I think you should be specific about your filtrations.
Is it really that both $X_n$ and $Y_n$ are in the same probability space $(\Omega, \mathscr F, \mathbb P)$?
Are they both $(\{\mathscr F_n\}_{(n \in \mathbb N)}, \mathbb P)-$martingales for the same filtration $\{\mathscr F_n\}_{(n \in \mathbb N)}$?
I believe you mean to say that $X_n$ is a $(\{\sigma(U_1, ..., U_n)\}_{(n \in \mathbb N)}, \mathbb P)-$martingale and that $Y_n$ is a $(\{\sigma(\xi_1, ..., \xi_n)\}_{(n \in \mathbb N)}, \mathbb P)-$martingale.
Check your proof in saying that each is a $(\cdot, \mathbb P)-$martingale. You may have used the facts that for $m < n$,
$2^{n-m} \prod_{k=m+1}^{n} U_k$ is independent of $\sigma(U_1, ..., U_m)$
$p^{m-n} \prod_{k=m+1}^{n} \xi_k$ is independent of $\sigma(\xi_1, ..., \xi_m)$.
$W_n$ is not necessarily a $(\{\sigma(U_1, ..., U_n)\}_{(n \in \mathbb N)}, \mathbb P)-$martingale or a $(\{\sigma(\xi_1, ..., \xi_n)\}_{(n \in \mathbb N)}, \mathbb P)-$martingale.
I guess we can suppose that there's some $\{\mathscr F_n\}_{(n \in \mathbb N)}$ that works for both $(*)$ s.t. $X_n$ and $Y_n$ are $(\{\mathscr F_n\}_{(n \in \mathbb N)}, \mathbb P)-$martingales.
So let us try to see if $W_n$ is a $(\{\mathscr F_n\}_{(n \in \mathbb N)}, \mathbb P)-$martingale.
Rewrite $W = (W_n)_{n \ge 0}$ using indicator functions:
$$W_n = X_n1_{A_n} + Y_n1_{A_n^C}$$
where $A_n = \{X_n \ge Y_n\}$ and $0 < P(A_n) < 1$
We have:
$$E[W_n | \mathscr F_m] = E[X_n1_{A_n} + Y_n1_{A_n^C} | \mathscr F_m]$$
$$ = E[X_n1_{A_n} + Y_n1_{A_n^C} | \mathscr F_m]$$
$$ = E[X_n1_{A_n}| \mathscr F_m] + E[Y_n1_{A_n^C} | \mathscr F_m]$$
$$ = X_m E[2^{n-m} \prod_{k=m+1}^{n} U_k 1_{A_n}| \mathscr F_m] + Y_m E[p^{m-n} \prod_{k=m+1}^{n} \xi_k 1_{A_n^C} | \mathscr F_m]$$
It does not necessarily follow that
$$E[2^{n-m} \prod_{k=m+1}^{n} U_k 1_{A_n}| \mathscr F_m] = 1_{A_m}$$
And
$$E[p^{m-n} \prod_{k=m+1}^{n} \xi_k 1_{A_n^C} | \mathscr F_m] = 1_{A_m^C}$$
Just because at time n, we have $A_n$ doesn't mean that at time m, we had $A_m$.
Hence, we have our counterexample.
Now if $P(A_j) = 0$ or $1 \ \forall j \in \mathbb N$, then $W_n$ is $X_n$ or $Y_n$ a.s., then yes, it is a $(\{\mathscr F_n\}_{(n \in \mathbb N)}, \mathbb P)-$martingale because such implies $1_{A_1} = 1_{A_2} = ...$ a.s., in particular, $1_{A_n} = 1_{A_m}$ a.s..
This might be relevant: Is a probability of 0 or 1 given information up to time t unchanged by information thereafter?
However, $0 < P(A_j) < 1$ based on $0 < p < 1$:
$$P(A_j) = P(X_j \ge Y_j) = P(\xi_1 = 0 or \xi_2 = 0 or ... or \xi_j = 0)$$
$$= 1-P(\xi_1=\xi_2=...=\xi_j=1)$$
$$= 1-\prod_{i=1}^{j} (1-p)$$
$$= 1- (1-p)^j$$
Edit to address edit (omitting $\mathbb P$'s):
Given that $X_n$ is a $\mathscr F_n^X$-martingale, $Y_n$ is a $\mathscr F_n^Y$-martingale $(**)$ and there is some filtration $\mathscr F_n$ $(*)$ s.t. $X_n$ and $Y_n$ are $\mathscr F_n$-martingales, show that $W_n$ is a $\mathscr F_n^W$-martingale.
