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
    \mathbb{E} \{AB\} \leq (\mathbb{E}\{A^2\})^{1/2} (\mathbb{E}\{B^2\})^{1/2}
    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
    \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}

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
    f_n(x,y) = \sum_{0 \leq m \leq \lfloor n/2 \rfloor} C_{n,m} x^{n-2m}y^m
    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
X_n=\sum_{k=0}^n Z_k^2\;.

  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
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)
is a Martingale.

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