Change of Measure/Girsanov’s Theorem Explained

Change of Measure or Girsanov’s Theorem is such an important theorem in Real Analysis or Quantitative Finance. Unfortunately, I never really understood it until much later after having left school. I blamed it to the professors and the textbook authors, of course.  The textbook version usually goes like this.

Given a probability space {\Omega,\mathcal{F},P}, and a non-negative random variable Z satisfying \mathbb{E}(Z) = 1 (why 1?). We then defined a new probability measure Q by the formula, for all A in \mathcal{F}.

Q(A) = \int _AZ(\omega)dP(w)

Any random variable X, a measurable process adapted to the natural filtration of the \mathcal{F}, now has two expectations, one under the original probability measure P, which denoted as \mathbb{E}_P(X), and the other under the new probability measure Q, denoted as \mathbb{E}_Q(X). They are related to each other by the formula

\mathbb{E}_Q(X) = \mathbb{E}_P(XZ)

If P(Z > 0) = 1, then P and Q agree on the null sets. We say Z is the Radon-Nikodym derivatives of Q with respect to P, and we write Z = \frac{dQ}{dP}. To remove the mean, μ, of a Brownian motion, we define

Z=\exp \left ( -\mu X - \frac{1}{2} \mu^2 \right )

Then under the probability measure Q, the random variable Y = X + μ is standard normal. In particular, \mathbb{E}_Q(X) = 0 (so what?).

This text made no sense to me when I first read it in school. It was very frustrated that the text was filled with unfamiliar terms like probability space and adaptation, and scary symbols like integration and \frac{dQ}{dP}. (I knew what \frac{dy}{dx} meant when y was a function and x a variable. But what on earth were dQ over dP?)

Now after I have become a professor to teach students in finance or financial math, I would get rid of all the jargon and rigorousness. I would focus on the intuition rather than the math details (traders are not mathematicians). Here is my laymen version.

Given a probability measure P. A probability measure is just a function that assigns numbers to a random variable, e.g., 0.5 to head and 0.5 to tail for a fair coin. There could be another measure Q that assigns different numbers to the head and tail, say, 0.6 and 0.4 (an unfair coin)! Assume P and Q are equivalent, meaning that they agree on what events are possible (positive probabilities) and what events have 0 probability. Is there a relation between P and Q? It turns out to be a resounding yes!

Let’s define Z=\frac{Q}{P}. Z here is a function as P and Q are just functions. Z is evaluated to be 0.6/0.5 and 0.4/0.5. Then we have

\mathbb{E}_Q(X) = \mathbb{E}_P(XZ)

This is intuitively true when doing some symbol cancellation. Forget about the proof even though it is quite easy like 2 lines. We traders don’t care about proof. Therefore, the distribution of X under Q is (by plugging in the indicator function in the last equation):

\mathbb{E}_Q(X \in A) = \mathbb{E}_P(I(X \in A)Z)

Moreover, setting X = 1, we have (Z here is a random variable):

\mathbb{E}_Q(X) = 1 = \mathbb{E}_P(Z)

These results hold in general, especially for the Gaussian random variable and hence Brownian motion. Suppose we have a random (i.e., stochastic) process generated by (adapted to) a Brownian motion and it has a drift μ under a probability measure P. We can find an equivalent measure Q so that under Q, this random process has a 0 drift. Wiki has a picture that shows the same random process under the two different measures: each of the 30 paths in the picture has a different probability under P and Q.

The change of measure, Z, is a function of the original drift (as would be guessed) and is given by:

Z=\exp \left ( -\mu X - \frac{1}{2} \mu^2 \right )

For a 0 drift process, hence no increment, the expectation of the future value of the process is the same as the current value (a laymen way of saying that the process is a martingale.) Therefore, with the ability to remove the drift of any random process (by finding a suitable Q using the Z formula), we are ready to do options pricing.

Now, if you understand my presentation and go back to the textbook version, you should have a much better understanding and easier read, I hope.