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.


1 Answer 1


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.

  • $\begingroup$ Then what is the function of $Z$. In discrete time, it connects two measures. Here the function should also be the same. $\endgroup$ Dec 2, 2014 at 11:28
  • $\begingroup$ @user3001408: I assume that by function you mean purpose. RN derivatives are always used to construct one measure from another, e.g. here $\mathrm d\hat P := Z\,\mathrm dP$. so in your terms $Z$ connects $P$ and $\hat P$. Of course, in particular it also connects their marginal (or finite-dimensional) distributions $P_t$ and $\hat P_t$ in a sense that $\mathrm d\hat P_t = h(Z)\,\mathrm dP_t$ for some function $h$. My point is that it's quite hard to deduce the shape of function $h$. $\endgroup$
    – Ulysses
    Dec 2, 2014 at 11:40
  • $\begingroup$ I am little lost - if its hard then how would I know the distribution of the new probaibilty measure!!? Upvoted you for the answer :) $\endgroup$ Dec 2, 2014 at 11:43
  • $\begingroup$ @user3001408: how would I know the distribution of the new probaibilty measure - why do you think it is possible? As I told you, we know distribution of GBM at any point of time, and we can obtain $\mathrm dX_t = (X_t-\sqrt{|X_t|})\mathrm dt + X_t\mathrm dW_t$ through the GBM using Girsanov's theorem, but that does really mean we have analytical formula for distribution of $X_t$, right? $\endgroup$
    – Ulysses
    Dec 2, 2014 at 11:52
  • $\begingroup$ @user3001408: see my edit $\endgroup$
    – Ulysses
    Dec 2, 2014 at 15:04

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