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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
\[ Z_t = {\rho}\, W_t +\sqrt{1-\rho^2}\, B_t\]
- Why is \(Z_t\) a standard Brownian Motion ?
- Calculate the cross-quadratic variations \([ Z,W]_t\) and \([ Z,B]_t\) .
- For what values of \(\rho\) is \(W_t\) independent of \(Z_t\) ?
- ** 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:
- \(\int_0^t B_t \circ dB_t\) (Explain why this agrees with the answer you obtained here).
- 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 ?
- (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.
- Find a function \(f(t)\) so that \(W_t^2 -f(t)\) is a Martingale.
- * 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 \(\omega\).
- 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 \(k\). - 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.
