Girsanov Change of Measure and the Density of a Process at One Time
Summary
The document clarifies the distinction between a probability measure on entire paths and the distribution of a process observed at a fixed time. Knowing the pathwise Radon–Nikodym process does not generally give the new one-time density by evaluating it at a single endpoint. To obtain the density change for an observed variable, the answer expresses the new density as the old density multiplied by the conditional expectation of the Radon–Nikodym process given that variable.
This conditional expectation accounts for all paths consistent with the observed value. The response notes that computing it can be difficult and may require nontrivial conditional expectations. It uses this difficulty as a sanity check against the idea that one could easily derive the density of almost any drift-modified process from the geometric Brownian motion density. The discussion gives conceptual guidance rather than a worked calculation for the specific interval in the question.
Key ideas
- A probability measure on paths is distinct from a process distribution at a fixed time.
- The new density of an observed variable is its old density multiplied by a conditional expectation of the pathwise likelihood ratio.
- That conditional expectation is taken given the observed process value.
- Computing the required conditional expectation can be difficult and may lack a simple analytic form.
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# Girsanov's Theorem - Change of Measure
# Girsanov's Theorem - Change of Measure
I have trouble understanding Girsanov's theorem. The Radon Nikodym process $Z$ is defined by:
$$Z(t)=\exp\left(-\int_0^t\phi(u) \, \mathrm dW(u) - \int_0^t\frac{\phi^2(u)}{2} \, \mathrm du\right)$$
Now $\hat P$ is a new probability measure. The trouble is I am not understanding how to go from old $P$ to the new one. The old $P$ is normally distributed with mean $0$ and variance $t$. Now say I want to know the new probability for an infinitesimally small interval around $0.2$. For that I need to know the value of $Z$ at this interval (event you may say). And then I can multiply (integrate) the value of $Z$ with old $P$, and get new $\hat P$.
Assume $t$ is fixed.
I have no idea how to calculate the value of $Z$ for this interval/event. Help would be appreciated.
## Answer by Ulysses (score 4)
https://quant.stackexchange.com/a/15724
Let's distinguish: measure $P$ gives probability over the paths, hence you can't really say that it has certain mean and variance: that would apply to a measure $P_t$ which restricts $P$ to the time instance $t$. Now, if you know that $P_t$ has density $f$ (with respect to Lebesgue measure) and $\frac{\mathrm d\hat P_t}{\mathrm d P_t} = g$ then $\hat P_t$ has density $$ \frac{\mathrm d\hat P_t}{\mathrm d \lambda} = \frac{\mathrm d\hat P_t}{\mathrm d P_t}\cdot \frac{\mathrm d P_t}{\mathrm d \lambda} = g\cdot f. $$ Unfortunately, I do not know whether you can get $g$ directly out of $Z$ - in fact, it does not seem to be that you can always do this in some nice analytical way since it involves computing rather peculiar conditional expectations. Another sanity check: if there would be an easy way to find $g$ out of $Z$, then just by knowing the density $f$ for the Geometric Brownian motion $\mathrm dX_t = \sigma X_t\,\mathrm dW_t$ would allow you to know densities for any process of the form $\mathrm dX_t = \mu_t\,\mathrm dt + \sigma X_t\,\mathrm dW_t$ for pretty much any adapted process $\mu_t$. I'm pretty sure that even if $\mu_t = \mu(X_t)$ there are a lot of cases where the densities are still not known.
Edit: to support the intuition above, I've computed $g$ in terms of $Z$, and indeed it looks pretty simple $g(x) = \Bbb E[Z_t|X_t = x]$ but its computation would be rather hard in general.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.