# Category Archives: Ito Integrals

## 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*}

## Paley-Wiener-Zygmund Integral

##### Definition of stochastic integrals by integration by parts

In 1959, Paley, Wiener, and Zygmund gave a definition of the stochastic integral based on integration by parts. The resulting integral will agree with the Ito integral when both are defined. However the Ito integral will have a much large domain of definition. We will now follow the develop the integral as outlined by Paley, Wiener, and Zygmund:

1. Let $$f(t)$$ be a deterministic function with $$f'(t)$$ continuous. Prove that \begin{align*} \int_0^1 f(t)dW(t) = f(1)W(1) – \int_0^1 f'(t)W(t) dt\end{align*}
where the first integral is the Ito integral and the last integral is defined path-wise as the standard Riemann integral since the integrands are a.s. continuous.
2. Now let $$f$$ we as above with in addition $$f(1)=0$$ and “define” the stochastic integral $$\int_0^1 f(t) * dW(t)$$ by the relationship
\begin{align*}
\int_0^1 f(t) *dW(t) = – \int_0^1 f'(t) W(t) dt\;.
\end{align*}
Where the integral on the right hand side is the standard Riemann integral.

If the condition $$f(1)=0$$ seems unnatural to you, what this is really saying is that $$f$$ is supported on $$[0,1)$$. In many ways it would be most natural to consider $$f$$ on $$[0,\infty)$$ with compact support. Then $$f(\infty)=0$$. We consider the unit interval for simplicity.

3. Show by direct calculation (not by the Ito isometry) that
\begin{align*}
\mathbf E \left[ \left(\int_0^1 f(t)* dW(t)\right)^2\right]=\int_0^1 f^2(t) dt\;,
\end{align*}
Paley, Wiener, and Zygmund then used this isometry to extend the integral to any deterministic function in $$L^2[0,1]$$. This can be done since for any $$f \in L^2[0,1]$$, one can find a sequence of deterministic functions in $$\phi_n \in C^1[0,1]$$ with $$\phi_n(1)=0$$ so that
\begin{equation*}
\int_0^1 (f(s) – \phi_n(s))^2ds \rightarrow 0 \text{ as } n \rightarrow 0\,.
\end{equation*}

## Stratanovich integral

Let $$X_t$$ be an Ito processes with
\begin{align*}
dX_t&=f_tdt + g_tdW_t
\end{align*}
and $$B_t$$ be a second (possibly correlated with $$W$$ ) Brownian
motion. We define the Stratanovich integral $$\int X_t \circ dB_t$$  by
\begin{align*}
\int_0^T X_t \circ dB_t = \int_0^T X_t dB_t + \frac12 \int_0^T \;d\langle X, B \rangle_t
\end{align*}
Recall that if $$B_t=W_t$$ then $$d\langle B, W \rangle_t =dt$$ and it is zero if they are independent. Use this definition to calculate:

1. $$\int_0^t B_t \circ dB_t$$ (Explain why this agrees with the answer you obtained here).
2. Let $$F$$ be a smooth function. Find equation satisfied by $$Y_t=F(B_t)$$ written in terms of Stratanovich integrals. (Use Ito’s formula to find the equation for $$dY_t$$ in terms of Ito integrals and then use the above definition to rewrite the Ito integrals as Stratanovich integrals“$$\circ dB_t$$”.) How does this compare to classical calculus ?
3. (Integration by parts) Let $$Z_t$$ be a second Ito process satisfying
\begin{align*}
dZ_t&=b_tdt + \sigma_tdW_t\;.
\end{align*}
Calculate $$d(X_t Z_t)$$ using Ito’s formula and then write it in terms of Stratanovich integrals. Why is this part of the problem labeled integration by parts ? (Write the integral form of the expression you derived for $$d(X_t Z_t)$$ in the two cases. What are the differences ?)

## A simple Ito Integral

Let $$\mathcal F_t$$ be a filtration of $$\sigma$$-algebra and $$W_t$$ a standard Brownian Motion adapted to the filtration. Define the adapted stochastic process $$X_t$$ by

$X_t = \alpha_0 \mathbf 1_{[0,\frac12]}(t) + \alpha_{\frac12} \mathbf 1_{(\frac12,1]}(t)$

where $$\alpha_0$$ is a random variable adapted to  $$\mathcal F_0$$ and  $$\alpha_{\frac12}$$ is a random variable adapted to  $$\mathcal F_{\frac12}$$.

