I need to show whether or not the maximum of two martingales is also a martingale. Originally, I thought yes. But supposedly the answer is no. So as a counter-example, let $U_i$ be $iid$ $unif(0,1)$, $X_0 = 1$, and
$$X_n = 2^n\prod_{k=1}^n U_k.\tag{1}$$
I already know that $X_n$ is a martingale.
For my second martingale, let $\xi_i$ be $iid$ $Bern(p),0<p<1$, $Y_0 = 1$,
and
$$Y_n = p^{-n}\prod_{k=1}^n \xi_i.\tag{2}$$
I already know that $Y_n$ is a martingale. So let $W_n = \max(X_n,Y_n)$.
What I did was I tried to get the distribution of $W_n$ using the CDF and I ended up with
$$F_W(w) = \left(\frac{w}{2}\right)^n(1-p)^n$$
But then I got stuck. I'm not sure what to do with this. I'm not even actually sure if I am on there right track. How can I proceed to show that $W_n$ is not a martingale? Any help is appreciated. Thanks.
EDIT: I forgot to mention that $X_n,Y_n$ are martingales with respect to the same filtration. However, in my class, we had a rudimentary treatment of martingales. The main definition I know/we use is:
A stochastic process $\{X_n;n = 0,1,\dotsc\}$ is a martingale if for $n = 0,1,2,\dotsc,$
- $E[|X_n|] <\infty$, and
- $E[X_{n+1}|X_0,\dotsc,X_n] = X_n$.
This is the level of detail I need. This is what did to "show" that the sum of martingales is also a martingale:
Given that $(X_n)$ and $(Y_n)$ are martingales with respect to the same filtration, $E[|X_n|]<\infty$ and $E[|Y_n|]<\infty$. Then
$$E[|Z_n|] = E[|X_n+Y_n|] \leq E[|X_n|] + E[|Y_n|] <\infty,$$
and
\begin{align*}
E[Z_{n+1}|\mathcal F_n] &= E[X_{n+1}+Y_{n+1}|\mathcal F_n] \\
&= E[X_{n+1}|\mathcal F_n]+E[Y_{n+1}|\mathcal F_n] \\
&= X_n + Y_n = Z_n.
\end{align*}
Therefore $(Z_n)$ is a martingale.
So, this is the level of detail I'm expected to know. So I'm stuck showing that the maximum of two martingales is not (necessarily) a martingale.
Best Answer
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:
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