for the help $\endgroup$ – mskanyal Apr 29 '19 at 3:53

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