Examples of Martingale

Note

The project was the WorldQuant University in MScFE 622 Continuous-time Stochastic Processes.

Question 1

Let $W = \{W_t : t \geq 0\}$ be a Brownian motion on $(\omega, \mathcal{F}, \mathbb{F} = (\mathcal{F}_t)_{t \geq 0}, \mathbb{P})$

Question 1.1

Show that $W$ is an $\mathbb{F}$-martingale

Ans

We consider the information at the state $s$ such that $s \leq t$. From the assumption $W$ is Brownian motion, that imply $W_t \sim N(0,t)$ and $W_s \sim N(0,s)$. The conditional expectation given the information state $s$ of the Brownian motion is,

$$ \begin{align*} \mathbb{E}[W_t|\mathcal{F}_s] &= \mathbb{E}[(W_t - W_s + W_s)|\mathcal{F}_s] & \text{Include and exclude rules} \\ & = \mathbb{E}[(W_t - W_s)|\mathcal{F}_s] + \mathbb{E}[W_s|\mathcal{F}_s] & \text{Linear property of the expectation } \\ & = \mathbb{E}[(W_t - W_s)] + W_s & \text{$W_t - W_s \sim N(0,t-s)$ and it is independent of $\mathcal{F}_s$}\\ & = \mathbb{E}[W_t] - \mathbb{E}[W_s] + W_s & \text{Linear property of the expectation} \\ & = 0 - 0+ W_s & \text{$W_t \sim N(0,t)$ and $W_s \sim N(0,s)$} \\ & = W_s. \end{align*} $$

Therefore, $W$ is an $\mathbb{F}-$martingale.

Question 1.2

Show that for every $\alpha \in \mathbb{R}$, the process

$$X_t^{\alpha} =: \exp(\alpha W_t - \frac{1}{2}\alpha^2t)$$

is an $\mathbb{F}$-martingale

Ans

We know that $W_t \sim N(0,t)$. From the moment generating of the normal distribution ($X \sim N(\mu,\sigma^2)$ then $M_X(\alpha) = \exp(\alpha\mu + \frac{1}{2}\sigma^2\alpha^2)$, the relationship between the $W_t-W_s \sim N(0,s-t)$ and its moment generating function is given by

$$ \begin{align} M_{W_t-W_s}(\alpha) &= \mathbb{E}(\exp(\alpha(W_t-W_s))) \nonumber\\ &= \exp(\alpha\mu + \frac{1}{2}\sigma^2\alpha^2) & W_t - W_s \sim N(0,s-t) \nonumber\\ & = \exp(\alpha (\mathbb{E}(W_t) - \mathbb{E}(W_s)) + \frac{1}{2}(s-t)\alpha^2) & \mu = \mathbb{E}(x),\ \sigma^2 = s-t \nonumber\\ & = \exp(\frac{1}{2}(s-t)\alpha^2) & \mathbb{E}(W_t) = \mathbb{E}(W_s) = 0. \ (E3) \end{align} $$

Again, we consider the conditional expected value of the process $X_t^\alpha$,

$$ \begin{align*} \mathbb{E}[X_t^{\alpha}|\mathcal{F}_s] &= \mathbb{E}\left[\exp(\alpha W_t - \frac{1}{2}\alpha^2t)\Bigg|\mathcal{F}_s\right] &\text{Given information} \\ & = \mathbb{E}\left[\exp(\alpha(W_t - W_s) + \alpha W_s- \frac{1}{2}\alpha^2t) \Bigg|\mathcal{F}_s\right] & \text{Include and exclude rules} \\ & = \mathbb{E}\left[\exp(\alpha(W_t - W_s))\exp(\alpha W_s -\frac{1}{2}\alpha^2t) \Bigg|\mathcal{F}_s\right] &\text{Exponential function's property} \\ & = \exp(\alpha W_s -\frac{1}{2}\alpha^2t) \mathbb{E}\left[\exp(\alpha(W_t - W_s)) \Bigg|\mathcal{F}_s\right] &\text{$W_s$ is measurable with respect to $\mathcal{F}_s$} \\ & = \exp(\alpha W_s -\frac{1}{2}\alpha^2t) \mathbb{E}[\exp(\alpha(W_t - W_s))]&\text{$W_t - W_s \sim N(0,t-s)$ independent of $\mathcal{F}_s$}\\ & = \exp(\alpha W_s -\frac{1}{2}\alpha^2t)\exp(\frac{1}{2}(s-t)\alpha^2) &\text{Using MGF (E3)} \\ & = \exp(\alpha W_s - \frac{1}{2}\alpha^2s) &\text{Simplifying} \\ & = X_s^\alpha &\text{From Definition of $X_s^\alpha$} \end{align*} $$

