The Radon–Nikodym Derivative

Two people can disagree about probabilities without disagreeing about what is possible. A pessimist and an optimist may price the same coin differently, yet both agree the coin can land heads and can land tails. When two measures \mathbb{P} and \mathbb{Q} agree on exactly which events are impossible — the same null sets — we call them equivalent and write \mathbb{P} \sim \mathbb{Q}.

Equivalence is the precise licence to translate one measure into the other. The dictionary is a single non-negative random variable, the density (or Radon–Nikodym derivative)

Z = \frac{d\mathbb{Q}}{d\mathbb{P}} \ge 0,

which reweights \mathbb{P} into \mathbb{Q} outcome by outcome. Where Z > 1 the optimist counts an outcome as more likely than the pessimist; where 0 < Z < 1, less likely. Formally, \mathbb{Q} measures a set A by summing Z over it under \mathbb{P}:

\mathbb{Q}(A) = \int_A Z \, d\mathbb{P} \qquad \text{for every event } A.

It helps to think of Z as a currency exchange rate between measures. Just as a currency converter turns a price in dollars into a price in euros by multiplying by the day's rate, Z turns a \mathbb{P}-expectation into a \mathbb{Q}-expectation by multiplying, outcome by outcome, before averaging. Every risk-neutral price you will ever compute is secretly this trick in disguise: somewhere behind the scenes sits a density Z converting the real-world measure into a fictional one in which discounted asset prices are fair games — and Girsanov's theorem is precisely the machine that manufactures Z for continuous-time models.

From sets to expectations, derived line by line

The defining relation \mathbb{Q}(A) = \int_A Z\,d\mathbb{P} only talks about probabilities of sets. We will upgrade it into the full change-of-measure formula for expectations,

\mathbb{E}_{\mathbb{Q}}[X] = \mathbb{E}_{\mathbb{P}}[XZ],

by the standard ladder of integration theory: start with indicators, climb to simple functions, then to limits.

Step 1 — read the definition as an expectation. The integral of Z over A is exactly the \mathbb{P}-expectation of Z masked by the indicator \mathbf{1}_A (which is 1 on A and 0 off it):

\mathbb{Q}(A) = \int_A Z\,d\mathbb{P} = \mathbb{E}_{\mathbb{P}}[\mathbf{1}_A\, Z].

Step 2 — the indicator case of the formula. But the \mathbb{Q}-probability of A is itself the \mathbb{Q}-expectation of its indicator, \mathbb{Q}(A) = \mathbb{E}_{\mathbb{Q}}[\mathbf{1}_A]. Comparing with Step 1, the formula \mathbb{E}_{\mathbb{Q}}[X] = \mathbb{E}_{\mathbb{P}}[XZ] already holds for every X = \mathbf{1}_A:

\mathbb{E}_{\mathbb{Q}}[\mathbf{1}_A] = \mathbb{E}_{\mathbb{P}}[\mathbf{1}_A\, Z].

Step 3 — extend to simple functions by linearity. A simple random variable is a finite sum X = \sum_{k} c_k\,\mathbf{1}_{A_k}. Both expectations are linear, so applying Step 2 term by term,

\mathbb{E}_{\mathbb{Q}}[X] = \sum_k c_k\,\mathbb{E}_{\mathbb{Q}}[\mathbf{1}_{A_k}] = \sum_k c_k\,\mathbb{E}_{\mathbb{P}}[\mathbf{1}_{A_k} Z] = \mathbb{E}_{\mathbb{P}}\Big[\Big(\textstyle\sum_k c_k \mathbf{1}_{A_k}\Big) Z\Big] = \mathbb{E}_{\mathbb{P}}[XZ].

Step 4 — pass to the limit. Any non-negative measurable X is an increasing limit of simple functions X_n \uparrow X. Step 3 holds for each X_n, and monotone convergence lets us take the limit inside both expectations (note X_n Z \uparrow XZ since Z \ge 0):

\mathbb{E}_{\mathbb{Q}}[X] = \lim_{n\to\infty} \mathbb{E}_{\mathbb{Q}}[X_n] = \lim_{n\to\infty} \mathbb{E}_{\mathbb{P}}[X_n Z] = \mathbb{E}_{\mathbb{P}}[XZ].

