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## Quadratic Variation of Ito Integrals

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.

- 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 \(V_n\). - \[\mathbf E V_n[M]= \int_0^T \mathbf E f_t^2 dt\]
- * \(V_n[M]\rightarrow \int_0^T f_t^2 dt\) 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 \(M\). - 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 \(N\).
- * 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 \( \langle M+N,M \rangle_T = \langle M, M+N\rangle_T= \langle M \rangle_T + \langle M, N\rangle_T\).

## Covariance of Ito Integrals

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:

- \( \mathbf E (X_t X_s ) \)
- \( \mathbf E (X_t Y_t ) \)

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. - 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.)

## Gaussian Ito Integrals

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

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

- 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

\[ \int_0^T \phi(t)^2 dt \] - * 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.