# Maximum of martingales

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 $n = 0,1,2,\dotsc,$

1. $E[|X_n|] <\infty$, and
2. $E[X_{n+1}|X_0,\dotsc,X_n] = X_n$.

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.

• Take two independent simple random talks $S_1(n)$ and $S_2(n)$. If they are NOT equal to each other at time $n$, then $E(M_{n+1}|M_n)=M_n$. However if $S_1(n)=S_2(n)$, then $E(M_{n+1}|M_n)=M_n+\frac 3 4 - \frac 1 4\ne M_n$ - hence $M_n$ is not a martingale.
– A.S.
Dec 7, 2015 at 16:12
• I think I get it now. We know $X_n$ and $Y_n$ are martingales w/rt their natural filtrations (by definition). It is additionally given that $X_n$ and $Y_n$ are martingales w/rt to some filtration $\mathscr F_n$. Now we must show that $W_n$ is a martingale w/rt to its natural filtration. Is that right?
– BCLC
Dec 7, 2015 at 19:40
• @BCLC Yes, that's correct. Now I must prove or disprove that $W_n$ is a martingale. My TA said that $W_n$ does not have to be a martingale. So I am trying to disprove it with a counterexample. Dec 8, 2015 at 5:17
• @Oscar How is our $W_n$ not a counterexample?
– BCLC
Dec 8, 2015 at 7:02
• @BCLC I don't know how to show that it is a counterexample. That is where I'm stuck. Dec 8, 2015 at 7:06

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..

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.

$$E[W_n | \mathscr F_{n-1}] = E[X_n1_{A_n} + Y_n1_{A_n^C} | \mathscr F_{n-1}]$$

$$= E[X_n1_{A_n} + Y_n1_{A_n^C} | \mathscr F_{n-1}]$$

$$= E[X_n1_{A_n}| \mathscr F_{n-1}] + E[Y_n1_{A_n^C} | \mathscr F_{n-1}]$$

$$= X_{n-1} E[2^{n-(n-1)} \prod_{k=(n-1)+1}^{n} U_k 1_{A_n}| \mathscr F_{n-1}] + Y_{n-1} E[p^{(n-1)-n} \prod_{k=(n-1)+1}^{n} \xi_k 1_{A_n^C} | \mathscr F_{n-1}]$$

$$= X_{n-1} E[2 U_n 1_{A_n}| \mathscr F_{n-1}] + Y_{n-1} E[p^{-1} \xi_n 1_{A_n^C} | \mathscr F_{n-1}]$$

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:

1. $\sigma(\sigma(U_1, ..., U_n) \cup \sigma(\xi_1, ..., \xi_n))$
2. $\sigma(\sigma(X_1, ..., X_n) \cup \sigma(Y_1, ..., Y_n))$
3. $\sigma(W_1, ..., W_n)$

I think $(3) \subseteq (2) \subseteq (1)$

I think $\sigma(A_1, ..., A_n) \subseteq (2), \subseteq (1), \subsetneq (3)$

$(**)$

FYI

$$\sigma(U_1, ..., U_n) \supseteq \mathscr F_n^X$$

$$\sigma(\xi_1, ..., \xi_n) \supseteq \mathscr F_n^Y$$

• Thanks for your response, but I don't have to go into this much detail I think. I'm in an introductory course. We didn't cover $\sigma-$algebras or filtrations in any depth. Dec 7, 2015 at 19:00
• @Oscar What don't you understand? Just ignore $\mathbb P$ I guess. Also you can replace $m$ with $n-1$. Anyway, the main idea is rewriting $W_n$ using indicator functions. Going to edit.
– BCLC
Dec 7, 2015 at 19:49
• @Oscar Oh right, edited
– BCLC
Dec 7, 2015 at 21:04
• @Oscar Edited..
– BCLC
Dec 8, 2015 at 13:37

Oscar, are you sure you know why your examples are 'martingales'?

(proofs in first and second parts are just formalities)

Let $(\Omega, \mathscr F, \mathbb P)$ be our probability space.

Let $Y_0, Y_1, Y_2, ...$ be iid random variables s.t. $P(Y_i = 1) = 1/2 = P(Y_i = -1)$. Define $W^Y_n := \sum_{i=0}^{n} Y_i$.

Let $X_0, X_1, X_2, ...$ be iid random variables s.t. $P(X_i = 1) = 1/2 = P(X_i = -1)$. Define $W^X_n := \sum_{i=0}^{n} X_i$.

Firstly, note that $W^X_n$ and $W^Y_n$ martingales w/rt their natural filtrations (this corresponds to your definition of martingale) ie

$$E[W^X_n|W^X_0, W^X_1, ..., W^X_{n-1}] = E[W^X_n|\mathscr F_{n-1}^{W^X}] = W^X_{n-1}$$

$$E[W^Y_n|W^Y_0, W^Y_1, ..., W^Y_{n-1}] = E[W^Y_n|\mathscr F_{n-1}^{W^Y}] = W^Y_{n-1}$$

They are also martingales w/rt to the natural filtrations of X and Y, resp ie

$$E[W^X_n|X_0, X_1, ..., X_{n-1}] = E[W^X_n|\mathscr F_{n-1}^{X}] = W^X_{n-1}$$

$$E[W^Y_n|Y_0, Y_1, ..., Y_{n-1}] = E[W^Y_n|\mathscr F_{n-1}^{Y}] = W^Y_{n-1}$$

Actually, the latter equations are easier to prove than the former equations.