Splitting a general X = X^+ - X^- into positive and negative parts extends it to any integrable X. The formula is proved.

The density integrates to one

Step 5 — take A = \Omega. The whole-space case of the defining relation is the normalisation that pins Z down. Since \mathbf{1}_\Omega = 1,

\mathbb{E}_{\mathbb{P}}[Z] = \int_\Omega Z\,d\mathbb{P} = \mathbb{Q}(\Omega) = 1,

because \mathbb{Q}, being a probability measure, assigns total mass 1. So a valid density is any non-negative Z with \mathbb{P}-mean exactly 1: it is a reweighting that neither creates nor destroys total probability. For the new measure to be equivalent (not merely absolutely continuous) we need a touch more — Z > 0 almost surely — so that the dictionary runs both ways and 1/Z = d\mathbb{P}/d\mathbb{Q}.

Let \mathbb{P} and \mathbb{Q} be probability measures on (\Omega, \mathcal{F}). Then:

The two notions are easy to confuse but do different jobs. \mathbb{Q} \ll \mathbb{P} (absolute continuity) is one-directional: anything \mathbb{P} rules out, \mathbb{Q} rules out too — but \mathbb{Q} may have extra impossible events of its own (those are the places where Z = 0). Equivalence is the two-sided version, \mathbb{Q} \ll \mathbb{P} and \mathbb{P} \ll \mathbb{Q} together: the same null sets, no exceptions, so Z > 0 everywhere and you can divide by it.

Statisticians know Z by another name: a likelihood ratio. If \mathbb{P} and \mathbb{Q} have densities p and q against some common reference, then Z(\omega) = \frac{d\mathbb{Q}}{d\mathbb{P}}(\omega) = \frac{q(\omega)}{p(\omega)}, the ratio of how plausible \omega is under the two hypotheses. Every time you compute a likelihood ratio you are silently writing down a Radon–Nikodym derivative.

Keep this dictionary in your back pocket: a "likelihood ratio," a "Radon–Nikodym derivative," and "the density that reweights \mathbb{P} into \mathbb{Q}" are three names for exactly the same object.

Worked example 1 — a two-outcome coin, checked by hand

Let \Omega = \{H, T\} with \mathbb{P}(H) = 0.6, \mathbb{P}(T) = 0.4, and a second measure \mathbb{Q}(H) = 0.75, \mathbb{Q}(T) = 0.25. Both measures put positive weight on both outcomes, so \mathbb{P} \sim \mathbb{Q}.

Step 1 — build the density outcome by outcome. On a finite space the density is simply the ratio of the two probabilities at each point:

Z(H) = \frac{\mathbb{Q}(H)}{\mathbb{P}(H)} = \frac{0.75}{0.6} = 1.25, \qquad Z(T) = \frac{\mathbb{Q}(T)}{\mathbb{P}(T)} = \frac{0.25}{0.4} = 0.625.

Step 2 — check the normalisation \mathbb{E}_{\mathbb{P}}[Z] = 1.

\mathbb{E}_{\mathbb{P}}[Z] = \mathbb{P}(H)Z(H) + \mathbb{P}(T)Z(T) = (0.6)(1.25) + (0.4)(0.625) = 0.75 + 0.25 = 1. \checkmark

Step 3 — verify the change-of-measure formula on a real random variable. Let X(H) = 10, X(T) = 2 (say, a payoff). Compute both sides independently:

\mathbb{E}_{\mathbb{Q}}[X] = (0.75)(10) + (0.25)(2) = 7.5 + 0.5 = 8, \mathbb{E}_{\mathbb{P}}[XZ] = (0.6)(10)(1.25) + (0.4)(2)(0.625) = 7.5 + 0.5 = 8. \checkmark

The two computations, done completely independently, land on the same number. That is the entire content of the theorem, seen with no machinery at all: reweighting the outcomes with Z before averaging under \mathbb{P} gives exactly the same answer as averaging under \mathbb{Q} directly.

