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
I was somewhat surprised that the answer is negative. (However, see the answer by John Dawkins, who notes that $Z_n$ is a Martingale with respect to some filtration.)
Let $\{\xi_n\}$ and $\{\eta_n\}$ be i.i.d. variables of mean zero, taking values $\pm 1$. Consider the partial sums $X_n:=\sum_{k=1}^n \xi_k$ and $Y_n:=\sum_{k=1}^n \eta_k$. Write $\gamma_j=\xi_j \eta_j$, and observe that $\gamma_j$ is independent of $\xi_j$. Denote by $\mathcal F_n$ the $\sigma$-field generated by the $2n+1$ variables $\{\xi_j\}_{j=1}^{n} , \{\eta_j\}_{j=1}^{n}$ and $\gamma_{n+1}$.
Then $$E[X_{n+1} -X_n |\mathcal F_n]=E[\xi_{n+1} |\mathcal F_n]=E[\xi_{n+1} |\gamma_{n+1}]=0 \,,$$ so $\{X_n\}$ is a $\{\mathcal F_n\}$-martingale. Similarly, $\{Y_n\}$ is a $\{\mathcal F_n\}$-martingale. However, the products $Z_n=X_n Y_n$ satisfy $$E[Z_{n+1} -Z_n |\mathcal F_n] =E[\xi_{n+1}Y_n+\eta_{n+1}X_n+ \gamma_{n+1} |\mathcal F_n]=0+0+\gamma_{n+1} \,,$$ so $\{Z_n\}$ is not a $\{\mathcal F_n\}$-martingale.