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Let $U \sim U(0,1)$ and $V_1,V_2,...$ be a sequence of i.i.d $U(0,1)$ random variables independent of $U$. We define $X_k = 1$ if $V_k \le U$ else $X_k =0$.

If $\displaystyle S_n = \sum_{i=1}^{n} X_i$, show that $M_n =\frac{S_n +1}{n+2}$ is a martingale with respect to the filtration $\mathcal{F}_n=\sigma(X_1,...,X_n)$.

My approach :

Firstly, I calculated $E[X_{n+1}| \mathcal{F}_n]= P[V_{n+1} \le U | \mathcal{F}_n]=\frac{1}{2}$

Now, I start off with $E[M_{n+1}| \mathcal{F}_n]=E[\frac{S_{n}+X_{n+1}+1}{n+3} | \mathcal{F}_n]$

But that is not coming out to be $M_n$. Where am I going wrong?

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1 Answer 1

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$X_{k}=1_{\{V_{k}\leq U\}}$, despite $V_{k}$ is iid sequence, $X_{k}$ is no longer iid since it contains a common realization of $U$. Intuitively, if we observe $X_{k}$ being zero for many $k$, it may be caused by a small realization of $U$, so $X_{k+1}$ is more likely to be $0$. Therefore, $\mathbb{E}[X_{n+1}|\mathcal{F}_{n}]\neq\mathbb{E}[X_{n+1}]$. In fact, you can show $\mathbb{E}[X_{n+1}|\mathcal{F}_{n}]=M_{n}$ hence is naturally a martingale.

By law of iterated expectation, $$ \begin{align*} \mathbb{E}[X_{n+1}|\mathcal{F}_{n}]&=\mathbb{E}\bigl[\mathbb{E}[X_{n+1}|\sigma(X_{1},\ldots,X_{n},U)]\big|\mathcal{F}_{n}\bigr]=\mathbb{E}[U|\mathcal{F}_{n}]\\ &=\int_{0}^{1}u\cdot\frac{u^{S_{n}}(1-u)^{n-S_{n}}}{\Gamma(S_{n}+1)\Gamma(n+S_{n}-1)/\Gamma(n+2)}du\\ &=\frac{S_{n}+1}{n+2}=M_{n}. \end{align*}$$

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  • $\begingroup$ I am not convinced why $U|\mathcal{F}_n \sim \beta(S_n +1 , n-S_n+1)$? $\endgroup$ Dec 7, 2021 at 5:41
  • $\begingroup$ $\mathbb{P}(X_{1}=x_{1},\ldots,X_{n}=x_{n}|U=u)=u^{\sum_{i=1}^{n}x_{i}}(1-u)^{n-\sum_{i=1}^{n}x_{i}}$, $U\sim\mathrm{Uniform}(0,1)$, then use Bayes theorem. $\endgroup$
    – Q9y5
    Dec 7, 2021 at 6:38

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