## Lecture 18. Square integrable martingales and quadratic variations

It turns out that stochastic integrals may be defined for other stochastic processes than Brownian motions. The key properties that were used in the above approach were the martingale property and the square integrability of the Brownian motion.

As above, we consider a filtered probability space $(\Omega,(\mathcal{F}_t)_{t \ge 0},\mathcal{F},\mathbb{P})$ that satisfies the usual conditions. A martingale $(M_t)_{t \ge 0}$ defined on this space is said to be square integrable if for every $t \geq 0$, $\mathbb{E}\left( M_t^2 \right) < + \infty$.

For instance, if $(B_t)_{t \ge 0}$ is a Brownian motion on $(\Omega, (\mathcal{F}_t)_{t \ge 0},\mathcal{F},\mathbb{P})$ and if $(u_t)_{t \ge 0}$ is a process which is progressively measurable with respect to the filtration $(\mathcal{F}_t)_{t \ge 0}$ such that for every $t \ge 0$, $\mathbb{E} \left( \int_0^t u_s^2 ds \right)<+\infty$ then, the process $M_t=\int_0^t u_s dB_s, \quad t \ge 0$ is a square integrable martingale.

The most important theorem concerning continuous square integrable martingales is that they admit a quadratic variation. Before proving this theorem, we state a preliminary lemma.

Lemma. Let $(M_t)_{0\le t \le T}$ be a continuous martingale such that
$\sup_{\Delta_n[0,T]} \sum_{k=0}^{n-1} | M_{t_{k+1}^n} -M_{t_{k}^n}|<+\infty.$
Then $(M_t)_{0\le t \le T}$ is constant.

Proof.
We may assume $M_0=0$. For $N \ge 0$, let us consider the stopping time
$T_N=\inf \left\{ s \in [0,T], |M_s| \ge N, \sup_{\Delta_n[0,s]} \sum_{k=0}^{n-1} | M_{t_{k+1}^n} -M_{t_{k}^n}| \ge N \right\}\wedge T.$
The stopped process $(M_{t \wedge T_N})_{ 0 \le t \le T }$ is a martingale and therefore for $s \le t$,
$\mathbb{E}((M_{t \wedge T_N} -M_{s \wedge T_N})^2)=\mathbb{E}(M_{t \wedge T_N}^2) -\mathbb{E}(M_{s \wedge T_N}^2).$
Consider now a sequence of subdivisions $\Delta_n[0,T]$ whose mesh tends to 0. By summing up the above inequality on the subdivision, we obtain
$\mathbb{E}(M_{ T_N}^2) =\sum_{k=0}^{n-1} \left( M_{t^n_k\wedge T_N}-M_{t^n_{k-1}\wedge T_N}\right)^2$
$\le \sup | M_{t^n_k\wedge T_N}-M_{t^n_{k-1}\wedge T_N} | \mathbb{E}\left( \sum_{k=0}^{n-1} \left| M_{t^n_k\wedge T_N}-M_{t^n_{k-1}\wedge T_N}\right|\right)$
$\le N \sup | M_{t^n_k\wedge T_N}-M_{t^n_{k-1}\wedge T_N} |.$
By letting $n \to +\infty$, we get $\mathbb{E}(M_{ T_N}^2)=0$. This implies $M_{ T_N}=0$. Letting now $N \to \infty$, we conclude $M_T=0$. $\square$

Theorem. Let $(M_t)_{t \geq 0}$ be a martingale on $(\Omega (\mathcal{F}_t)_{t \ge 0},\mathcal{F},\mathbb{P})$ which is continuous and square integrable and such that $M_0=0$.There is a unique continuous and increasing process denoted $(\langle M \rangle_t)_{t \geq 0}$ that satisfies the following properties:

• $\langle M \rangle_0=0$;
• The process $(M_t^2 - \langle M \rangle_t)_{t \geq 0}$ is a martingale.

Actually for every $t \ge 0$ and for every sequence of subdivisions $\Delta_n [0,t]$ such that $\lim_{n \rightarrow +\infty}\mid\Delta_n [0,t]\mid=0,$
the following convergence takes place in probability:
$\lim_{n \rightarrow +\infty} \sum_{k=1}^{n} \left( M_{t^n_k}-M_{t^n_{k-1}}\right)^2=\langle M \rangle_t.$
The process $(\langle M \rangle_t)_{t \geq 0}$ is called the quadratic variation process of $(M_t)_{t \geq 0}$.

