# Category Archives: Martingales

## BDG Inequality

Consider $$I(t)$$ defined by $I(t)=\int_0^t \sigma(s,\omega)dB(s,\omega)$ where $$\sigma$$ is adapted and $$|\sigma(t,\omega)| \leq K$$ for all $$t$$ with probability one. Inspired by   problem “Homogeneous Martingales and Hermite Polynomials”  Let us set
\begin{align*}Y(t,\omega)=I(t)^4 – 6 I(t)^2\langle I \rangle(t) + 3 \langle I \rangle(t)^2 \ .\end{align*}

1. Quote  the problem “Ito Moments” to show that $$\mathbb{E}\{ |Y(t)|^2\} < \infty$$ for all $$t$$. Then  verify that $$Y_t$$ is  a martingale.
2. Show that $\mathbb{E}\{ I(t)^4 \} \leq 6 \mathbb{E} \big\{ \{I(t)^2\langle I \rangle(t) \big\}$
3. Recall the Cauchy-Schwartz inequality. In our language it states that
\begin{align*}
\mathbb{E} \{AB\} \leq (\mathbb{E}\{A^2\})^{1/2} (\mathbb{E}\{B^2\})^{1/2}
\end{align*}
Combine this with the previous inequality to show that\begin{align*}\mathbb{E}\{ I(t)^4 \} \leq 36 \mathbb{E} \big\{\langle I \rangle(t)^2 \big\} \end{align*}
4. We know that  $$I^4$$ is a submartingale (because $$x \mapsto x^4$$ is convex). Use the Kolmogorov-Doob inequality and all that we have just derived to show that
\begin{align*}
\mathbb{P}\left\{ \sup_{0\leq s \leq T}|I(s)|^4 \geq \lambda \right\} \leq ( \text{const}) \frac{ \mathbb{E}\left( \int_0^T \sigma(s,\omega)^2 ds\right)^2 }{\lambda}
\end{align*}

## Homogeneous Martingales and Hermite Polynomials

1. Let $$f(x,y):\mathbb{R}^2 \rightarrow \mathbb{R}$$ be a twice differentiable function in both $$x$$ and $$y$$. Let $$M(t)$$ be defined by $M(t)=\int_0^t \sigma(s,\omega) dB(s,\omega)$. Assume that $$\sigma(t,\omega)$$ is adapted and that $$\mathbf{E} M^2 < \infty$$ for all $$t$$ a.s. .(Here $$B(t)$$ is standard Brownian Motion.) Let $$[M]_t$$ be the quadratic variation process of $$M(t)$$. What equation does $$f$$ have to satisfy so that $$Y(t)=f(M(t),[M]_t)$$ is again a martingale if we assume that $$\mathbf E\int_0^t \sigma(s,\omega)^2 ds < \infty$$.
2. Set
\begin{align*}
f_n(x,y) = \sum_{0 \leq m \leq \lfloor n/2 \rfloor} C_{n,m} x^{n-2m}y^m
\end{align*}
here $$\lfloor n/2 \rfloor$$ is the largest integer less than or equal to $$n/2$$. Set $$C_{n,0}=1$$ for all $$n$$. Then find a recurrence relation for $$C_{n,m+1}$$ in terms of $$C_{n,m}$$, so that $$Y(t)=f_n(B(t),t)$$ will be a martingale.Write out explicitly $$f_1(B(t),t), \cdots, f_4(B(t),t)$$ as defined in the previous item.
3. Again let $$M(t)=\int_0^t \sigma(s,\omega) dB(s,\omega)$$ with $$|\sigma(t,\omega)| < K$$ almost surely. Show that $$f_n(M(t),[M]_t)$$ is again a martingale where $$[M]_t$$ is the quadratic variation of $$M(t)$$ and $$f_n$$ is the function found above.
4. * Do you recognize the recursion relation you obtained above for $$f_n$$ as being associated to a famous recursion relation ? (Hint: Look at the title of the problem)

## Exponential Martingale

Let $$\sigma(t,\omega)$$ be adapted to the filtration generated by a standard Brownian Motion $$B_t$$ such that $$|\sigma(x,\omega)| < K$$ for some  bound  $$K$$ . Let $$I(t,\omega)=\int_0^t \sigma(s,\omega) dB(s,\omega)$$.