Therefore, $X_t^\alpha$ is an $\mathbb{F}-$martingale for all $\alpha \in \mathbb{R}$. \

Question 2

Define the polynomials $H_n(x,y); \ n = 0,1,2,\dots$ by

$$H_n(x,y) = \frac{\partial^n}{\partial \alpha^n} \exp(\alpha x - \frac{1}{2}\alpha^2y) \ \text{at} \ \alpha = 0$$

For example,

$$H_0(x,y) = 1, \ H_(x,y) =x, \ H_2(x,y) = x^2-y, \ H_3(x,y) = x^3-3xy,\ H_4(x,y) = x^4-6x^2y+3y^3, \ \text{etc.}$$

It can be shown (using Taylor series) that

$$X_t^\alpha = \exp(\alpha W_t - \frac{1}{2}\alpha^2t) = \sum_{n = 0}^{\infty} \frac{\alpha^n}{n!}H_n(W_t,t)$$

We now show that $H_n(W_t,t)$ is a martingale for each $n$

Question 2.1

Let $0 \leq s \leq t$ and $\alpha \in \mathbb{R}$. Explain why for each $F \in \mathcal{F}_s$

$$\int_{F} X_t^\alpha d \mathbb{P} = \int_{F} X_s^\alpha d\mathbb{P}$$

Ans

This statement can be easily explained by the definition of the conditional expectation,

$$\mathbb{E}(X|F) = \int_F X d\mathbb{P},$$

for $F \in \mathcal{F}$. We use the conditional expectation of the process $X_t^\alpha$,

$$ X_s^\alpha = \mathbb{E}[(X_t^\alpha|F)]= \int_{F} X_s^\alpha d\mathbb{P} \ (E1.), $$

because of an $\mathbb{F}$-martingale of $X_t^\alpha$ and the definition of the conditional expectation. We also know that $\mathbb{E}(X_t^\alpha) < \infty$. Then we consider again $X_s^\alpha$,

$$ X_s^\alpha = \mathbb{E}[(X_s^\alpha|F)]= \int_{F} X_s^\alpha d\mathbb{P} \ (E2.), $$

due to $X_s$ is $\mathcal{F}_s$-measurable and the definition of the conditional expectation. Setting $E1 = E2$,

$$\int_{F} X_t^\alpha d \mathbb{P} = \int_{F} X_s^\alpha d\mathbb{P}$$

Question 2.2

By differentiating (3a) on both sides $n$ times with respect to $\alpha$ and interchanging the derivative with the integral (no need to justify this step), show that

$$\mathbb{E}(H_n(W_t,t)|\mathcal{F}_s) = H_n(W_s,s),$$

Ans

Taking $n$ time derivative respect to $\alpha$, we have

$$ \begin{align*} \pdv[n]{}{\alpha}\int_{F} X_t^\alpha d \mathbb{P} &= \pdv[n]{}{\alpha}\int_{F} X_s^\alpha d\mathbb{P} &\text{From (3a)} \\ \int_{F} \pdv[n]{}{\alpha} X_t^\alpha d \mathbb{P} &=\int_{F} \pdv[n]{}{\alpha} X_s^\alpha d\mathbb{P} & \text{Interchange between} \\ && \text{derivative and integral} \\ \int_{F} \pdv[n]{}{\alpha}\exp(\alpha W_t - \frac{1}{2}\alpha^2t) d \mathbb{P} &=\int_{F} \pdv[n]{}{\alpha} \exp(\alpha W_s - \frac{1}{2}\alpha^2s) d \mathbb{P} &\text{Definition of $X_{\cdot}^\alpha$}\\ \int_{F} H_n(W_t,t) d \mathbb{P} &=\int_{F} H_n(W_s,s) d\mathbb{P} &\text{Definition of $H_n(x,y)$}\\ \mathbb{E}(H_n(W_t,t)|F) &= \mathbb{E}(H_n(W_s,s)|F) &\text{Definition of}\\ &&\text{the conditional expectation}\\ \mathbb{E}(H_n(W_t,t)|F) &= H_n(W_s,s) &\text{$W_s$ is $F$-measurable} \end{align*} $$

Since $F \in \mathcal{F}_s$,

$$ \mathbb{E}(H_n(W_t,t)|\mathcal{F}_s) = H_n(W_s,s) $$

Question 2.3

Conclude that $\{ H_n(W_t,t) : t \geq 0 \}$ is a martingale.

Ans

From the equation (\ref{4}), the process $\{ H_n(W_t,t) : t \geq 0 \}$ is an $\mathcal{F}_s$-martingale.