Sequence of random variables depending on another random variable

Solution 1:

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 $$E[(S_{n+1}+1)/(n+3) \mid \mathcal{F}_n] = \frac{1}{n+3} \left(S_n + 1 + E[X_{n+1} \mid\mathcal{F}_n]\right) = \cdots.$$