I am working on the following problem:
Suppose that $U\sim\rm Unif[0,1]$ and consider a sequence of random variables $X_i$ (which are iid when the value of $U$ is given) with $X_i \sim \rm Ber(U)$ (i.e., the Bernoulli distribution). Let $\mathcal{F}_n = \sigma(X_1,…,X_n)$.
- Find $\mathbb{E}[X_{n+1}|\mathcal{F}_n]$.
- Let $S_n = \sum_1^n X_i$. Show that $(S_n + 1)/(n + 2)$ is a martingale w.r.t. the filtration $\mathcal{F}_n$.
I got stuck on question 1 and below is my attempt:
For a fixed value $U = u$ and a specific sequence $(x_i)_{i=1}^n\in \{0, 1\}^n$ with $m < n$ zeros and $(n-m)$ ones, $\mathbb{P}((X_i)_i = (x_i)_i) = u^m(1-u)^{n-m}$. Thus
\begin{align*}
\mathbb{E}[X_{n+1} | \mathcal{F}_n]
&= \mathbb{E}[X_{n+1}|X_1,…,X_n] = \mathbb{E}[X_{n+1} | (X_i)_{i=1}^n = (x_i)_{i=1}^n]\\
&= \mathbb{P}(X_{n+1} = 1 | (X_i)_{i=1}^{n} = (x_i)_{i=1}^n) = \frac{\mathbb{P}((X_1,…, X_n, X_{n+1}) = (x_1,…,x_n, 1))}{\mathbb{P}((X_1,…, X_n) = (x_1,…,x_n))}.
\end{align*}
But it was at this step that I realized my computation would simply give $u$, for any particular $U = u$. I am confused because generally, $X_i$'s are not independent. Did I make any mistake in my attempt? If so, what is a proper scheme to deal with such questions? Any of your help will be greatly appreciated! 🙂
Best Answer
Conditioning on $U$ along with $\mathcal{F}_n$ yields $$E[X_{n+1} \mid X_1, \ldots, X_n, U] = E[X_{n+1} \mid U] = U$$ where the first equality is due to $X_{n+1}$ being conditionally independent of $X_1, \ldots, X_n$ given $U$. I'm guessing this is your "my computation would simply give $u$ for any particular $U=u$."
However, the question is really asking for $E[X_{n+1} \mid X_1=x_1, \ldots, X_n=x_n] = \frac{P(X_1=x_1, \ldots, X_n=x_n, X_{n+1}=1)}{P(X_1=x_1, \ldots, X_n=x_n)}$. You've written down this expression correctly but computed it incorrectly. The mistake is that you need to integrate over $U$ rather than considering some fixed value of $U$.
For instance, the denominator is $$P(X_1=x_1, \ldots, X_n=x_n) = \int_0^1 P(X_1=x_1, \ldots, X_n=x_n, U=u) \, du = \int_0^1 u^{s_n} (1-u)^{n-s_n} \, du = \frac{s_n! (n-s_n)!}{(n+1)!},$$ where $s_n = \sum_{i=1}^n x_i$. You can compute the numerator similarly.
See if you can then arrive at the final result $E[X_{n+1} \mid X_1=x_1, \ldots, X_n=x_n] = \frac{s_n+1}{n+2}$, i.e. $E[X_{n+1} \mid X_1, \ldots, X_n] = \frac{S_n+1}{n+2}$.
From here, question 2 is simple, since $$E[(S_{n+1}+1)/(n+3) \mid \mathcal{F}_n] = \frac{1}{n+3} \left(S_n + 1 + E[X_{n+1} \mid\mathcal{F}_n]\right) = \cdots.$$