Proof of the former equations (I'll just use X):

$$\mathscr F_{n-1}^{X} \supseteq \mathscr F_{n-1}^{W^X} \tag{*}$$

Then we have

$$E[W^X_n|\mathscr F_{n-1}^{W^X}] = E[X_0 + X_1 + ... + X_{n-1} + X_n|\mathscr F_{n-1}^{W^X}]$$

$$= E[X_0 + X_1 + ... + X_{n-1}|\mathscr F_{n-1}^{W^X}] + E[X_n|\mathscr F_{n-1}^{W^X}]$$

Consider the first term.

$$E[X_0 + X_1 + ... + X_{n-1}|\mathscr F_{n-1}^{W^X}] = \sum_{k=0}^{n-1} E[X_{i}|\mathscr F_{n-1}^{W^X}]$$

Consider the summand. Convince yourself that $X_i$ is $\mathscr F_{i}^{W^X}$-measurable (if we know $X_0, X_0 + X_1, ..., X_0 + X_1 + ... + X_i$, then we know $X_i$). Then $\because \mathscr F_{i}^{W^X} \subseteq \mathscr F_{n-1}^{W^X}$, we have that $X_i$ is $\mathscr F_{n-1}^{W^X}$-measurable. Hence,

$$E[X_{i}|\mathscr F_{n-1}^{W^X}] = X_i$$

Consider the second term $$E[X_n|\mathscr F_{n-1}^{W^X}]$$

By $(*)$ and the the tower property of conditional expectation, we have

$$E[X_n|\mathscr F_{n-1}^{W^X}] = E[E[X_n|\mathscr F_{n-1}^{W^X}]|\mathscr F_{n-1}^{X}]$$

Essentially, the tower property of conditional expectation states that the smaller filtration wins. Hence, it is also true that

$$E[X_n|\mathscr F_{n-1}^{W^X}] = E[E[X_n|\mathscr F_{n-1}^{X}]|\mathscr F_{n-1}^{W^X}]$$

Thus, we have

$$E[X_n|\mathscr F_{n-1}^{W^X}] = E[E[X_n|\mathscr F_{n-1}^{W^X}]|\mathscr F_{n-1}^{X}] = E[E[X_n|\mathscr F_{n-1}^{X}]|\mathscr F_{n-1}^{W^X}]$$

By independence, we have:

$$= E[E[X_n]|\mathscr F_{n-1}^{W^X}] = E[X_n] = 0$$

Hopefully, you're convinced that it's much easier to prove the latter than the former equations.

Secondly, consider

$$M_n = W_n^X 1_{A_n} + W_n^Y 1_{A_n^C}$$

where $A_n = \{W_n^X \ge W_n^Y\}$ and $0 < P(A_n) < 1$

Let us try to show (and hopefully fail in trying) that

$$E[M_n|\mathscr F_{n-1}^M] = M_{n-1}$$

where $\mathscr F_{n-1}^M = \sigma(M_0, M_1, ..., M_{n-1})$

Hopefully, you understand that it's possible to do the following:

$$E[M_n|\mathscr F_{n-1}^M] = E[E[M_n|\mathscr F_{n-1}]|\mathscr F_{n-1}^M]$$

where $\mathscr F_{n} := \sigma(\mathscr F_{n}^X, \mathscr F_{n}^Y) \supseteq \sigma(\mathscr F_{n}^{W^X}, \mathscr F_{n}^{W^Y}) \supseteq \mathscr F_n^M$

Thus, if $E[M_n|\mathscr F_{n-1}] = M_{n-1}$, then $E[M_n|\mathscr F_{n-1}^M] = M_{n-1}$. So hopefully, $E[M_n|\mathscr F_{n-1}] \ne M_{n-1}$.

Finally, let's compute

$$E[M_n|\mathscr F_{n-1}] = E[W_n^X 1_{A_n} + W_n^Y 1_{A_n^C}|\mathscr F_{n-1}]$$

$$= E[W_n^X 1_{A_n}|\mathscr F_{n-1}] + E[W_n^Y 1_{A_n^C}|\mathscr F_{n-1}]$$.

Now consider this. If at time 1, we have that $W_1^X = 2$ and $W_1^Y = 1$, it's possible that $W_2^X = 1$ and $W_2^Y = 2$. Thus, $M_2 = W_2^Y$ but $M_1 = W_1^X$

Formally:

If at time n, we have $\omega \in A_n$ (ie $W_n^X(\omega) \ge W_n^Y(\omega)$), then

$$E[M_n|\mathscr F_{n-1}] = E[W_n^X 1_{A_n}|\mathscr F_{n-1}] + 0$$

$$= E[W_n^X|\mathscr F_{n-1}] = W_{n-1}^X$$

But is it that $W_{n-1}^X = M_{n-1}$?

Not necessarily. It could be that $\omega \in A_{n-1}^C$ (ie $W_{n-1}^X(\omega) \le W_{n-1}^Y(\omega)$). Hence, $M_{n-1} = W_{n-1}^Y$

$\therefore, \because$ it is not necessarily true that $E[M_n|\mathscr F_{n-1}] = M_{n-1}$, it does not follow that $E[M_n|\mathscr F_{n-1}^M] = M_{n-1}$ QED