This post is a collection of notes about Martingales.
A Martingales is a stochastic process that is driftless (aka it is pure randomness or just volatility). We also say that martingales are constant mean stochastic process.
Continuous time martingales
A continuous time stochastic process \(\{ M_t: t \in \mathbb{R}^+ \}\) such that \(M_t\) is adapted to \(\mathcal{F}_t\) (or is \(\mathcal{F}_t\) measurable) is a martingale if:
- integrability condition: \(\mathbb{E}[M_t] \lt \infty\)
- conditional expectation condition: \(\mathbb{E}_s[M_{t}|\mathcal{F}_s] = M_s, \space 0 \leq s \leq t\)
Link between itô integrales and martingales
Let \(X\) be a stochastic process and \(F = X^2(t)\). Recall then \[F(t) = t + 2 \int_0^t X_{\tau} \space d{\tau}\] \[X^2(t) = t + 2 \int_0^t X_{\tau} \space d{\tau}\] Taking expectation on both side \[\mathbb{E} \left[ X^2(t) \right] = t + \mathbb{E} \left[ 2 \int_0^t X_{\tau} \space d{\tau} \right]\] With the quadratic variation We already know that \[\mathbb{E} \left[ X^2(t) \right] = t\] Which means that \[\mathbb{E} \left[ 2 \int_0^t X_{\tau} \space d{\tau} \right]\] should be equal to 0. Which means that the itô integral \[\mathbb{E} \left[ 2 \int_0^t X_{\tau} \space d{\tau} \right]\] is a martingale.
Itô integrals are martingales.
Let \(g(t, X_t)\) be a function of a stochastic process, then \(\mathbb{E} \left[ \int_0^t g(\tau, x_\tau) \space dX_\tau \right] = 0\)
A continuous time stochastic process is a martingale
\(Yt)\) is a stochastic process that satisfies the following Stochastic Differential Equation \[dY(t) = f(Y_t, t) \space dt + g(Y_t, t) \space dX(t) \tag{1}\] with initial condition \(Y(0) = 0\).
How to tell if \(Y(t)\) is martingale? We will use the fact, from above, that Itô integrals are martingales. For this, by definition, we need \[\mathbb{E}_s[Y_t | \mathcal{F}_s] = Y_s \space, \space 0 \leq s \leq t\]
Integrating Equation 1 both side, we can get an exact form for \(Y(t)\) \[Y(t) = Y(s) + \int_s^t f(Y_u, u) \space du + \int_s^t g(Y_u, u) \space dX(u)\]
Taking the expectation on both side: \[\mathbb{E}(Y_t | \mathcal{F}_s)= \mathbb{E} \left[Y(s) + \int_s^t f(Y_u, u) \space du + \int_s^t g(Y_u, u) \space dX(u) \space | \mathcal{F} \right]\]
\[\mathbb{E}(Y_t | \mathcal{F}_s) = Y(s) + \mathbb{E} \left[\int_s^t f(Y_u, u) \space du \space | \mathcal{F_s} \right]\]
This is because, see above, ito integrals are martingales and \(\int_s^t g(Y_u, u) \space dX(u)\) is an ito integral. Hence, its expectation is 0 ==> \(\mathbb{E} \left[ \int_s^t g(Y_u, u) \space dX(u) \right] = 0\)
In order for \(\mathbb{E}(Y_t | \mathcal{F}_s) = Y(s)\), we now need \(\mathbb{E} \left[\int_s^t f(Y_u, u) \space du \space | \mathcal{F}_s \right] = 0\). This means that \(f(Y_t, t) = 0, \space \forall t\).
Going back to the SDE, we can say that \(dY(t)\) is a martingale iff \[dY(t) = g(Y_t, t) \space dX(t)\]
Exponential martingales
LEt’s consider a stochastic process \(Y(t)\) that satisfies the following Stochastic Differential Equation: \[dY(t) = f(t) \space dt + g(t) \space dX(t)\] with initial condition \(Y(0) = 0\). \(X(t)\) is a Brownian Motion, \(f(t)\) and \(g(t)\) are time-dependent functions.
We can now define a new process such that \[Z(t) = e^{Y(t)}\]
How can we ensure \(Z(t)\) is a martingale? How should we choose \(f(t)\) such that \(Z(t)\) is a martingale?