# Category Archives: Stochastic Calculus

## 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.

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

## Solving a class of SDEs

Let us try a systematic procedure for solving SDEs which works for a class of SDEs. Let
\begin{align*}
X(t)=a(t)\left[ x_0 + \int_0^t b(s) dB(s) \right] +c(t) \ .
\end{align*}
Assuming $$a$$, $$b$$, and $$c$$ are differentiable, use Ito’s formula to find the equation for $$dX(t)$$ of the form
\begin{align*}
dX(t)=[ F(t) X(t) + H(t)] dt + G(t)dB(t)
\end{align*}
were $$F(t)$$, $$G(t)$$, and $$H(t)$$ are some functions of time depending on $$a,b$$ and maybe their derivatives. Solve the following equations by matching the coefficients. Let $$\alpha$$, $$\gamma$$ and $$\beta$$ be fixed numbers.

Notice that
\begin{align*}
X(t)=a(t)\left[ x_0 + \int_0^t b(s) dB(s) \right] +c(t)=F(t,Y(t)) \ .
\end{align*}
where $$dY(t)=b(t) dB(t)$$. Then you can apply Ito’s formula to this definition to find $$dX(t)$$.

1. First consider
$dX_t = (-\alpha X_t + \gamma) dt + \beta dB_t$
with $$X_0 =x_0$$
. Solve this for $$t \geq 0$$
2. Now consider
$dY(t)=\frac{\beta-Y(t)}{1-t} dt + dB(t) ~,~~ 0\leq t < 1 ~,~~Y(0)=\alpha.$
Solve this for $$t\in[0,1]$$.
3. \begin{align*}
dX_t = -2 \frac{X_t}{1-t} dt + \sqrt{2 t(1-t)} dB_t ~,~~X(0)=\alpha
\end{align*}
Solve this for $$t\in[0,1]$$.

## 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*}

## Associated PDE

Show that if

1. $I(t,\omega)=\int_0^t \sigma(s,\omega) dB(s,\omega)$is a stochastic integral then $I^2(t)-\int_0^t \sigma^2(s)ds$ is a martingale.
2. What equation must $$u(t,x)$$ satisfy so that
$t \mapsto u(t,B(t))e^{\int_0^t V(B(s))ds}$
is a martingale? Here $$V$$ is a bounded function. Hint: Set $$Y(t)=\int_0^t V(B(s))ds$$ and apply It\^0’s formula to $$Z(t,B(t),Y(t))=u(t,B(t))\exp(Y(t))$$.

## Exponential Martingale Bound

Let $$\sigma(t,\omega)$$ be nonanticipating with $$|\sigma(x,\omega)| < M$$ for some  bound  $$M$$ . Let $$I(t,\omega)=\int_0^t \sigma(s,\omega) dB(s,\omega)$$. Use the exponential martingale $\exp\big\{\alpha I(t)-\frac{\alpha^2}{2}\int_0^t \sigma^2(s)ds \big\}$ (see the problem here)  and the Kolmogorov-Doob inequality to get the estimate
$P\Big\{ \sup_{0\leq t\leq T}|I(t)| \geq \lambda \Big\}\leq 2 \exp\left\{\frac{-\lambda^2}{2M^2 T}\right\}$
First express the event of interest in terms of the exponential martingale, then use the Kolmogorov-Doob inequality and after this choose the parameter $$\alpha$$ to get the best bound.

