# Martingale constructed with Bernoulli random variables

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?

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