Deriving Shreve’s Girsanov Result with Quadratic Covariation

# Girsanov Theorem and Quadratic Variation


# Girsanov Theorem and Quadratic Variation












Girsanov theorem seems to have many different forms. I've got a problem matching the form in wiki to the one in Shreve's book, due to the difficulty of quadratic variation calculation.

Below is the Girsanov Theorem from wiki:

> Let $\{W_t\}$ be a Wiener process on the Wiener probability space $\{\Omega,\mathcal{F},P\}$. Let $X_t$ be a measurable process adapted to the natural filtration of the Wiener process $\{\mathcal{F}^W_t\}$. Given an adapted process $X_t$ with $X_0 = 0$, define $Z_t=\mathcal{E}(X)_t,\,$ where $\mathcal{E}(X)$ is the stochastic exponential (or Doléans exponential) of X with respect to W, i.e. $\mathcal{E}(X)_t=\exp \left ( X_t - \frac{1}{2} [X]_t \right )$, where $[X]_t$ is a quadratic variation for $X_t$. Thus $Z_t$ is a strictly positive local martingale, and a probability measure $Q$ can be defined on $\{\Omega,\mathcal{F}\}$ such that we have Radon–Nikodym derivative $\frac{d Q}{d P} |_{\mathcal{F}_t} = Z_t = \mathcal{E} (X)_t$. Then for each $t$ the measure $Q$ restricted to the unaugmented sigma fields $\mathcal{F}^W_t$ is equivalent to $P$ restricted to $\mathcal{F}^W_t.\,$ Furthermore if $Y$ is a local martingale under $P$ then the process $\tilde Y_t = Y_t - \left[ Y,X \right]_t$ is a $Q$ local martingale on the filtered probability space $\{\Omega,F,Q,\{F^W_t\}\}$.

Below is the Girsanov Theorem from Shreve's book "Stochastic calculus for finance II"

> Theorem 5.2.3 (Girsanov, one dimension). Let $W(t)$, $0 \leq t \leq T$, be a Brownian motion on a probability space $(\Omega, \mathscr F, \mathbb P)$, and let $\mathscr F(t)$, $0 \leq t \leq T$, be a filtration for this Brownian motion. Let $\Theta(t)$, $0 \leq t \leq T$, be an adapted process. Define $$Z(t) = \text{exp} \left\{ -\int_0^t \Theta(u)dW(u) - \frac{1}{2} \int_0^t \Theta^2(u) du \right \}, \tag{5.2.11}$$ $$\widetilde W(t) = W(t) + \int_0^t \Theta(u) du, \tag{5.2.12}$$ and assume that $$\mathbb E \int_0^T \Theta^2(u) Z^2(u) du < \infty \tag{5.2.13}$$ Set $Z = Z(T)$. Then $\mathbb E Z = 1$ and under the probability measure $\widetilde P$ given by (5.2.1), the process $\widetilde W(t)$, $0 \leq t \leq T$, is a Brownian motion.

Seems the Girsanov theorem form wiki is more general than the one on Shreve's book.

Now my questions is: How to derive the latter from the former?

It seems only need to take $Y(t) = W(t)$ and $X(t) = \int_0^t \Theta(u) du$ in the wiki definition. This left to prove

$$[W(t), \int_0^t \Theta(u) du]_t = - \int_0^t \Theta(u) du$$

, but how to calculate the quadratic variation?

Quadratic variation definition is $$[X,Y](T) := \lim_{\|\Pi\|\to 0} \sum_{j=0}^{n-1} \left[ X(t_{j+1}) - X(t_j) \right] \left[ Y(t_{j+1}) - Y(t_j) \right]$$ , where $\Pi := \{ t_0, t_1, \cdots, t_n \}$ . But I'm a bit stuck here.

Could you please kindly give me some hint how to proceed?

## Answer by emcor (score 11, accepted)

https://quant.stackexchange.com/a/14795

Shreve's theorem also called "Girsanov II" indeed represents a special case of the general "Girsanov I" from Wiki above, with $$Y_t:=W_t,$$$$X_t:=-\int_0^t\Theta_udW_u$$

We can show: $$[Y,X]=-\int_0^t\Theta_udu$$ by using general Stochastic Calculus rules (e.g. p.37, 6.6 here):

$$[Y,X]=[W_t,-\int_0^t\Theta_udW_u]=-\int_0^t\Theta_ud[W_u,W_u]=-\int_0^t\Theta_udu$$

as $[W,W]=[W]=t$.