The Radon-Nikodym Derivative

Options are priced with probabilities that nobody believes. The textbook says to value a call by taking an expected payoff — but under a set of probabilities in which every asset drifts at the risk-free rate, which is plainly not how anyone forecasts a stock. Two obvious questions follow. How can you legitimately swap one set of probabilities for another? And if you do, what exactly is the exchange rate? The Radon-Nikodym derivative is the answer to both, and it is a smaller idea than the name suggests.

Two forecasters, one world

Two analysts watch the same stock. They agree completely on what can happen: up 20, flat, down 20. They disagree on the odds. The first says 50/30/20. The second says 25/50/25 — she is more cautious, so she shifts weight away from the good outcome.

You want to convert any calculation done in the first analyst's world into the second's. You do not need to redo the work. You just need, for each scenario, a conversion factor: how much more or less the second analyst weights it. Up 20 gets a factor of $0.25 / 0.50 = 0.5$ (halved). Flat gets $0.50/0.30 = 1.667$ (boosted). Down 20 gets $0.25/0.20 = 1.25$.

That table of factors is the Radon-Nikodym derivative. It is a currency exchange rate, except the rate depends on which scenario you are in. Everything below is that, written carefully.

Writing it down

Let $P$ and $Q$ be two probability measures on the same sample space, meaning two consistent assignments of odds to the same set of scenarios. Suppose $Q$ is absolutely continuous with respect to $P$, written $Q \ll P$, which means:

$$P(A) = 0 \;\Longrightarrow\; Q(A) = 0 \quad \text{for every event } A .$$

In words: anything $P$ rules out, $Q$ rules out too — $Q$ cannot invent possibilities that $P$ says never happen. That is the one condition you need, and it is the condition the theorem below cannot do without.

The Radon-Nikodym theorem then says there is a single random variable $Z \ge 0$, written $Z = \dfrac{dQ}{dP}$, such that

$$Q(A) = \int_A Z \, dP \qquad \text{for every event } A .$$

Here $A$ is any event, $\int_A \dots dP$ is the Lebesgue integral over that event under $P$, and $Z$ is the exchange rate. Read out loud: to get $Q$'s probability of an event, walk over the scenarios in that event, and add up their $P$-probabilities after re-weighting each one by $Z$.

The version you actually use is the one for expectations. For any random variable $X$,

$$E_Q[X] = E_P\!\left[ X \, Z \right] .$$

An average under the new measure is the average under the old measure of the quantity multiplied by the exchange rate. Two facts fall straight out. Setting $X = 1$ gives $E_P[Z] = 1$: the weights must average to one, or the new probabilities would not sum to one. And $Z$ is a random variable, not a number — it takes a different value in each scenario.

Worked example 1: pricing a call two ways

Take the three-outcome world above, with a stock at 100 today and a zero interest rate. Under $P$ the probabilities are $(0.50,\, 0.30,\, 0.20)$ for final prices $(120,\, 100,\, 80)$.

First check that $Q = (0.25,\, 0.50,\, 0.25)$ deserves to be called the pricing measure. It must make the stock a fair game:

$$E_Q[S] = 0.25(120) + 0.50(100) + 0.25(80) = 30 + 50 + 20 = 100 ,$$

which equals today's price, so yes — under $Q$ the stock is a martingale. Now build $Z$ scenario by scenario, dividing $Q$ by $P$:

scenario	$P$	$Q$	$Z = Q/P$
$S = 120$	0.50	0.25	0.500
$S = 100$	0.30	0.50	1.667
$S = 80$	0.20	0.25	1.250

Verify the weights average to one under $P$:

$$E_P[Z] = 0.50(0.500) + 0.30(1.667) + 0.20(1.250) = 0.25 + 0.50 + 0.25 = 1.000 . \;\checkmark$$

Now price a call struck at 100. Its payoffs across the three scenarios are $20$, $0$, $0$.

