Martingale representation theorem

Remarks

As a consequence, every square-integrable martingale $M$ for $(\mathcal{F}_t)$ has the form $M_t = M_0 + \int_0^t H_s \, dB_s$, and so has a continuous version. For the proof, the random variables $\mathcal{E}\left(\int_0^{\cdot} f \, dB\right)_T$ with deterministic $f$ are stochastic exponentials with explicit representations. Their linear span is dense in $L^2(\mathcal{F}_T)$, and the Itô isometry passes the representation to limits. Example: Itô's lemma gives $B_T^2 = T + \int_0^T 2 B_t \, dB_t$, so $H_t = 2 B_t$. The filtration must be generated by $B$: if $\mathcal{F}_0$ contains a nonconstant variable $\xi$ independent of $B$ with $E[\xi] = 0$, then $\xi$ has no such representation, because $E\left[\int_0^T H \, dB \mid \mathcal{F}_0\right] = 0 \neq \xi$.

Going deeper

For a student

The theorem is the mathematical basis of hedging in complete markets, such as the Black–Scholes model. Combined with the Girsanov theorem, it shows that each claim can be replicated by trading in the stock. The Clark–Ocone formula of Malliavin calculus identifies the integrand $H$ for differentiable $X$. The result also holds for a $d$-dimensional Brownian motion, with a vector-valued integrand.