# Tag Archives: JCM_math545_HW5_S14

## Ito Integration by parts

Recall that if $$u(t)$$ and $$v(t)$$ are deterministic functions which are once differentiable then the classic integration by parts formula states that
$\int_0^t u(s) (\frac{dv}{ds})(s)\,ds = u(t)v(t) – u(0)v(0) – \int_0^t v(s) (\frac{du}{ds})(s)\,ds$

As is suggested by the formal relations

$(\frac{dv}{ds})(s)\,ds=dv(s) \qquad\text{and}\qquad (\frac{du}{ds})(s)\, ds=du(s)$

this can be rearranged  to state

$u(t)v(t)- u(0)v(0)= \int_0^t u(s) dv(s) + \int_0^t v(s) du(s)$

which holds for more general Riemann–Stieltjes integrals. Now consider two Ito processes $$X_t$$ and $$Y_t$$ given by

$dX_t=b_s ds + \sigma_s dW_t \qquad\text{and}\qquad dY_t=f_s ds + g_s dW_t$

where $$W_t$$ is a standard Brownian Motion. Derive the “Integration by Parts formula” for Ito calculus by applying Ito’s formula to $$X_tY_t$$. Compare this the the classical formula given above.

## Cross-quadratic variation: correlated Brownian Motions

Let $$W_t$$ and $$B_t$$ be two independent standard Brownian Motions. For $$\rho \in [0,1]$$ define
$Z_t = {\rho}\, W_t +\sqrt{1-\rho^2}\, B_t$

1. Why is $$Z_t$$ a standard Brownian Motion ?
2. Calculate  the cross-quadratic variations $$[ Z,W]_t$$ and $$[ Z,B]_t$$ .
3. For what values of $$\rho$$ is $$W_t$$ independent of $$Z_t$$ ?
4. ** Argue that two standard Brownian motions  are independent if and only if  their cross-quadratic variation is zero.

## 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 ?)

## 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.

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

## 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.