# functional analytic interpretation of the (co)variation and the doob decompostion

I have a question concerning the covariation of two time-discrete stochastic processes.

Let $(\mathcal{F}_i)_{i\in T}$ be a filtration. We call a time-discrete, real-valued, adapted process $X$ (e.i. $X : \mathbb{N} \to \mathbb{R}^{\Omega}$ and $X_i$ is $\mathcal{F}_i$-measurable) a discrete semi-martingale.

A theorem from Doob states, that every discrete semi-martingale $X$ can be uniquely represented as $X = M + A + X_0$, where $M$ is a Martingale, $A$ is a predictable process and $M_0 = A_0 = 0$. This representation is called the Doob decomposition of $X$. The process $A$ is called the compensator of $X$.

For two discrete semi-martingales $X,Y$ we call $[X,Y]_n := \sum_{i=1}^n \Delta X_i\Delta Y_i$ the variation of $X$ and $Y$. The compensator of the discrete semi-martingale $[X,Y]:=([X,Y]_n)_{n\in \mathbb{N}}$ is called the covariation of $X$ and $Y$ andis denoted by $\langle X,Y \rangle$.

One can show that the variation as well as the covariation are symmetric and bilinear. The variation maps a tuple of discrete semi-martingales onto a discrete semi-martingale. The covariation maps a tuple of discrete semi-martingales onto a predictable process.

Here is my first question: Is there any connection between a scalar product (or even a Hilbertspace) and the variation or covariation?

Here is my second question: Is there any functional analytic interpretation of the Doob decomposition (e.g. in words of projections) ?(Answered, thank you A Blumenthal!)

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Note: This question came up when I looked at the Galtschuk-Kunita-Watanabe decomposition of a Martingale. There the $L_2$ length of a residual vector of a projection looks similar to the "hilbertspace case" if one reads the "stochastic integral dot" as a multiplication, the covariation brackets as scalar product brackets and predictable processes as scalars. –  Mat Feb 14 '13 at 17:06

Here we have a filtration $\mathcal{F}_n$ on a probability space $(\Omega, \mathcal{F},\mathbb{P})$. Letting $H = L^2(\mathcal{F},\mathbb{P})$, we can see the sigma-algebra filtration as giving rise to a filtration of subspaces $H_n = L^2(\Omega,\mathcal{F}_n, \mathbb{P})$ (all this means is that $H_1 \subset H_2 \subset \cdots$ is an increasing sequence of closed subspaces of $H$ ordered by inclusion).

Since we're working with a single measure $\mathbb{P}$, from here I'll write $L^2(\mathcal{F})$ as the space of $L^2$ random variables measurable with respect to $\mathcal{F}$.

It is a fact that the conditional expectation of $L^2$ functions/random variables can be thought of as orthogonal projection onto a subspace. To wit, if $X \in L^2(\mathcal{F})$ and $\mathcal{G} \subset \mathcal{F}$ is a sub-sigma-algebra, then the mapping $X \mapsto \mathbb{E}[X \mid \mathcal{G}]$ is the orthogonal projection of $L^2(\mathcal{F}) \rightarrow L^2(\mathcal{G})$.

We can paint a geometrical picture of martingales by using this fact. Let $\{X_n\}$ be an $\mathcal{F}_n$, square-integrable martingale. Then the martingale property asserts that each martingale difference $X_{n+1}-X_n$ is the orthogonal component to $L^2(\mathcal{F}_n)$ of $X_{n+1}$; thus such a martingale is 'built up' to the present time $n$ from $X_1$ by, at each time step, adding an orthogonal piece to the previous timesteps. This should remind you of $\mathbb{R}^n$, where you can think of a vector $(x_1,\cdots,x_n)$ as being 'built up' by $(x_1,0,\cdots,0)$, then $(0,x_2,0,\cdots,0)$ and so on.

Now let $X_n$ be a 'semimartingale' (I've never heard that terminology; I call this an adapted process). The meaning of the predictable compensator $A_n$ in this context is the component of the difference $X_{n} - X_{n-1}$ lying in $L^2(\mathcal{F}_{n-1})$ (and so it is the 'nonorthogonal obstruction' to the martingale property). It should be no surprise then that it's useful to take $A_n$ to be predictable- it'll always be the part of the difference that 'spills over' into the previous subspace.

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Thank you for this nice idea. It gives us a easy proof of the existence and uniqueness of the Doob decomposition: Let $P_n$ denote the orthogonal projection on $L^2(\mathcal{F}_{n})$. We have $\Delta X_n = (I - P_{n-1})X_n + P_{n-1} X_n - X_{n-1}$. Summing up gives us the Doob decomposition. The uniqueness follows from the uniqueness of the orthogonal projections. Do you have any thoughts about my first question? With best regards –  Mat Feb 14 '13 at 20:33
@Mat I have discovered what it is you're looking for. The idea is that you can think of $E(\langle X, Y \rangle_n)$ as an actual, genuine inner product (for fixed $n$) on certain spaces of adapted processes. This picture becomes interesting in the continuous time case, where the norm $\|X\|_t = \sqrt{E\langle X \rangle_t}$ defined for a $X$ square integrable process gives rise to a function space of processes with nice properties. A reference is Karatzas and Shreve, Brownian Motion and Stochastic Calculus, Thm 1.5.22. –  A Blumenthal Mar 11 '13 at 21:31
Hello! Thank you for your answer :) I will have a look at this! With best regards Mat –  Mat Mar 13 '13 at 12:37