## Ballistic Growth

Consider the SDE
$dX(t)=b(X(t))dt +\sigma(X(t))dB(t)$
with $$b(x)\to b_0 >0$$ as $$x\to\infty$$ and with $$\sigma$$ bounded and positive. Suppose that $$b$$ and $$\sigma$$ are such that
$\lim_{t\to\infty}X(t)=\infty$, with probability one for any starting point. Show that
$P_x\Big\{\lim_{t\to\infty}\frac{X(t)}{b_0 t}=1\Big\}=1 \ .$
From
$X(t)=x+\int_0^{t}b(X(s))ds +\int_0^{t}\sigma(X(s))dB(s)$
and the hypotheses, note that the result follows from showing that
\begin{align*}
\mathbf P_x\Big\{\lim_{t\to\infty}\frac{1}{t}\int_0^{t}\sigma(X(s))dB(s)=0\Big\}=1 \ .
\end{align*}

There are a number of ways of thinking about this. In the end they all come down to essentially the same calculations. One way is to show that for some fixed $$\delta \in(0,1)$$ the following statement holds with probability one:

There exist a constants $$C(\omega)$$ so that
\begin{align*}
\int_0^{t}\sigma(X(s))dB(s) \leq Ct^\delta
\end{align*}
for all $$t >0$$.

To show this partition $$[0,\infty]$$ into blocks and use the Doob-Kolmogorov inequality to estimate the probability that the max of $$\int_0^{t}\sigma(X(s))ds$$ on each block excess $$t^\delta$$ on that block. Then use the Borel-Cantelli to show that this happens only a finite number of times.

A different way to organize the same calculation is to estimate
$\mathbf P_x\Big\{\sup_{t>a}\frac{1}{t}|\int_0^t \sigma(X(s))dB(s)|>\epsilon\Big\}$
by breaking the interval $$t>a$$ into the union of intervals of the form $$a2^k <t\leq a2^{k+1}$$ for $$k=0,1,\dots$$ and using Doob-Kolmogorov Martingale inequality. Then let $$a\to\infty$$.

## Entry and Exit through boundaries

Consider the following one dimensional SDE.
\begin{align*}
dX_t&= \cos( X_t )^\alpha dW(t)\\
X_0&=0
\end{align*}
Consider the equation for $$\alpha >0$$. On what interval do you expect to find the solution at all times ? Classify the behavior at the boundaries.

For what values of $$\alpha < 0$$ does it seem reasonable to define the process ? any ? justify your answer.

## 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 $$b \rightarrow \infty$$.)

## Around the Circle

Consider the equation
\begin{align}
dX_t &= -Y_t dB_t – \frac12 X_t dt\\
dY_t &= X_t dB_t – \frac12 Y_t dt
\end{align}
Let $$(X_0,Y_0)=(x,y)$$ with $$x^2+y^2=1$$. Show that $$X_t^2 + Y_t^2 =1$$ for all $$t$$ and hence the SDE lives on the unit circle. Does this make intuitive sense ?

## Shifted Brownian Motion and a PDE

Let $$f \in C_0^2(\mathbf R^n)$$ and $$\alpha(x)=(\alpha_1(x),\dots,\alpha_n(x))$$ with $$\alpha_i \in C_0^2(\mathbf R^n)$$ be given functions and consider the partial differential equations
\begin{align*}
\frac{\partial u}{\partial t} &= \sum_{i=1}^n \alpha_i(x)
\frac{\partial u}{\partial x_i} + \frac{1}{2} \frac{\partial^2
u}{\partial x_i^2} \ \text{ for } t >0 \text{ and }x\in \mathbf R^n \\
u(0,x)&=f(x) \ \text{ for } \ x \in \mathbf R^n
\end{align*}
Use the Girsonov theorem to show that the unique bounded solution $$u(t,x)$$ of this equation can be expressed by
\begin{align*}
u(t,x) = \mathbf E_x \left[ \exp\left(\int_0^t \alpha(B(s))\cdot dB(s) –
\frac{1}{2}\int_0^t |\alpha(B(s))|^2 ds \right)f(B(t))\right]
\end{align*}

where $$\mathbf E_x$$ is the with respect to $$\mathbf P_x$$ when the Brownian Motion starts at $$x$$. (Note there maybe a sign error in the above exponential term. Use what ever sign is right.) For the remainder, assume that $$\alpha$$
is a fixed constant $$\alpha_0$$. Now using what you know about the distribution of $$B_t$$ write the solution to the above equation as an integral kernel integrated against $$f(x)$$. (In other words, write $$u(t,x)$$ so that your your friends who don’t know any probability might understand it. ie $$u(t,x)=\int K(x,y,t)f(y)dy$$ for some $$K(x,y,t)$$)

