4
$\begingroup$

Summary: long-story cut short, the question is asking for what types of functions $f(.)$, the Cameron-Martin-Girsanov theorem can be used as follows:

$$ \mathbb{E}^{\mathbb{P}^2}[f(W_t)]=\mathbb{E}^{\mathbb{P}^1}\left[\frac{d\mathbb{P}^2}{d\mathbb{P}^1}f(W_t)\right] $$

Long story: the Radon-Nikodym when changing from risk-neutral to Stock measure is:

$$\frac{dN^{S}}{dN^{Q}}=\frac{N^{Q}_{t_0}}{N^{Q}_{t}} \frac{N^{S}_{t}}{N^{S}_{t_0}}=\frac{1}{e^{rt}}\frac{S_t}{S_{t_0}}=e^{-0.5\sigma^2t+\sigma W_t}$$

The following type of calc is often seen in finance:

$$\mathbb{E}^{N^S}\left[S_t \right]=\mathbb{E}^{N^Q}\left[S_t^Q \frac{dN^{S}}{dN^{Q}} \right]=\\=\mathbb{E}[S_t^Q*e^{-0.5\sigma^2t+\sigma W_t}]=\\=S_0e^{rt-0.5\sigma^2t+\sigma W_t}*e^{-0.5\sigma^2t+\sigma W_t}=\\=S_0e^{rt+\sigma^2t}$$

The CMG theorem tells us that the Radon-Nikodym derivative $e^{-0.5\sigma^2t+\sigma W_t}$ can be applied to $W_t$ directly to modify it's drift and to create some new measure under which $W_t$ will no longer be a Standard Brownian motion. If we step-away from finance and denote the measure under which $W_t$ is standard Brownian as $\mathbb{P}^1$, the new measure under which $W_t$ is a Brownian with a drift as $\mathbb{P}^2$, and the radon-nikodym as $\frac{d\mathbb{P}^2}{d\mathbb{P}^1}$, we can write:

$$\mathbb{P}^2(W_t<a):=\mathbb{E}^{\mathbb{P}^1}\left[\frac{d\mathbb{P}^2}{d\mathbb{P}^1} * I_{\{W_t<a\}} \right] =\mathbb{E}^{\mathbb{P}^1}\left[e^{-0.5\sigma^2t+\sigma W_t} * I_{\{W_t<a\}} \right] $$

The above is basically the definition of $\mathbb{P^2}$ via the implicit definition of the Radon-Nikodym derivative. An extension of the above definition is that:

$$ \mathbb{E}^{\mathbb{P}^2}[W_t]=\mathbb{E}^{\mathbb{P}^1}\left[\frac{d\mathbb{P}^2}{d\mathbb{P}^1}W_t\right] $$

Question: in our finance case of stock, the stock price process is actually a function of $W_t$, so we could write $S_t=f(W_t)$. In the equation $\mathbb{E}^{N^S}\left[S_t \right]=\mathbb{E}^{N^Q}\left[S_t^Q \frac{dN^{S}}{dN^{Q}} \right]$, we are actually using the fact that:

$$ \mathbb{E}^{\mathbb{P}^2}[f(W_t)]=\mathbb{E}^{\mathbb{P}^1}\left[\frac{d\mathbb{P}^2}{d\mathbb{P}^1}f(W_t)\right] $$

Is there an easy way to prove that we can do that? Obviously it does work as shown in the case of the stock price process above, because it produces the correct result. But for what $f(.)$ does the result hold? I am sure there must be some restrictions on the types of functions $f(.)$ for which the result holds true.

$\endgroup$
0
3
$\begingroup$

(I might not be answering your question, but I feel this clarification is needed.)

A random variable $X$ of $(\Omega, \mathcal{F})$ is a $\mathcal{F}$-measurable function $X : \Omega → \mathbf{R}$. So, $X$ depends on $\Omega$ and $\mathcal{F}$, but does not depend on the probability measure put on $(\Omega, \mathcal{F})$. It is the distribution of $X$ that depends on the measure.

Given $P1$ and $P_2$ probability measures on $(\Omega, \mathcal{F})$, where $P_2$ is $P_1$-absolutely continuous on $\mathcal{F}$ and $$ L = \frac{dP_2}{dP_1} $$ is the Radon-Nicodym derivative ($\mathcal{F}$-measurable, $\mathcal{P_1}$-integrable), we have: $$X\in L^1(\Omega, P_2) \iff XL\in L^1(\Omega, P_1).$$ In that case, we then have: $$ \mathbf{E}^{P_2}[X] = \mathbf{E}^{P_1}[XL] $$

or in its integral form:

$$ \int_\Omega X dP_2 = \int_\Omega X \frac{dP_2}{dP_1} dP_1 $$

(Note that there is no need to introduce notation $X^{P_2}$ competing with $X$.)

For your question:

$$ \mathbf{E}^{P_2}[f(W_t)] = \mathbf{E}^{P_2}[f(W_t^\theta -\int_0^t \theta_u du)] $$

if $P_2$ is the Girsanov measure built from process $\theta$ and $W_t^\theta = W_t +\int_0^t \theta_u du$ is the induced Brownian motion under $P_2$ ($W_t$ is a Brownian motion under $P_1$). One can compute the expectation under $P_2$. Or go back to $P_1$ as you said:

$$ \mathbf{E}^{P_2}[f(W_t)] = \mathbf{E}^{P_1}\left[f(W_t)\frac{dP_2}{dP_1} \right]. $$

In your case $\theta_t = \sigma$ and

$$ \frac{dP_2}{dP_1} =\exp\left(-\frac{\sigma^2}{2} t + \sigma W_t \right). $$

$\endgroup$
3
  • $\begingroup$ I see: so there is no big deal applying the result to the function of the Brownian as opposed to just plain Brownian motion...? $\endgroup$ – Jan Stuller Jul 23 '20 at 17:49
  • 1
    $\begingroup$ I don't think so. Integrability to be able to get the expectation. $\endgroup$ – ir7 Jul 23 '20 at 18:02
  • 3
    $\begingroup$ Of course, for $f(W_t)$ or $g(W_t^\theta)$ dynamics (SDE's) calculation, we need Ito's conditions (or extended Ito/Ito-Tanaka) on $f$ and $g$, to be able to apply those theorems. $\endgroup$ – ir7 Jul 23 '20 at 18:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.