Proof.
We first assume that the martingale $(M_t)_{t \geq 0}$ is bounded and prove that if $\Delta_n [0,t]$ is a sequence of subdivisions of the interval $[0,t]$ such that
$\lim_{n \rightarrow +\infty}\mid\Delta_n [0,t]\mid=0,$
then the limit
$\lim_{n \rightarrow +\infty} \sum_{k=1}^{n} \left( M_{t^n_k}-M_{t^n_{k-1}}\right)^2$
exists in $L^2$ and thus in probability.

Toward this goal, we introduce some notations. If $\Delta [0,T]$ is a subdivision of the time interval $[0,T]$ and if $(X_t)_{t\ge 0}$ is a stochastic process, then we denote
$S_t^{\Delta [0,T]}(X)=\sum_{i=0}^{k-1}\left( X_{t_{i+1}} -X_{t_i} \right)^2 +(X_t-X_{t_k})^2,$
where $k$ is such that $t_k \le t \le t_{k+1}$,

An easy computation on conditional expectations shows that if $(X_t)_{t\ge 0}$ is a martingale, then the process $X_t^2-S_t^{\Delta [0,T]}(X), \quad t \le T$ is also a martingale. Also, if $\Delta [0,T]$ and $\Delta' [0,T]$ are two subdivisions of the time interval $[0,T]$, we will denote by $\Delta \vee \Delta' [0,T]$ the subdivision obtained by putting together the points $\Delta [0,T]$ and the points of $\Delta' [0,T]$. Let now $\Delta_n [0,T]$ be a sequence of subdivisions of $[0,T]$ such that
$\lim_{n \rightarrow +\infty}\mid\Delta_n [0,T]\mid=0.$
Let us show that the sequence $S_T^{\Delta_n [0,T]}(M)$ is a Cauchy sequence in $L^2$. Since the process $S^{\Delta_n [0,T]}(M)-S^{\Delta_p [0,T]}(M)$ is a martingale (as a difference of two martingales), we deduce that
$\mathbb{E}\left( \left(S_T^{\Delta_n [0,T]}(M)-S_T^{\Delta_p [0,T]}(M) \right)^2 \right)$
$= \mathbb{E}\left(S_T^{\Delta_n \vee \Delta_p [0,T]}(S^{\Delta_n [0,T]}(M)-S^{\Delta_p [0,T]}(M))\right)$
$\le 2 \left( \mathbb{E}\left(S_T^{\Delta_n \vee \Delta_p [0,T]}(S^{\Delta_n [0,T]}(M))\right)+\mathbb{E}\left(S_T^{\Delta_n \vee \Delta_p [0,T]}(S^{\Delta_p [0,T]}(M))\right) \right).$
Let us denote by $s_k$‘s the points of the subdivision $\Delta_n \vee \Delta_p [0,T]$ and for fixed $s_k$, we denote by $t_l$ the point of $\Delta_n [0,T]$ which is the closest to $s_k$ and such that $t_l \le s_k \le t_{l+1}$. We have
$S_{s_{k+1}}^{\Delta_n [0,T]}(M)-S_{s_{k}}^{\Delta_n [0,T]}(M)=(M_{s_{k+1}} -M_{t_l})^2-(M_{s_{k}} M_{t_l})^2$
$=(M_{s_{k+1}} -M_{s_k})(M_{s_{k+1}} +M_{s_k}-2M_{t_l}).$
Therefore, from Cauchy-Schwarz inequality,
$\mathbb{E}\left(S_T^{\Delta_n \vee \Delta_p [0,T]}(S^{\Delta_n [0,T]}(M))\right)\le \mathbb{E} \left( \sup_k (M_{s_{k+1}} +M_{s_k}-2M_{t_l})^4\right)^{1/2}\mathbb{E} \left( \left(S_T^{\Delta_n \vee \Delta_p [0,T]}(M) \right)^2\right)^{1/2}.$
Since the martingale $M$ is assumed to be continuous, when $n,p \rightarrow +\infty$,
$\mathbb{E} \left( \sup_k (M_{s_{k+1}} +M_{s_k}-2M_{t_l})^4\right) \rightarrow 0.$
Thus, in order to conclude, it suffices to prove that $\mathbb{E} \left( \left(S_T^{\Delta_n \vee \Delta_p [0,T]}(M) \right)^2\right)$ is bounded. This fact is an easy consequence of the fact that $M$ is assumed to be bounded. Therefore, in the $L^2$ sense the following convergence holds
$\langle M \rangle_t =\lim_{n \rightarrow +\infty} \sum_{k=1}^{n}\left( M_{t^n_k} -M_{t^n_{k 1}}\right)^2.$
The process $(M_t^2 - \langle M \rangle_t)_{t \geq 0}$ is seen to be a martingale because for every $n$ and $T \ge 0$, the process $M_t^2-S_t^{\Delta_n [0,T]}(M), \quad t \le T$ is a martingale. Let us now show that the obtained process $\langle M \rangle$ is a continuous process. From Doob’s inequality, for $n,p \ge 0$ and $\varepsilon > 0$,
$\mathbb{P}\left( \sup_{0 \le t \le T} \left(S_t^{\Delta_n[0,T]}(M)-S_t^{\Delta_p [0,T]}(M) \right) \ge \varepsilon \right)\le \frac{\mathbb{E}\left( \left(S_T^{\Delta_n[0,T]}(M)-S_T^{\Delta_p[0,T]}(M) \right)^2\right)}{\varepsilon^2}.$