## Probability Bridge

For fixed $$\alpha$$ and $$\beta$$ consider the stochastic differential equation
$dY(t)=\frac{\beta-Y(t)}{1-t} dt + dB(t) ~,~~ 0\leq t < 1 ~,~~Y(0)=\alpha.$
Verify that $$\lim_{t\to 1}Y(t)=\beta$$ with probability one. ( This is called the Brownian bridge from $$\alpha$$ to $$\beta$$.)
Hint: In the problem “Solving a class of SDEs“,  you found that this equation had the solution
\begin{equation*}
Y_t = a(1-t) + bt + (1-t)\int_0^t \frac{dB_s}{1-s} \quad 0 \leq t <1\; .
\end{equation*}
To answer the question show that
\begin{equation*}
\lim_{t \rightarrow 1^-} (1-t) \int_0^t\frac{dB_s}{1-s} =0 \quad \text{a.s.}
\end{equation*}

## 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
$X_t=B_t^2\qquad Y_t=\int_0^t B_s ds \qquad Z_t=\int_0^t B_s^2 ds$

## Correlated SDEs

Let $$B_t$$ and $$W_t$$ be standard Brownian motions which are
independent. Consider
\begin{align*}
dX_t&= (-X_t +1)dt + \rho dB_t + \sqrt{1-\rho^2} dW_t\\
dY_t&= -Y_t dt + dB_t \ .
\end{align*}
Find the covariance of $$\text{Cov}(X_t,Y_t)=\mathbf{E} (X_t Y_t) – \mathbf{E} (X_t) \mathbf{E}( Y_t)$$.

## Hyperbolic SDE

Consider
\begin{align*}
dX_t=& Y_t dB_t + \frac12 X_t dt\\
dY_t=& X_t dB_t + \frac12 Y_t dt
\end{align*}
Show that $$X_t^2-Y_t^2$$ is constant for all $$t$$.

## Diffusion and Brownian motion

Let $$B_t$$ be a standard Brownian Motion  starting from zero and define

$p(t,x) = \frac1{\sqrt{2\pi t}}e^{-\frac{x^2}{2t} }$

Given any $$x \in \mathbf R$$, define $$X_t=x + B_t$$ . Of course $$X_t$$ is just a Brownian Motion stating from $$x$$ at time 0. Fixing a smooth, bounded, compactly supported function $$f:\mathbf R \rightarrow \mathbf R$$, we define the function $$u(x,t)$$ by

$u(x,t) = \mathbf E_x f(X_t)$

where we have decorated the expectation with the subscript $$x$$ to remind us that we are starting from the point $$x$$.

1. Explain why $u(x,t) = \int_{\infty}^\infty f(y)p(t,x-y)dy$
2. Show by direct calculation using the formula from the previous question that for $$t>0$$, $$u(x,t)$$ satisfies the diffusion equation
$\frac{\partial u}{\partial t}= c\frac{\partial^2 u}{\partial x^2}$
for some constant $$c$$. (Find the correct $$c$$ !)
3. Again using the formula from part 1), show that
$\lim_{t \rightarrow 0} u(t,x) = f(x)$
and hence the initial condition for the diffusion equation is $$f$$.

## Levy’s construction of Brownian Motion

Let $$\{ \xi_k^{(n)} : n =0,1,\dots ; k =1,\dots,2^n\}$$ be a collection of independent Gaussian random variables with  $$\xi_k^{(n)}$$ having mean zero and variance $$2^{-n}$$. Define the random variable $$\eta_k^{(n)}$$ recursively by

$\eta_1^{(0)} = Z \qquad\text{with}\quad Z\sim N(0,1) \quad\text{and independent of the $$\xi$$’s}$

$\eta_{2k}^{(n+1)} = \frac12\eta_{k}^{(n)} -\frac12 \xi_{k}^{(n)}$

$\eta_{2k-1}^{(n+1)} = \frac12\eta_{k}^{(n)} +\frac12 \xi_{k}^{(n)}$

For any time $$t \in [0,1]$$ of the form $$t=k 2^{-n}$$ define

$W^{(n)}_t = \sum_{j=1}^k \eta_{j}^{(n)}$

For $$t \in [0,1]$$ not of this form we connect the two nearest defined points with a line.

