15 Martingales
15.1 Definitions
Definition 15.1 (Filtration) A collection of sigma algebras (\mathcal F_i)_{i \in T} indexed by some set T \subset \mathbb R such that \mathcal F_s \subset \mathcal F_t for s< t \in I.
Definition 15.2 (Adaptation) A sequence (X_i) is adapted to (\mathcal F_i) if X_i \in \mathcal F_i for all i \in T.
Definition 15.3 (Natural filtration) For some given stochastic process (X_t)_{t \in T}, the natural filtration is:
\mathcal F_t = \sigma \{ X_s : s \leq t\}
Definition 15.4 (Martingale) Say we have a stochastic process (X_t) adapted to some filtration (\mathcal F_t).
We say the process is a martingale if for all s > t \in T, \mathbb E[X_s | \mathcal F_t] = X_t
A submartingale if for all s > t, \mathbb E[X_s | \mathcal F_t] \geq X_t
and a supermartingale if
\mathbb E[X_s | \mathcal F_t] \leq X_t
For discrete time space, the equivalent condition is to check \mathbb E[X_{t+1} | \mathcal F_t] for all t.
Definition 15.5 A process (H_n) is predictable wrt a filtration (\mathcal F_n) if H_n \in \mathcal F_{n-1} for all n.
Definition 15.6 (Betting on a martingale) Let (X_n) (stock price) be a adapted and (H_n) (number of shares held between n-1 and n) be predictable wrt (\mathcal F_n). Define the (profit) process Y_n = \sum_{i \leq n} H_i \cdot (X_i - X_{i-1})
Then (Y_n) is adapted to (\mathcal F_n)
If (X_n) is a martingale, then so is (Y_n).
If (X_n) is a submartingale and H_n \geq 0 for all n, then (Y_n) is a submartingale.
Proof
\begin{align*} Y_{n+1} &= Y_n + H_{n+1} (X_{n+1} - X_{n})\\ \mathbb E[Y_{n+1}|\mathcal F_n] &= Y_n + H_{n+1} \mathbb E[ (X_{n+1} - X_{n}) |\mathcal F_n]\\ &= Y_n + H_{n+1} (\mathbb E[ X_{n+1} |\mathcal F_n]- X_{n}) \end{align*}
15.2 Convergence
15.2.1 Bounded martingale convergence theorem
If (X_n) is a submartingale and \mathbb E[X_n^+] is bounded above i.e. \sup_{n \in T} \mathbb E[X_n^+] < \infty then there exists an integrable random variable X_\infty with \mathbb E X_\infty < \infty such that X_n \stackrel{a.s.}{\to} X_\infty
Similarly, for a supermartingale (X_n), if \mathbb E[X_n^-] is bounded then it converges almost surely. A special case of this is that non negative supermartingales (and thus martingales) converge almost surely to a finite limit.
Proof
We prove for the supermatingale case. Let R = \lim \inf X_n and S= \lim \sup X_n. We first verify that R \stackrel{a.s.}{=} S, for which we need to rule out oscillating tail behaviour.
If there exist a< b, such that R < a < b < s, then U_\infty (a, b) = \infty.
We know by Lemma 15.1 and by the assumption that the expected negative part is bounded:
\begin{align*} \mathbb E U_n(a,b) &\leq \frac{\mathbb E[(X_n-a)^-]}{b-a} \leq M(a, b)\\ \mathbb E U_\infty(a,b) &\leq M(a, b) \qquad \text{monotone convergence}\\ &\implies \mathbb P(\{ U_\infty(a,b) = \infty\}) = 0\\ &\implies \mathbb P(B_{(a,b)} ) = 0 \end{align*}
Now this is true for any fixed a and b. Now R<S \iff \exists a, b \in \mathbb Q ; R< a< b < S.
\mathbb P(R < S) = \mathbb P( \cup_{a, b \in \mathbb Q} B_{(a,b)}) \leq \sum_{a, b \in \mathbb Q} \mathbb P( B_{(a,b)}) = 0
Hence R \stackrel{a.s.}{=} S so there exists a limiting random variable.
