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
dX_t&=f_tdt + g_tdW_t
and \(B_t\) be a second (possibly correlated with \(W\) ) Brownian
motion. We define the Stratanovich integral \(\int X_t \circ dB_t\)  by
\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
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
    dZ_t&=b_tdt + \sigma_tdW_t\;.
    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
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)
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.