1. Follow given steps to show that for fixed $$n$$, $$W^{(n)}_t$$ is random walk on $$\mathbf R$$ with Gaussian steps.
1. Show $$\mathbf E \eta_{k}^{(n)} = 0$$ and  $$\mathbf E \big[ (\eta_{k}^{(n)})^2\big] = 2^{-n}$$
2. Argue that $$\eta_{k}^{(n)}$$ is Gaussian and that for any fixed $$n$$,
$\{ \eta_{k}^{(n)} : k=1,\dots, 2^n\}$
are a collection of mutually independent random variables. (To show independence show that they are mean zero Gaussians  with correlation $$\mathbf E [\eta_{k}^{(n)}\eta_{j}^{(n)}]=0$$ when $$j\neq k$$.)
2. To understand the relationship between $$W^{(n)}$$ and $$W^{(n+1)}$$, simulate a collection of random $$\xi_k^{(n)}$$ and plot $W^{(0)}, W^{(1)}, W^{(2)}, W^{(3)}, W^{(4)}$
over the time interval $$[0,1]$$. Notice that at $$n$$ increases the functions seem to converge. Try a few different realizations to get a feeling for how the limiting function might look.

## 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. * $\lim_{N\rightarrow \infty}\mathbf E \big( I_N^\alpha(t)\big)^2$
3. * For which choice of $$\alpha$$ is $$I_N^\alpha(t)$$ a martingale ?

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

## Calculating with Brownian Motion

Let $$W_t$$ be a standard brownian motion. Fixing an integer $$n$$ and a terminal time $$T >0$$, let $$\{t_i\}_{i=1}^n$$ be a partition of the interval $$[0,T]$$ with

$0=t_0 < t_1< \cdots< t_{n-1} < t_n=T$

Calculate the following two expressions:

1. $\mathbf{E} \Big(\sum_{k=1}^n W_{t_k} \big[ W_{t_{k}} – W_{t_{k-1}} \big] \Big)$
Hint: you might want to do the second part of the problem first and then return to this question and write
$W_{t_k} \big[ W_{t_{k}} – W_{t_{k-1}} \big]= W_{t_{k-1}} \big[ W_{t_{k}} – W_{t_{k-1}} \big]+ \big[W_{t_{k}} -W_{t_{k-1}}\big]\big[ W_{t_{k}} – W_{t_{k-1}}\big]$
2. $\mathbf{E} \Big(\sum_{k=1}^n W_{t_{k-1}} \big[ W_{t_{k}} – W_{t_{k-1}} \big] \Big)$

## Simple Numerical Exercise

Let $$\omega_i$$ , $$i=1,\cdots$$ be a collection of mutually independent, uniform on $$[0,1]$$ random variables. Define

$\eta_i(\omega)= \omega_i -\frac12$

and

$X_n(\omega) = \sum_{i=1}^n \eta_i(\omega)\,.$

1. What is $$\mathbf{E}\,X_n$$ ?
2. What is $$\mathrm{Var}(X_n)$$ ?
3. What is $$\mathbf{E}\,X_{n+k} | X_n$$ for $$n, k >0$$ ?
4. What is $$\mathbf{E}(\,X_5^2 \,|\, X_3)$$ ?
5. [optional] Write a computer program to simulate some realizations of this process viewing $$n$$ as time. Plot some plots of $$n$$ vs $$X_n$$.
6. [optional] How do you simulations agree with the first two parts ?

## 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.