Worked example 2 — Girsanov's kernel, derived in one completed square

Let \mathbb{P} give X a standard normal density p(x) = \tfrac{1}{\sqrt{2\pi}} e^{-x^2/2}, and let \mathbb{Q} give it the shifted density q(x) = \tfrac{1}{\sqrt{2\pi}} e^{-(x-\theta)^2/2} — the same bell curve, slid sideways to mean \theta. What is Z = d\mathbb{Q}/d\mathbb{P}?

Step 1 — take the ratio of densities (the constants 1/\sqrt{2\pi} cancel):

Z(x) = \frac{q(x)}{p(x)} = \exp\!\left(-\frac{(x-\theta)^2}{2} + \frac{x^2}{2}\right).

Step 2 — expand the square in the exponent. (x-\theta)^2 = x^2 - 2\theta x + \theta^2, so

-\frac{(x-\theta)^2}{2} + \frac{x^2}{2} = -\frac{x^2 - 2\theta x + \theta^2}{2} + \frac{x^2}{2} = \theta x - \frac{\theta^2}{2}.

Step 3 — read off the density.

Z(x) = e^{\theta x - \theta^2/2}.

This is exactly the tilt used in the chart below, and it is precisely the exponential martingale that Girsanov's theorem builds in continuous time to shift a Brownian motion's drift. The whole of Girsanov's theorem is, in essence, this one completed-square calculation done at every instant along a path.

When the derivative fails to exist

The theorem's \mathbb{Q} \ll \mathbb{P} hypothesis is not red tape — drop it and there is no density to speak of. Let \Omega = [0, 1], let \mathbb{P} be the uniform (Lebesgue) measure, and let \mathbb{Q} put all its probability on the single point \{0.5\} (a point mass, or "Dirac" measure). Take A = \{0.5\}: since a single point has zero length, \mathbb{P}(A) = 0 — but by construction \mathbb{Q}(A) = 1 \ne 0. A \mathbb{P}-null set is not \mathbb{Q}-null, so \mathbb{Q} \ll \mathbb{P} fails outright, and no function Z with \mathbb{Q}(A) = \int_A Z\, d\mathbb{P} can possibly exist — any such integral over a single point is 0, never 1. Some pairs of measures simply refuse to be exchanged.

Four points about the density that are easy to state and easy to misuse:

Suppose you need to estimate the probability that a stock crashes 40% in a day, or that a server farm gets hit by an outage during its rarest failure mode — events so unlikely under the real-world measure \mathbb{P} that a plain Monte Carlo simulation would need billions of draws before seeing even one. The trick, called importance sampling, is a direct application of this page: simulate instead from a friendlier measure \mathbb{Q} under which the rare event is common, then undo the distortion by reweighting each simulated path with the density running the other way,

\mathbb{E}_{\mathbb{P}}[h(X)] = \mathbb{E}_{\mathbb{Q}}\!\left[h(X)\,\frac{d\mathbb{P}}{d\mathbb{Q}}(X)\right] = \mathbb{E}_{\mathbb{Q}}\!\left[h(X)\, \frac{1}{Z(X)}\right].

Concretely: to estimate \mathbb{P}(X > 4) for a standard normal X, don't draw a billion standard normals hoping a handful exceed 4. Draw from \mathbb{Q} = N(4, 1) instead — now crossing 4 is an everyday event — and multiply each hit by 1/Z(X) to convert the count back into a \mathbb{P}-probability. Same idea, same formula, wildly fewer simulations.

Reweighting a density, live

Take a base measure \mathbb{P} with a standard-normal density and tilt it by an exponential factor Z(x) = e^{\theta x - \theta^2/2} — the very density derived by completing the square in worked example 2 above, and exactly the shape Girsanov will use. This Z has \mathbb{P}-mean 1 (so total probability is preserved), and multiplying it onto the \mathbb{P} bell produces the \mathbb{Q} density — a normal shifted to mean \theta. Slide \theta to watch the mass slosh sideways while the area stays 1.

The faint fixed curve is the original \mathbb{P} density; the bold curve is \mathbb{Q} = Z\cdot\mathbb{P}. At \theta = 0 the tilt is Z \equiv 1 and the two coincide — no reweighting at all.