Route A — average under $Q$ directly:

$$E_Q[\text{payoff}] = 0.25(20) + 0.50(0) + 0.25(0) = 5 .$$

Route B — stay under $P$ and carry the weights:

$$E_P[\text{payoff} \times Z] = 0.50(20 \times 0.500) + 0.30(0) + 0.20(0) = 0.50 \times 10 = 5 .$$

Both give a fair value of $5. That is the whole trick: you never have to leave your own probabilities, you just multiply by $Z$ inside the expectation.

Worked example 2: shifting the drift of a bell curve

Now the continuous case, which is Girsanov in miniature. Let the annual return $X$ be normal with mean $\mu = 0.08$ and standard deviation $\sigma = 0.20$ under $P$. You want a measure $Q$ under which the mean is the risk-free rate $\nu = 0.02$, with the standard deviation unchanged.

Divide one normal density by the other and simplify — the $x^2$ terms cancel, which is the reason this works at all:

$$Z(x) = \frac{dQ}{dP}(x) = \exp\!\left( \frac{(\nu - \mu)\,x}{\sigma^2} \;-\; \frac{\nu^2 - \mu^2}{2\sigma^2} \right).$$

Fill in the numbers. $\sigma^2 = 0.20^2 = 0.04$. The first coefficient is $(0.02 - 0.08)/0.04 = -0.06/0.04 = -1.5$. The constant is $-(0.0004 - 0.0064)/(2 \times 0.04) = 0.006/0.08 = 0.075$. So

$$Z(x) = \exp(-1.5\,x + 0.075).$$

In English: every scenario is re-weighted by an exponential that falls as the return rises — good outcomes get marked down, bad outcomes get marked up. Evaluate it at four returns:

return $x$	$-1.5x + 0.075$	$Z(x)$
$-0.20$ (bad year)	$0.375$	1.455
$0.02$ (risk-free)	$0.045$	1.046
$0.08$ (P-mean)	$-0.045$	0.956
$0.40$ (great year)	$-0.525$	0.592

A 40% year is counted at 59% of its real-world weight; a $-20\%$ year at 146%. That single sentence is what "risk-neutral" means in practice.

Two checks you can do by hand. First, the weights must average to one. Using the normal moment formula $E[e^{aX}] = \exp(a\mu + \tfrac{1}{2}a^2\sigma^2)$ with $a = -1.5$:

$$E_P[Z] = e^{0.075} \exp\!\left(-1.5(0.08) + \tfrac{1}{2}(2.25)(0.04)\right) = e^{0.075} e^{-0.12 + 0.045} = e^{0.075} e^{-0.075} = 1 . \;\checkmark$$

Second, confirm the ratio really is the ratio of the two densities at, say, $x = 0.40$. The exponent of the $P$-density there is $-(0.40 - 0.08)^2 / (2 \times 0.04) = -0.1024/0.08 = -1.280$, giving $e^{-1.280} = 0.2780$. For the $Q$-density: $-(0.40 - 0.02)^2/0.08 = -0.1444/0.08 = -1.805$, giving $e^{-1.805} = 0.1645$. Their ratio is $0.1645 / 0.2780 = 0.592$ — exactly the table entry. ✓

What this means in practice

• Risk-neutral pricing. A derivative's value is $E_Q[\text{payoff}]$ discounted. Girsanov's theorem is worked example 2 applied to a whole Brownian path: the change of measure shifts the drift to the risk-free rate and leaves the volatility untouched. That is why implied volatility is meaningful and "implied drift" is not.
• Importance sampling. Simulating a rare loss under the true measure wastes almost every path. Simulate under a tilted measure that makes the loss common, then multiply each sample by $dP/dQ$ to remove the bias. Same identity, run backwards. See importance sampling.
• Likelihood ratios. In statistics, the likelihood ratio between two models is a Radon-Nikodym derivative. Neyman-Pearson tests, sequential tests and Bayes factors are all built on the same object.
• Densities themselves. "The density of $X$" means $dF_X / d\lambda$: the Radon-Nikodym derivative of the distribution with respect to Lebesgue measure. A random variable has a density exactly when its law is absolutely continuous — which is why an outcome with an atom, like a capped payoff, has none.