Now to rule out it doesn’t go to \infty, we can use Fatou’s lemma, which states that for a sequence of non-negative RVs Y_n
\begin{align*} \mathbb E[\lim \inf Y_n ] &\leq \lim \inf \mathbb E[ Y_n]\\ \mathbb E[\lim \inf X_n^-] &\leq \lim \inf \mathbb E[ X_n^-] \qquad \text{which is bounded} \\ \mathbb E[\lim \inf X_\infty^-] &\leq c \end{align*}
Now
\begin{align*} X_n &= X_n^+ - X_n^-\\ X_n^+ &= X_n + X_n^-\\ \mathbb E [ X_n^+] &= \mathbb E[(X_n + X_n^-)] \\ &\leq \mathbb E[X_0] + \sup_n \mathbb E X_n^- \\ \mathbb E[X_\infty^+] &= \mathbb E[\lim X_n^+] = \mathbb E[\lim \inf X_n^+] \leq \lim \inf \mathbb E[X_n^+] < \infty\\ \end{align*}
Thus \mathbb E[|X_\infty|]| = \mathbb E[X_\infty^+] + \mathbb E[X_\infty^-] < \infty.Definition 15.7 (Upcrossing) The number of complete trips the price makes from a lower threshold to an upper one.
U_{n}^X (a,b) = \sup \{m: \exists s_1 < t_1 \dots s_m < t_m \leq n, X_{s_i} \leq a, X_{t_i} \geq b \}
To ensure uniqueness, we define s_i to be first time step after s_{i-1} where X_i \leq a and similarly for t_i.
Lemma 15.1 (Doob’s upcrossing lemma) If X_n is a supermartingale, and a < b then \mathbb E[U_n(a,b) ] \leq \frac{\mathbb E[(X_n-a)^-]}{b-a}
Proof
Let Y_n be the betting profit process. Z_n = \sum_{i \leq n} H_i (X_i - X_{i-1}) where H_n \in \{0, 1\}. Then Y_n is also a supermartingale and \mathbb E[Y_n] \leq 0 since Y_0 = 0 by convention.
Now observe that the profit is bounded below by the number of upcrossings times width, minus the final loss (if there is in fact a loss in the end). \begin{align*} Y_n &\geq (b-a) U_n(a,b) - (X_n - a)^- \\ (b-a) \mathbb E [U_n(a,b)] - \mathbb E[(X_n - a)^-] &\leq \mathbb E[Y_n] \leq 0 \\ \end{align*}15.2.2 L^p martingale convergence theorem
If (X_n) is a submartingale and is bounded in L^p for some p >1 i.e. there exists c such that \sup_n \|X_n\|_p < c for some c then there exists an integrable random variable X_\infty with \mathbb E |X_\infty|^p < \infty such that
X_n \stackrel{a.s.}{\to} X_\infty and \|X_n - X_\infty\|_p \to 0 i.e X_n \stackrel{L^p}{\to} X_\infty
15.3 Stopping times
Definition 15.8 (Stopping time) A random variable \tau : \Omega \to T \cup \{\infty\} is a stopping time wrt a filtration (\mathcal F_t) if for all t \in T, the set \{\tau \leq t\} \in \mathcal F_t.
Theorem 15.1 If (X_n) is a submartingale wrt (\mathcal F_n) and \tau is a stopping time, then the stopped process X_{n \wedge \tau} is also a submartingale wrt (\mathcal F_n).
Similarly for martingales and supermartingales.
Proof
Define the non-negative predictable process H_i = \mathbb I(i \leq \tau).
We can use it to define a betting process. X_{n\wedge \tau} = \sum H_i (X_i - X_{i-1}) + X_0\\
If (X_n) is a submartingale, then by ?lem-bettingmtg, (X_{n \wedge \tau}) is also a submartingale.Let \tau be a bounded stopping time such that \mathbb P(\tau \leq m) = 1 for some finite m. If (X_n) is a submartingale then
\mathbb E[X_0] \leq \mathbb E[X_\tau] \leq \mathbb E[X_m]
Proof
We know Y_n = X_{n \wedge \tau} is a submartingale by Theorem 15.1.
\begin{align*} \mathbb E[Y_m|\mathcal F_0] &\geq \mathbb E[Y_0] \\ \implies \mathbb E[Y_m] &\geq \mathbb E[Y_0] \\ \implies \mathbb E[X_\tau] &\geq \mathbb E[X_0] \end{align*}
Now for n \geq \tau define the process Z_n = X_n - X_\tau = \sum_{i=1}^n 1_{[\tau+1, n]}(i) \cdot (X_i - X_{i-1}). By ?lem-bettingmtg, (Z_n)_{n \geq \tau} is a submartingale. Thus, \mathbb E[Z_m] = \mathbb E[X_m - X_\tau] \geq \mathbb E[Z_\tau]= 0