From Borel-Cantelli lemma, there exists therefore a sequence $n_k$ such that the sequence of continuous stochastic processes $\left( S_t^{\Delta_{n_k} [0,T]}(M)\right)_{0 \le t \le T}$ almost surely uniformly converges to the process $\left( \langle M \rangle_t\right)_{0 \le t \le T}$. This proves the existence of a continuous version for $\langle M \rangle$. Finally, to prove that $\langle M \rangle$ is increasing, it is enough to consider a an increasing sequence of subdivisions whose mesh tends to $0$. Let us now prove that $\langle M \rangle$ is the unique process such that $M^2-\langle M \rangle$ is a martingale. Let $A$ and $A'$ be two continuous and increasing stochastic processes such that $A_0=A'_0=0$ and such that $(M_t^2 -A_t)_{ t\ge 0}$ and $(M_t^2 -A'_t)_{ t\ge 0}$ are martingales. The process $(N_t)_{t\ge 0}=(A_t -A'_t)_{t\ge 0}$ is then seen to be a martingale that has a bounded variation. From the previous lemma, this implies that $(N_t)_{t\ge 0}$ is constant and therefore equal to 0 due to its initial condition.

We now turn to the case where $(M_t)_{t \ge 0}$ is not necessarily bounded. Let us introduce the sequence of stopping times:
$T_N=\inf \{ t \ge 0, |M_t | \ge N \}.$
According to the previous arguments, for every $N \ge 0$, there is an increasing process $A^N$ such that $(M_{t\wedge T_N}^2-A^N_t)_{t \ge 0}$ is a martingale. By uniqueness of this process, it is clear that $A^{N+1}_{t\wedge T_N}=A^N_t$, therefore we can define a process $A_t$ by requiring that $A_t(\omega)= A^N_t(\omega)$ provided that $T_N(\omega)\ge t$. By using convergence theorems, it is then checked that $(M_t^2-A_t)_{t \ge 0}$ is a martingale.

Finally, let $\Delta_n[0,t]$ be a sequence of subdivisions whose mesh tends to $0$. We have for every $\varepsilon >0$,
$\mathbb{P} \left( \left|A_t - \sum_{k=1}^{n} \left( M_{t^n_k}-M_{t^n_{k-1}}\right)^2 \right|\ge \varepsilon \right)$
$\mathbb{P} (T_N \le t)+\mathbb{P} \left( \left|A^N_t - \sum_{k=1}^{n} \left( M_{t^n_k\wedge T_N}-M_{t^n_{k-1}\wedge T_N}\right)^2 \right|\ge \varepsilon \right).$
This easily implies the announced convergence in probability of the quadratic variations to $A_t$ $\square$

Exercise. Let $(M_t)_{t \geq 0}$ be a square integrable martingale on a filtered probability space $(\Omega,(\mathcal{F}_t)_{t \ge 0},\mathcal{F},\mathbb{P})$. Assume that $M_0=0$. If $\Delta [0,T]$ is a subdivision of the time interval $[0,T]$ and if $(X_t)_{t\ge 0}$ is a stochastic process, we denote
$S_t^{\Delta [0,T]}(X)=\sum_{i=0}^{k-1}\left( X_{t_{i+1}} -X_{t_i} \right)^2 +(X_t-X_{t_k})^2,$
where $k$ is such that $t_k \le t . Let $\Delta_n [0,T]$ be a sequence of subdivisions of $[0,T]$ such that
$\lim_{n \rightarrow +\infty}\mid\Delta_n [0,T]\mid=0.$latex
Show that the following convergence holds in probability,
$\lim_{n \rightarrow +\infty} \sup_{0\le t \le T} \left| S_t^{\Delta [0,T]}(M) - \langle M \rangle_t \right|=0.$
Thus, in the previous theorem, the convergence is actually uniform on compact intervals.