Write explicitly the Ito integral

$\int_0^t X_s dW_s$

and show by direct calculation that

$\mathbf E \Big( \int_0^t X_s dW_s\Big) = 0$

and

$\mathbf E \Big[\Big( \int_0^t X_s dW_s\Big)^2\Big] = \int_0^t \mathbf E X_s^2 ds$

## 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.

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 $$\langle M+N,M \rangle_T = \langle M, M+N\rangle_T= \langle M \rangle_T + \langle M, N\rangle_T$$.

## 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:

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.)

## 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

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

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.

## Homogeneous Martingales and BDG Inequality

### 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*}

## Ito Moments

Use Ito’s formula to show that if $$\sigma(t,\omega)$$ is a bounded nonanticipating functional, $$|\sigma|\leq M$$, then for the stochastic integral $$I(t,\omega)=\int_0^t \sigma(s,\omega) dB(s,\omega)$$ we have the moment estimates
$\mathbf E\{ |I(t)|^{2p}\}\leq 1\cdot 3\cdot 5 \cdots (2p-1) (M^2 t)^p$
for $$p=1,2,3,…$$. Follow the steps below to achieve this result.

1. First assume that
$\mathbf E \int_0^t |I_s|^{k} \sigma_s dB_s =0$
for all positive , integer $$k$$. Under this assumption, prove the result. Why don’t we know a priori that this expectation is zero ?
2. Now define  for positive $$L$$, define $$\chi^{p}_L(x)$$ as $$x^p$$ for $$|x| < L$$ and 0 if $$|x| > L+1$$ and connected monotonically in between such that the whole function is  smooth. Now define
$\psi^{(p)}_L(x) = p\int_0^x \chi^{(p-1)}_L(y) dy\qquad\text{and}\qquad\phi^{(p)}_L(x)= p\int_0^x \psi^{(p-1)}_L(y) dy$
Observe that  all three of the functions are globally bounded for a given $$L$$. Apply Ito’s formula to $$\phi^{(p)}_L(I_t)$$ and the fact that Fatou’s lemma implies that
$\mathbf E |I_t|^{2p} \leq \lim_{L \rightarrow \infty} \mathbf E \phi^{(2p)}_L(I_t)$
to prove the estimate started at the start by following a similar induction step as used above.

## Ito to Stratonovich

Let’s think about different ways to make sense of $\int_0^t W(s)dW(s)$ were $$W(t)$$ is a standard Brownian motion. Fix any $$\alpha \in [0,1]$$define

\begin{equation*}
I_N^\alpha(t)=\sum_{j=0}^{N-1} W(t_j^\alpha)[W(t_{j+1})-W(t_j)]
\end{equation*}
were $$t_j=\frac{j t}N$$ and $$t_j^\alpha=\alpha t_j + (1-\alpha)t_{j+1}$$.
Calculate

1. $\lim_{N\rightarrow \infty}\mathbf E I_N^\alpha(t) \ .$
2. * $\lim_{N\rightarrow \infty}\mathbf E \big( I_N^\alpha(t)\big)^2$

What choice of $$\alpha$$ is the standard It\^o integral ? What choice is the Stratonovich integral ?

## Moment Bounds on Ito Integrals

Use Ito’s formula to show that if $$\sigma(t,\omega)$$ is a
nonanticipating random function which is bounded. That is to say

$|\sigma(t,\omega)|\leq M$

for all $$t \geq 0$$ and all $$\omega$$.

1. Under this assumption show that the stochastic integral
$I(t,\omega)=\int_0^t \sigma(s,\omega) dB(s,\omega)$
satisfies  the following moment estimates
$\mathbf E\{ |I(t)|^{2p}\}\leq 1\cdot 3\cdot 5 \cdots (2p-1) (M^2 t)^p$
for $$p=1,2,3,…$$ if one assumes
$\mathbf E \int_0^t |I(s)|^k \sigma(s) dB(s) =0$
for any integer $$k$$.
2. Prove the above result without assuming that
$\mathbf E \int_0^t |I(s)|^k \sigma(s) dB(s) =0$
since this requires that
$\mathbf E \int_0^t |I(s)|^{2k} \sigma^2(s) ds < \infty$
which we do not know a priory.