1. Show that
$M_t=\exp\big\{\alpha I(t)-\frac{\alpha^2}{2}\int_0^t \sigma^2(s)ds \big\}$
is a martingale. It is called the  exponential martingale.
2. Show that $$M_t$$ satisfies the equation
$dM_t =\alpha M_t dI_t = \alpha M_t \sigma_t dB_t$

## Complex Exponential Martingale

Let $$W_t$$ be a standard Brownian Motion. Find $$\alpha \in \mathbb{R}$$ so that
$e^{i W_t + \alpha t}$
is a martingale (and show that it is a martingale).

## Martingale Brownian Squared

Let $$W_t$$ be standard Brownian Motion.

1. Find a function  $$f(t)$$ so that $$W_t^2 -f(t)$$ is a Martingale.
2. * Argue that in some sense this $$f(t)$$is unique among increasing functions with finite variation. Compare this with the problem here.

## Martingale of squares

Let $$\{ Z_n: n=0,1, \cdots\}$$ be a collection of mutually independent random variables with $$Z_n$$ distributed as a Gaussian with mean zero and variance $$\sigma_n^2$$ for $$\sigma_n \in \mathbb{R}$$. Define $$X_n$$ by
\begin{equation*}
X_n=\sum_{k=0}^n Z_k^2\;.
\end{equation*}

1. Find a stochastic process $$Y_n$$ so that $$X_n-Y_n$$ is a Martingale with respect to the filtration $$\mathcal{F}_n=\sigma(Z_0,\cdots,Z_n)$$.
2. Find a second process $$\tilde Y_n$$ so that $$X_n-Y_n$$ is again a Martingale with respect to the filtration $$\mathcal{F}_n$$ but$$Y_n \neq \tilde Y_n$$ almost surely.

## Martingale Exit from an Interval – I

Let $$\tau$$ be the first time that a continuous martingale $$M_t$$ starting from $$x$$ exits the interval $$(a,b)$$, with $$a<x<b$$. In all of the following, we assume that $$\mathbf P(\tau < \infty)=1$$. Let $$p=\mathbf P_x\{M(\tau)=a\}$$.

Find and analytic expression for $$p$$ :

1. For this part assume that $$M_t$$ is the solution to a time homogeneous SDE. That is that $dM_t=\sigma(M_t)dB_t.$ (with $$\sigma$$ bounded and smooth.) What PDE should you solve to find $$p$$ ? with what boundary data ? Assume for a moment that $$M_t$$ is standard Brownian Motion ($$\sigma=1$$). Solve the PDE you mentioned above in this case.
2. A probabilistic way of thinking: Return to a general martingale $$M_t$$. Let us assume that $$dM_t=\sigma(t,\omega)dB_t$$ again with $$\sigma$$ smooth and uniformly bounded  from above and away from zero. Assume that $$\tau < \infty$$ almost surely and notice that $\mathbf E_x M(\tau)=a \mathbf P_x\{M_\tau=a\} + b \mathbf P_x\{M_\tau=b.\}$ Of course the process has to exit through one side or the other, so $\mathbf P_x\{M_\tau=a\} = 1 – \mathbf P_x\{M_\tau=b\}$. Use all of these facts and the Optimal Stopping Theorem to derive the equation for $$p$$.
3. Return to the case when $dM_t=\sigma(M_t)dB_t$. (with $$\sigma$$ bounded and smooth.) Write down the equations that $$v(x)= \mathbf E_x\{\tau\}$$, $$w(x,t)=\mathbf P_x\{ \tau >t\}$$, and $$u(x)=\mathbf E_x\{e^{-\lambda\tau}\}$$ with $$\lambda > 0$$ satisfy. ( For extra credit: Solve them for $$M_t=B_t$$ in this one dimensional setting and see what happens as $$b \rightarrow \infty$$.)

## Making the Cube of Brownian Motion a Martingale

Let $$B_t$$ be a standard one dimensional Brownian
Motion. Find the function $$F:\mathbf{R}^5 \rightarrow \mathbf R$$ so that
\begin{align*}
B_t^3 – F\Big(t,B_t,B_t^2,\int_0^t B_s ds, \int_0^t B_s^2 ds\Big)
\end{align*}
is a Martingale.

Hint: It might be useful to introduce the processes
$X_t=B_t^2\qquad Y_t=\int_0^t B_s ds \qquad Z_t=\int_0^t B_s^2 ds$