Radon-Nikodym Derivative
Definition
RN derivative $Z$ is the link between two equavalent probability measure $P$ and $\tilde{P}$. $Z$ can be defined as below:
\[Z = \frac{d\tilde{P}}{dP}\]equavalent probability measure means these two probability measure must agree on what is possible and what is impossible
\[P(\omega) = \tilde{P}(\omega)= 0\]
It can be also written as:
\[\tilde P(A) = \int_{\omega \in A}Z(\omega)dP(\omega)\]Furthermore:
\[\tilde E(X) = \int_{\omega \in \Omega}X(\omega)d\tilde P(\omega) = E(XZ) = \int_{\omega \in \Omega}X(\omega)Z(\omega)dP(\omega)\]Properties of RN Derivative
RN derivative must satisfy the following two properties:
- $E(Z) = \int_{\omega \in \Omega}Z(\omega)dP(\omega) = 1$
- $Z(\omega) > 0$ for $\omega \in \Omega$ almost surely
The first propertity is to make sure that $\tilde P(\Omega) = 1$. This is one of the requirement that make $\tilde P$ to be a probability measure.
The second propertity is to make sure there is no negative probability.
Change Measure for Standard Normal Variable
Suppose that X is a normal distribution $N(\theta, \sigma^2)$ under measure $P$.
We can construct a RN derivative as below:
\[Z(\omega) = e^{-\theta X(\omega) - \frac{1}{2}\theta^2}\]The expectation under $\tilde P$ is calculated as below:
\[\tilde E(X) = \int x * \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}}*e^{-\theta x - \frac{1}{2}\theta^2}dx\\ =\int x * \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x+\theta)^2}{2\sigma^2}}dx = 0\]Under the new measure $\tilde P$, X is still normal distributed. But with 0 mean.
Radon-Nikodym Derivative Process
In previous discussion, it’s based on a single random variable. This scetion will talk about changing the measure for a whole process.
For $0 \le t \le T$, where $T$ is a fixed final time. Suppose $Z$ is an almost surely postive random variable satisfying $E(Z) = 1$. RN Derivative process can be defiend as:
\[Z(t) = E[Z|\mathcal{F}(t)]\]Girsanov Theorem
Define
\[Z(t) = exp(-\int_0^t\theta(u)dW(u)-\frac{1}{2}\int_0^t\theta^2(u)du)\] \[\tilde W(t) = W(t) + \int_0^t\theta(u)du\]It’s clearly in differential form :
\[d\tilde W(t) = dW(t) + \theta(t)dt\]Set $Z = Z(T)$. Using $Z(t)$ to change the measure, $\tilde W(t)$ is a brownian motion under new measure $\tilde P$.
