91Ó°ÊÓ

Since Y(t) is a function of the Brownian motion B(t) given by \(Y(t) = B^2(t) - t\), it is clearly adapted to the filtration generated by the Brownian motion B(t).

To show that Y(t) has finite expectation, we will compute \(E[|Y(t)|]\) and show that it is finite for all t ≥ 0. We have: \[ E[|Y(t)|] = E[|B^2(t) - t|] \leq E[|B^2(t)| + |t|] = E[|B^2(t)|] + E[|t|] \] We know that the expectation of the square of a standard Brownian motion B(t) is t, as \(E[B^2(t)] = t\). Therefore, we have: \[ E[|Y(t)|] \leq E[|B^2(t)|] + E[|t|] = t + t = 2t < \infty \] Thus, Y(t) has finite expectation for all t ≥ 0.

To show that Y(t) has the Martingale property, we will compute the conditional expectation of Y(t) given the information up to time s, denoted by \(E[Y(t) | B(s)]\), and show that it equals Y(s). We have: \[ E[Y(t) | B(s)] = E[B^2(t) - t | B(s)] \] Since B(t) is a standard Brownian motion, the increment \( B(t) - B(s) \) is normally distributed with mean 0 and variance t-s, i.e., \(N(0, t-s)\). Therefore, we can write: \[ B^2(t) = (B(s) + (B(t) - B(s))^2 \\ \] Taking expectation conditioned on B(s), we get: \[ E[B^2(t) | B(s)] = E[B^2(s) + 2B(s)(B(t) - B(s) + (B(t) - B(s))^2 | B(s)] \\ \] Note that \(B(s)\) is deterministic from the given information, and \(B(t) - B(s)\) is independent of \(B(s)\). Therefore, we have: \[ E[B^2(t) | B(s)] = B^2(s) + 2B(s)E[B(t)-B(s)] + E[(B(t)-B(s))^2] \\ \] Since \(E[B(t)-B(s)] = 0\), the second term vanishes and \(E[(B(t)-B(s))^2] = t - s\). Thus, we have: \[ E[B^2(t) | B(s)] = B^2(s) + t - s \] Substituting this back into our conditional expectation for Y(t), we get: \[ E[Y(t) | B(s)] = E[B^2(t) - t | B(s)] = (B^2(s) + t - s) - t = B^2(s) - s = Y(s) \] This shows that Y(t) has the Martingale property.

Now we will find the expected value of Y(t), denoted by \(E[Y(t)]\). We have: \[ E[Y(t)] = E[B^2(t) - t] \] Recall that \(E[B^2(t)] = t\), so we have: \[ E[Y(t)] = E[B^2(t) - t] = t - t = 0 \] Thus, the expected value of Y(t) is 0.