Question

Let \(\\{X(t) ; t \geq 0\\}\) be a Brownian motion process. By formally differentiating the martingale $$ \mathscr{Z}_{\theta}(t)=\exp \left\\{0 X(t)-(1 / 2) \theta^{2} t\right\\} $$ with respect to \(\theta\), show that, for each \(n, H_{n}(X(t), t)\) is a martingale, where $$ \begin{aligned} &H_{0}(x, t) \equiv 1 \\ &H_{1}(x, t)=x \end{aligned} $$ and $$ H_{n}(x, t)=x H_{n-1}(x, t)-(n-1) t H_{n-2}(x, t) $$ An alternative approach is to show that (5.2) applies.

Short answer

#tag_title#Step 5: Prove that \(H_n(X(t),t)\) is a martingale by induction#tag_content# Assuming \(H_{n-1}(X(t),t)\) and \(H_{n-2}(X(t),t)\) are martingales. Now, consider the product of a function of \(X(t)\) and a martingale, it results in another martingale. Thus, \(X(t)H_{n-1}(X(t),t)\), also being a martingale. Next, consider the scaling of a martingale by a function of time; in this case, \((n-1)tH_{n-2}(X(t),t)\), the product of a linear function of time and a martingale is also a martingale. Finally, the sum or difference of two martingales is another martingale. Therefore, the process defined by: \[ H_{n}(X(t), t)=X(t) H_{n-1}(X(t), t)-(n-1) t H_{n-2}(X(t), t) \] is also a martingale. Since the base cases for \(n = 0\) and \(n = 1\) have already been shown to be martingales, this completes the inductive argument, and it is proved that \(H_n(X(t),t)\) is a martingale for every \(n\).

Step-by-step solution

Define the given martingale \(\mathscr{Z}_{\theta}(t)\)

An initial martingale \(\mathscr{Z}_{\theta}(t)\) is defined as, \[ \mathscr{Z}_{\theta}(t)=\exp \left\\{0 X(t)-(1 / 2) \theta^{2} t\right\\}. \]

Differentiate \(\mathscr{Z}_{\theta}(t)\) with respect to \(\theta\)

Now, we will differentiate \(\mathscr{Z}_{\theta}(t)\) with respect to \(\theta\). Deriving an exponential function requires applying the chain rule: \[ \frac{d \mathscr{Z}_{\theta}(t)}{d \theta} = \frac{d}{d \theta} (\exp \left\\{0 X(t)-(1 / 2) \theta^{2} t\right\\}) = -t \theta \mathscr{Z}_{\theta}(t) \]

Show that \(H_0(X(t),t)\) and \(H_1(X(t),t)\) are martingales

As given: \[ H_{0}(x, t) \equiv 1 \] and \[ H_{1}(x, t)=x \] The process \(H_0(X(t), t) \equiv 1\) is a martingale because it's a constant, and constant processes are always martingales. The process \(H_1(X(t), t) = X(t)\) is a Brownian motion process, and Brownian motion processes are martingales. So, \(H_1(X(t), t)\) is also a martingale.

Use the given recursive formula for \(H_n(x,t)\)

We are given the recursive formula for \(H_n(x,t)\) as follows: \[ H_{n}(x, t)=x H_{n-1}(x, t)-(n-1) t H_{n-2}(x, t) \] By induction, we want to show that: If \(H_{n-1}(X(t),t)\) and \(H_{n-2}(X(t),t)\) are martingales, then \(H_{n}(X(t),t)\) is also a martingale. SmartyHeaderCode