Question

Show that \(\\{Y(t), t \geqslant 0\\}\) is a Martingale when $$ Y(t)=\exp \left\\{c B(t)-c^{2} t / 2\right\\} $$ where \(c\) is an arbitrary constant. What is \(E[Y(t)] ?\) An important property of a Martingale is that if you continually observe the process and then stop at some time \(T\), then, subject to some technical conditions (which will hold in the problems to be considered), $$ E[Y(T)]=E[Y(0)] $$ The time \(T\) usually depends on the values of the process and is known as a stopping time for the Martingale. This result, that the expected value of the stopped Martingale is equal to its fixed time expectation, is known as the Martingale stopping theorem.

Short answer

The process \(\{Y(t), t \geq 0\}\) given by \[ Y(t) = \exp\{cB(t) - \frac{c^2t}{2}\} \] is not a Martingale, as it does not satisfy the Martingale property \(E[Y(t) | Y(s)] = Y(s)\) for all \(0 \leq s \leq t\). However, we found that the expected value of the process, \(E[Y(t)]\), is given by \[E[Y(t)] = \exp\left(-\frac{c^2t}{2}\right)\] and is finite for all \(t \geq 0\).

Step-by-step solution

Check the expected value and Martingale property of Y(t)

To show that \(E[|Y(t)|]<\infty\), let's first find the expected value of \(Y(t)\) : \[E[Y(t)] = E\left[\exp\left\{cB(t) - \frac{c^2t}{2}\right\}\right]\] Now, we need to show that the process \(\{Y(t), t \geq 0\}\) satisfies the Martingale property. To do this, let's find the conditional expectation \(E[Y(t)| Y(s)]\) for \(0 \leq s \leq t\): \[E[Y(t) | Y(s)] = E\left[\exp\left\{cB(t) - \frac{c^2t}{2}\right\} | Y(s)\right]\]

Compute the conditional expectation of Y(t) given Y(s)

Let's find the conditional expectation of \(Y(t)\) given \(Y(s)\): \[E[Y(t) | Y(s)] = E\left[\exp\left\{cB(t-s) - \frac{c^2}{2}(t - s) +cB(s)\right\} | Y(s)\right]\] Notice that \(B(t) = B(t-s) + B(s)\), so the term inside the exponential becomes \(\exp\left\{c(B(t-s) + B(s)) - \frac{c^2}{2}(t - s) - \frac{c^2s}{2}\right\}\). Since \(B(t-s)\) is independent of \(B(s)\) and has a normal distribution with mean zero and variance \(t-s\), we can find the expected value of the product of exponentials by taking the product of their expected values: \[E[Y(t) | Y(s)] = E\left[\exp\left(cB(t-s) - \frac{c^2}{2}(t - s)\right)\right] \cdot Y(s)\] Since \(B(t-s)\) is normally distributed, we can use the moment-generating function of a normal distribution to find this expected value: \[E[Y(t) | Y(s)] = \exp\left\{-\frac{c^2}{2}(t-s)\right\} \cdot Y(s)\]

Check if the Martingale property holds for Y(t)

Now, we compare \(E[Y(t)| Y(s)]\) to \(Y(s)\) to check if the Martingale property holds for \(\{Y(t), t \geq 0\}\): \[E[Y(t) | Y(s)] = \exp\left\{-\frac{c^2}{2}(t-s)\right\} \cdot Y(s)\] If the Martingale property holds, then \(E[Y(t) | Y(s)]\) should equal \(Y(s)\). But we have found that \(E[Y(t) | Y(s)] = \exp\left\{-\frac{c^2}{2}(t-s)\right\} \cdot Y(s)\), which is not equal to \(Y(s)\). Therefore, the process \(\{Y(t), t \geq 0\}\) does not satisfy the Martingale property, and it is not a Martingale process. However, if we find the expected value of \[\exp\left\{cB(t) - \frac{c^2t}{2}\right\}\], using \(B(t)\) with mean zero and variance \(t\), we can say that: \[E[Y(t)] = E\left[\exp\left\{cB(t) - \frac{c^2t}{2}\right\}\right] = \exp\left(-\frac{c^2t}{2}\right)\] This result shows that the expected value of the given process is a function of time and is finite for all \(t \geq 0\).