Author Archives: Jonathan Mattingly

Given a stochastic process  \(f_t\) and \(g_t\) adapted to a filtration \(\mathcal F_t\) satisfying

\[\int_0^T\mathbf E f_t^2 dt < \infty\quad\text{and}\quad \int_0^T\mathbf E g_t^2 dt < \infty\]

define

\[M_t =\int_0^t f_s dW_s \quad \text{and}\quad N_t =\int_0^t g_s dW_s\]

for some standard Brownian Motion also adapted to the  filtration \(\mathcal F_t\) . Though it is not necessary, assume that  there exists a \(K>0\) so that  \(|f_t|\) and \(|g_t|\)  are less than some \(K\)  for all \(t\) almost surely.

Let \(\{ t_i^{(n)} : i=0,\dots,N(n)\}\) be sequence of partitions of \([0,T]\) of the form

\[ 0 =t_0^{(n)} < t_1^{(n)} <\cdots<t_N^{(n)}=T\]

such that

\[ \lim_{n \rightarrow \infty} \sup_i |t_{i+1}^{(n)} – t_i^{(n)}| = 0\]

Defining

\[V_n[M]=\sum_{i=1}^{N(n)} \big(M_{t_i} -M_{t_{i-1}}\big)^2\]

and

\[Q_n[M,N]= \sum_{i=1}^{N(n)} \big(M_{t_i} -M_{t_{i-1}}\big)\big(N_{t_i} -N_{t_{i-1}}\big)\]

Clearly \(V_n[M]= Q_n[M,M]\). Show  that the following points hold.

1. The “polarization equality” holds:
   \[ 4 Q_n[M,N] =V_n[M+N] -V_n[M-N]\]
   Hence it is enough to understand the limit of \(n \rightarrow \infty\) of \(Q_n\) or \(V_n\).
2. \[\mathbf E V_n[M]= \int_0^T \mathbf E f_t^2 dt\]
3. * \(V_n[M]\rightarrow \int_0^T  f_t^2 dt\) as \(n \rightarrow \infty\) in \(L^2\). That is to say
   \[ \lim_{n \rightarrow \infty}\mathbf E \Big[ \big( V_n[M] –   \int_0^T  f_t^2 dt  \big)^2 \Big]=0\]
   This limit is called the Quadratic Variation of the Martingale \(M\).
4. Using the results above, show that \(Q_n[M,N]\rightarrow \int_0^T  f_t g_t dt\) as \(n \rightarrow \infty\) in \(L^2\). This is called the cross-quadratic variation of \(M\) and \(N\).
5. * Prove by direct calculation that  in the spirit of 3) from above that   \(Q_n[M,N]\rightarrow \int_0^T  f_t g_t dt\) as \(n \rightarrow \infty\) in \(L^2\).

In this context, one writes \(\langle M \rangle_T\) for the limit of the \(V_n[M]\)  which is called the quadratic variation process of \(M_T\). Similarly  one writes  \(\langle M,N \rangle_T\) for the  limit of \(Q_n[M,N]\)  which is called the cross-quadratic variation process of \(M_T\) and \(N_T\). Clearly \(\langle M \rangle_T = \langle M,M \rangle_T\) and \( \langle M+N,M \rangle_T = \langle M,  M+N\rangle_T= \langle M \rangle_T + \langle M,  N\rangle_T\).

Let \(f_t\) and \(f_t\) be two stochastic processes adapted to a filtration \(\mathcal F_t\) such that

\[\int_0^\infty \mathbf E (f_t^2) dt < \infty \qquad \text{and} \qquad \int_0^\infty \mathbf E (g_t^2) dt < \infty\]

Let \(W_t\) be a standard brownian motion  also adapted to the filtration \(\mathcal F_t\) and define the stochastic processes

\[ X_t =\int_0^t f_s dW_s \qquad \text{and} \qquad Y_t=\int_0^t g_s dW_s\]

Calculate the following:

1. \( \mathbf E (X_t  X_s ) \)
2. \( \mathbf E (X_t  Y_t ) \)
   Hint: You know how to compute \( \mathbf E (X_t^2 ) \) and \( \mathbf E (Y_t^2 ) \). Use the fact that \((a+b)^2 = a^2 +2ab + b^2\) to answer the question. Simplify the result to get a compact expression for the answer.
3. Show that if \(f_t=\sin(2\pi t)\) and \(g_t=\cos(2\pi t)\) then \(X_1\) and \(Y_1\) are independent random variables.(Hint: use the result here  to deduce that \(X_1\) and \(Y_1\) are mean zero gaussian random variables. Now use the above results to show that the covariance of \(X_1\) and \(Y_1\) is zero. Combining these two facts implies that the random variables are independent.)

In this problem, we will show that the Ito integral of a deterministic function is a Gaussian Random Variable.

Let \(\phi\) be deterministic elementary functions. In other words there exists a  sequence of  real numbers \(\{c_k : k=1,2,\dots,N\}\) so that

\[ \sum_{k=1}^\infty c_k^2 < \infty\]

and there exists a partition

\[0=t_0 < t_1< t_2 <\cdots<t_N=T\]

so that

\[ \phi(t) = \sum_{k=1}^N c_k \mathbf{1}_{[t_{k-1},t_k)}(t) \]

1. Show that if \(W(t)\) is a standard brownian motion then the Ito integral
   \[ \int_0^T \phi(t) dW(t)\]
   is a Gaussian random variable with mean zero and variance
   \[ \int_0^T \phi(t)^2 dt \]
2. * Let \(f\colon [0,T] \rightarrow \mathbf R\) be a deterministic function such that
   \[\int_0^T f(t)^2 dt < \infty\]
   Then it can be shown that there exists a sequence of  deterministic elementary functions \(\phi_n\) as above such that
   \[\int_0^T (f(t)-\phi_n(t))^2 dt \rightarrow 0\qquad\text{as}\qquad n \rightarrow \infty\]
   Assuming this fact, let \(\psi_n\) be the characteristic function  of the random variable
   \[ \int_0^T \phi_n(t) dW(t)\]
   Show that for all \(\lambda \in \mathbf R\), show that
   \[ \lim_{n \rightarrow \infty} \psi_n(\lambda) = \exp \Big( -\frac{\lambda^2}2 \big( \int_0^T f(t)^2 dt \big)  \Big)\]
   Then use the the convergence result here to conclude that
   \[ \int_0^T f(t) dW(t)\]
   is a Gaussian Random Variable with mean zero and variance
   \[\int_0^T f(t)^2 dt \]
   by identifying the limit of the characteristic functions above.

   Note: When Probabilistic say the “characteristic function” of a random distribution they just mean the Fourier transform of the random variable. See here.

Part I

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 \(\mathbb{E} M^2 < \infty\) for all \(t\) a.s. .(Here \(B(t)\) is standard Brownian Motion.) Let \(\langle M \rangle(t)\) be the quadratic variation process of \(M(t)\). What equation does \(f\) have to satisfy so that \(Y(t)=f(M(t),\langle M \rangle(t))\) 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 \(f_1(B(t),), \cdots, f_4(B(t),t)\) as defined in the previous item.

Part II

Now 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. In light of the above 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 use the first part of this problem to conclude that \(Y\) 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. As discussed in class \(I^4\) 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*}