Question

Show that \(\\{Y(t), t \geqslant 0\\}\) is a Martingale when $$ Y(t)=B^{2}(t)-t $$ What is \(E[Y(t)] ?\) Hint: First compute \(E[Y(t) \mid B(u), 0 \leqslant u \leqslant s]\).

Short answer

In summary, the process \(Y(t) = B^2(t) - t\) is a martingale, as it satisfies the necessary conditions. The expected value \(E[Y(t)]\) is 0.

Step-by-step solution

Check Condition 1 - Finite expected value for all t

By definition, \(B(t)\) is a standard Brownian motion. Therefore, \(B(t)\) has expected value \(E[B(t)] = 0\) and its variance is \(Var(B(t)) = t\). So, \(E[B^2(t)] = t\), because it is a well-known property of Brownian motion. Now, we compute the expected value of \(Y(t)\): \[ E[Y(t)] = E[B^2(t) - t] \] Since the expected value operator is linear, we can rewrite the equation as: \[ E[Y(t)] = E[B^2(t)] - E[t] \] Given that we already know \(E[B^2(t)] = t\), we can easily find: \[ E[Y(t)] = t - t = 0 \] Thus, \(E[Y(t)]\) is indeed finite, and condition 1 holds.

Check Condition 2 - Conditional expectation

For all \(0 \leq s \leq t\), we need to verify if the following condition holds: \[ E[Y(t) \mid B(u), 0 \leq u \leq s] = Y(s) \] First, it's important to note that since \(B(t)\) is a martingale, then \(B(t) - B(s)\) is independent of \(B(u)\) for \(0 \leq u \leq s\). Now, let's calculate the conditional expectation: \[ E[Y(t) \mid B(u), 0 \leq u \leq s] = E[(B^2(t) - t) \mid B(u), 0 \leq u \leq s] \] Now we'll exploit the independence between the increments of Brownian motion, and apply the law of total expectation: \[ E[Y(t) \mid B(u), 0 \leq u \leq s] = E[((B(t) - B(s)) + B(s))^2 - t) \mid B(u), 0 \leq u \leq s] \] \[ E[Y(t) \mid B(u), 0 \leq u \leq s] = E[((B(t) - B(s))^2 + 2B(s)(B(t) - B(s)) + B(s)^2 - t) \mid B(u), 0 \leq u \leq s] \] Use linearity of conditional expectation: \[ E[Y(t) \mid B(u), 0 \leq u \leq s] = E[(B(t) - B(s))^2 \mid B(u), 0 \leq u \leq s] + 2B(s)E[B(t) - B(s) \mid B(u), 0 \leq u \leq s] \\ + E[B(s)^2 \mid B(u), 0 \leq u \leq s]- E[t \mid B(u), 0 \leq u \leq s] \] Since \(B(t) - B(s)\) is independent of \(B(u)\) for \(0 \leq u \leq s\): \[ E[Y(t) \mid B(u), 0 \leq u \leq s] = E[(B(t) - B(s))^2] + 2B(s)E[B(t) - B(s)] + B(s)^2 - t \] As mentioned previously, we know that \(E[B(t) - B(s)] = 0\), so now we just need the expected value of the squared increment: \[ E[(B(t) - B(s))^2] = Var(B(t) - B(s)) = t - s \] Now we can compute the conditional expectation: \[ E[Y(t) \mid B(u), 0 \leq u \leq s] = (t - s) + 0 + B(s)^2 - t \] \[ E[Y(t) \mid B(u), 0 \leq u \leq s] = B(s)^2 - s = Y(s) \] So, \(Y(t)\) satisfies condition 2. Since both conditions are met, we can conclude that \(Y(t)\) is a martingale.

Calculate E[Y(t)]

Now we will compute the expected value of Y(t) which we already found in Step 1: \[E[Y(t)] = 0\] So, the expected value of the process \(Y(t)=B^2(t) - t\) is 0.