Using n-1 instead of m:
$$E[W_n | \mathscr F_{n-1}] = E[X_n1_{A_n} + Y_n1_{A_n^C} | \mathscr F_{n-1}]$$
$$ = E[X_n1_{A_n} + Y_n1_{A_n^C} | \mathscr F_{n-1}]$$
$$ = E[X_n1_{A_n}| \mathscr F_{n-1}] + E[Y_n1_{A_n^C} | \mathscr F_{n-1}]$$
$$ = X_{n-1} E[2^{n-(n-1)} \prod_{k=(n-1)+1}^{n} U_k 1_{A_n}| \mathscr F_{n-1}] + Y_{n-1} E[p^{(n-1)-n} \prod_{k=(n-1)+1}^{n} \xi_k 1_{A_n^C} | \mathscr F_{n-1}]$$
$$ = X_{n-1} E[2 U_n 1_{A_n}| \mathscr F_{n-1}] + Y_{n-1} E[p^{-1} \xi_n 1_{A_n^C} | \mathscr F_{n-1}]$$
We still run into the same problem. How can we say that
$$E[2 U_n 1_{A_n}| \mathscr F_{n-1}] = 1_{A_{n-1}}$$
or
$$E[p^{-1} \xi_n 1_{A_n^C} | \mathscr F_{n-1}] = 1_{A_{n-1}^C}$$
?
Just because at time n, we have $A_n$ doesn't mean that at time n-1, we had $A_{n-1}$.
$(*)$
I think some candidates for $\mathscr F_n$ are:
- $\sigma(\sigma(U_1, ..., U_n) \cup \sigma(\xi_1, ..., \xi_n))$
- $\sigma(\sigma(X_1, ..., X_n) \cup \sigma(Y_1, ..., Y_n))$
- $\sigma(W_1, ..., W_n)$
I think $(3) \subseteq (2) \subseteq (1)$
I think $\sigma(A_1, ..., A_n) \subseteq (2), \subseteq (1), \subsetneq (3)$
$(**)$
FYI
$$\sigma(U_1, ..., U_n) \supseteq \mathscr F_n^X$$
$$\sigma(\xi_1, ..., \xi_n) \supseteq \mathscr F_n^Y$$
More info here: Prove Z is a martingale by defining it is a product of random variables
Best Answer
Note that since $(Y_n,\mathcal F_n^Y)$ is a martingale, then uniform integrability is equivalent with convergence in $L_1$. Moreover, if the limit exists (call it $Y$), then it must be the case that $\mathbb E[Y]=1$ since $\mathbb E[Y_n] = 1$ for every $n\in \mathbb N$. However, as you noticed, we know (since it is non-negative martingale) that $Y_n \to Y$ almost surely, for some $Y \in [0,\infty)$. We'll show that $Y = 0$ almost surely.
Note that there exist $\delta,\epsilon > 0$ such that $\mathbb P(|\xi_k-1| > \epsilon) =\delta >0 $
By that we have $\sum_{k=1}^\infty \mathbb P(|\xi_k - 1| > \epsilon) = \infty$, and by Borel Cantelli, there is set $\Omega_0$ of measure $1$ such that for every $\omega \in \Omega_0$ there exist subsequence $(k_m(\omega))_{m \in \mathbb N}$ such that $|\xi_{k_m(\omega)}(\omega) - 1 | > \epsilon$ for every $m \in \mathbb N$.
Now, we need to use some tool from analysis, note that if infinite product $\prod_{k} a_k$ of nonnegative numbers $a_k$ converges to some $a \in (0,\infty)$ then $\lim_{k \to \infty} a_k = 1$ (it is due to taking logarithm of that product and noticing that $\lim_{k \to \infty} \ln(a_k) = 0$ iff $\lim_{k \to \infty} a_k = 1)$
But we showed that in our case for $\omega \in \Omega_0$ (so almost surely) $\xi_k(\omega) \not \to 1 $, which means that $Y$ must be $0$ or $\infty$ almost surely (the product must diverge). However, by martingale convergence theorem, we know that limit is finite almost surely, so it must be that $Y=0$ a.s. Hence it cannot converge in $L_1$, since $\mathbb E[|Y_n - Y|] = \mathbb E[|Y_n|] = 1 \not \to 0$