We have already pointed out that stochastic integrals with respect to Brownian motion provide an example of square integrable martingale, they therefore have a quadratic variation. The next proposition explicitly computes this variation.

Proposition. Let $(B_t)_{t \ge 0}$ be a Brownian motion on a filtered probability space $(\Omega, (\mathcal{F}_t)_{t \ge 0},\mathcal{F},\mathbb{P})$ that satisfies the usual conditions. Let $(u_t)_{t \ge 0}$be a progressively measurable process such that for every $t \ge 0$, $\mathbb{E} \left( \int_0^t u_s^2 ds \right)<+\infty$. For $t \ge 0$:
$\left\langle \int_0^{\cdot} u_s dB_s \right\rangle_t=\int_0^t u_s^2ds.$

Proof.
Since the process $\left( \int_0^t u_s^2ds \right)_{t \ge 0}$ is continuous, increasing and equals $0$ when $t=0$, we just need to prove that
$\left( \int_0^{t} u_s dB_s \right)^2 - \int_0^t u_s^2ds$
is a martingale.

If $u \in \mathcal{E}$, is a simple process, it is easily seen that for $t \ge s$:
$\mathbb{E} \left( \left( \int_0^{t} u_v dB_v \right)^2 \mid\mathcal{F}_s \right)$
$=\mathbb{E} \left( \left( \int_0^{s} u_v dB_v +\int_s^{t} u_v dB_v \right)^2 \mid \mathcal{F}_s \right)$
$=\mathbb{E} \left( \left( \int_0^{s} u_v dB_v \right)^2 \mid \mathcal{F}_s \right)+\mathbb{E} \left( \left( \int_s^{t} u_v dB_v \right)^2 \mid \mathcal{F}_s \right)$
$=\left( \int_0^{s} u_v dB_v \right)^2 +\mathbb{E} \left( \int_s^{t} u_v^2 dv \mid \mathcal{F}_s \right).$
We may then conclude by using the density of $\mathcal{E}$ in $L^2 (\Omega,(\mathcal{F}_t)_{t \ge 0},\mathbb{P})$ $\square$

As a straightforward corollary of the existence of a quadratic variation for the square integrable martingales, we immediately obtain:

Corollary. Let $(M_t)_{t \geq 0}$ and $(N_t)_{t \ge 0}$ be two continuous square integrable martingales on $(\Omega, (\mathcal{F}_t)_{t \ge 0},\mathcal{F},\mathbb{P})$ such that $M_0=N_0=0$. There is a unique continuous process $(\langle M ,N \rangle_t)_{t \geq 0}$ with bounded variation that satisfies:

• $\langle M ,N\rangle_0=0$;
• The process $(M_t N_t - \langle M ,N \rangle_t)_{t \geq 0}$ is a martingale.

Moreover, for $t \ge 0$ and for every sequence $\Delta_n [0,t]$ such that $\lim_{n \rightarrow +\infty}\mid\Delta_n [0,t]\mid=0$, the following convergence holds in probability:
$\lim_{n \rightarrow +\infty} \sum_{k=1}^{n} \left( M_{t^n_k} -M_{t^n_{k-1}}\right)\left( N_{t^n_k} -N_{t^n_{k-1}}\right)=\langle M,N \rangle_t.$
The process $(\langle M ,N\rangle_t)_{t \geq 0}$ is called the quadratic covariation process of $(M_t)_{t \geq 0}$ and $(N_t)_{t \geq 0}$.

Proof.
We may actually just use the formula
$\langle M, N \rangle =\frac{1}{4} \left( \langle M+N \rangle - \langle M- N \rangle \right),$
as a definition of the covariation and then check that the above properties are indeed satisfied $\square$

Exercise. Let $(B_t^1)_{t\ge 0}$ and $(B_t^2)_{t \ge 0}$ be two independent Brownian motions. Show that $\langle B^1, B^2 \rangle_t =0$.

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