$\max (X_t, a)$ is not a martingale, counterexample

Question

It is quite easy to prove that if $[X_t \}_{t \in [0,T]}$ is a martingale, then for any number $a \in \mathbb{R}$ $ \{\max (X_t,a)\} _t$ is a submartingale and $ \{\min (X_t, a)\} _t$ but I cannot come up with a counterexample which proves that those need not be martingales.

That is, I want to find a martingale $\{X_t\}_{t \in [0,T]}$ and a constant $a \in \mathbb{R}$ such that $$\mathbb{E}\left(\max (X_t,a) | F_s \right) < \max(X_s, a)$$ for all $s \le t$ and adequately for minimum: $$\mathbb{E}\left(\min (X_t,a) | F_s \right) > \min(X_s, a)$$ for all $s \le t$

Could you tell me how/where to look for such examples?

Answer 1

Hint: An easier way to show that $(Z_t)$ is not a martingale is to find some $t$ with $\mathbb{E}(Z_0)\neq \mathbb{E}(Z_t)$.

Answer 2

Let $(\Omega, \mathscr F, \{\mathscr F_t\}_{t \in [0,T]}, \mathbb P)$ be a filtered probability space where $T > 0$.

If we have two $(\{\mathscr F_t\}_{t \in [0,T]}, \mathbb P)-$martingales $X$ and $Y$, it does not follow that $\max(X,Y)$ is a $(\{\mathscr F_t\}_{t \in [0,T]}, \mathbb P)-$martingale. $\tag{*}$

Now, $\forall a \in \mathbb R$ the stochastic process $A = (A_t)_{t \in [0,T]}$ given by $a = A_t \ \forall t \in [0,T]$ is a $(\{\mathscr F_t\}_{t \in [0,T]}, \mathbb P)-$martingale.

Following the proof of $(*)$, we can show that it does not follow that $\max(X,A) = (\max(X_t,a))_{t \in [0,T]}$ is a $(\{\mathscr F_t\}_{t \in [0,T]}, \mathbb P)-$martingale.

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Pf based on answers here:

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

Let $X_0, X_1, X_2, ...$ be iid random variables s.t.

$$P(X_i = 1) = 1/2 = P(X_i = -1)$$

Define $W^X_n := \sum_{i=0}^{n} X_i$.

Consider the filtration $\mathscr F_{n} := \mathscr F_{n}^X$ for $n = 0,1,...$

Note that $W^X_n$ is a martingale w/rt $\mathscr F_{n}$.

Let $a \in \mathbb R$. Consider $M = (M_n)_{n \in \mathbb N}$ where

$$M_n := \max(W_n^X, a) = W_n^X 1_{A_n} + a 1_{A_n^C}$$

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

Let us try to show (and hopefully fail in trying) that $E[M_n|\mathscr F_{n-1}] = M_{n-1}$:

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

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

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

Now consider this. If at time 1, we have that $W_1^X = a + 0.5$ it's possible that at time 2, we have $W_2^X = a - 0.5$. Thus, $M_2 = a$ but $M_1 = W_1^X$

Formally:

If at time n, we have $\omega \in A_n$ (ie $W_n^X(\omega) \ge a$), 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 $W_{n-1}^X(\omega) \le a$). Hence, $M_{n-1} = a$

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

I hope you don't mind that I used $\mathbb N$ instead of $[0,T]$ :P

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$(**)$

Oh also $P(A_n) \notin \{0,1\}$ is important:

If $P(A_j) = 0$ or $1 \ \forall j \in \mathbb N$, then $M_n$ is $W_n^X$ or a a.s. Hence, $M$ is a $(\{\mathscr F_n\}_{(n \in \mathbb N)}, \mathbb P)-$martingale (at least almost surely) because such implies $1_{A_1} = 1_{A_2} = ...$ a.s., in particular, $1_{A_n} = 1_{A_{n-1}}$ a.s..

This might be relevant: Is a probability of 0 or 1 given information up to time t unchanged by information thereafter?