Question

Let $$ T=\operatorname{Min}\\{t: B(t)=2-4 t\\} $$ That is, \(T\) is the first time that standard Brownian motion hits the line \(2-4 t\). Use the Martingale stopping theorem to find \(E[T]\).

Short answer

The expected value of the stopping time \(T\) when the standard Brownian motion first hits the line \(2 - 4t\) is \(E(T) = \frac{1}{2}\).

Step-by-step solution

Define a Martingale

In order to apply the Martingale stopping theorem, we first need to define a martingale. A martingale is a stochastic process that satisfies two conditions: 1. \(E(\left| M(t)\right|) < \infty\) for all \(t\). 2. \(E(M(t)\mid\mathcal{F}_{s}) = M(s)\) for all \(0\leq s \leq t\) . Here, \(M(t)\) is one possible martingale involving the given Brownian motion, and \(\mathcal{F}_{s}\) is the filtration (information) up to time \(s\). For this problem, we choose our martingale as \(M(t) = B(t) + 2t\).

Verify the Martingale Conditions

Now, we need to verify that the chosen process \(M(t) = B(t) + 2t\) is a martingale. 1. First, we check that the expectation \(E(\left| M(t)\right|)\) is finite for all \(t\). Since \(B(t)\) is a standard Brownian motion, we know that it has mean \(0\) and variance \(t\). So, \(E(B(t)) = 0\) and \(E\left(B(t)^2\right) = t\). Also, \(E(2t) = 2t\). Therefore, we have \[ E(\left| M(t) \right|) = E(\left| B(t) + 2t \right|) \leq E\left(\left| B(t) \right|\right) + 2t < \infty. \] 2. Second, we need to show that \(E(M(t)\mid\mathcal{F}_{s}) = M(s)\) for all \(0\leq s \leq t\). According to the properties of Brownian motion, the increment \(B(t) - B(s)\) is independent of \(\mathcal{F}_{s}\) and normally distributed with mean \(0\) and variance \(t - s\). Therefore, we can write \[ E(M(t)\mid\mathcal{F}_{s}) = E(B(t) + 2t\mid\mathcal{F}_{s}) = E(B(t)\mid\mathcal{F}_{s}) + 2t. \] Since \(B(t\mid\mathcal{F}_{s}) = B(s) + (B(t) - B(s))\), we obtain \[ E(B(t)\mid\mathcal{F}_{s}) = B(s) + E(B(t) - B(s)) = B(s). \] Therefore, \[ E(M(t)\mid\mathcal{F}_{s}) = B(s) + 2t = M(s). \] Thus, \(M(t) = B(t) + 2t\) is a martingale.

Apply Martingale Stopping Theorem

Now, we can apply the Martingale stopping theorem on the martingale \(M(t)\) and the stopping time \(T\). The theorem states that if \(M(t)\) is a martingale and \(T\) is a stopping time with finite expectation, then \[ E(M(T)) = E(M(0)). \] At the stopping time \(T\), the process hits the line \(2 - 4t\), so we have \[ M(T) = B(T) + 2T = 2 - 4T. \] Thus, by the martingale stopping theorem, we have \[ E(2 - 4T) = M(0)= B(0) + 2\cdot 0 = 0. \] Solving for \(E(T)\), we get \[ E(T) = \frac{1}{2}. \] Therefore, the expected value of the stopping time \(T\) when the standard Brownian motion first hits the line \(2 - 4t\) is \(E(T) = \frac{1}{2}\).