Radon-Nikodym Derivative: Likelihood Ratios in Finance

What Is a Radon-Nikodym Derivative?

The Radon-Nikodym derivative $d\mathbb Q/d\mathbb P$ is a likelihood ratio between two measures on the same measurable space. It tells us how to calculate a $\mathbb Q$-probability or expectation while integrating under $\mathbb P$.

For an event $A$,

$$\mathbb Q(A) =\int_A\frac{d\mathbb Q}{d\mathbb P}\,d\mathbb P =\mathbb E^{\mathbb P}\!\left[ \mathbf 1_A\frac{d\mathbb Q}{d\mathbb P} \right].$$

The derivative is not an ordinary pointwise derivative. It is a measurable random variable defined up to $\mathbb P$-null sets. In quantitative finance it is the common language behind importance sampling, risk-neutral pricing, Girsanov's theorem and changes of numeraire.

Johann Radon proved a Euclidean version in 1913; Otto Nikodym established the general result in 1930. The Encyclopedia of Mathematics statement gives the measure-theoretic form.

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The Theorem and Its Assumptions

Let $(\Omega,\mathcal F)$ be a measurable space. For probability measures $\mathbb P$ and $\mathbb Q$, write $\mathbb Q\ll\mathbb P$ when

$$\mathbb P(A)=0\implies\mathbb Q(A)=0 \qquad\text{for every }A\in\mathcal F.$$

This is absolute continuity. The Radon-Nikodym theorem says that if $\mathbb Q\ll\mathbb P$, there is a non-negative $\mathcal F$-measurable variable $Z$ such that

$$\mathbb Q(A)=\int_A Z\,d\mathbb P \quad\text{for all }A\in\mathcal F.$$

We write $Z=d\mathbb Q/d\mathbb P$. It is unique $\mathbb P$-almost surely and

$$\mathbb E^{\mathbb P}[Z]=1.$$

Probability measures are finite, hence sigma-finite, so the standard theorem's sigma-finiteness requirements are automatic here. In the general measure-theory version the reference measure must satisfy the relevant sigma-finiteness condition.

If $Y$ is $\mathbb Q$-integrable, then

$$\mathbb E^{\mathbb Q}[Y] =\mathbb E^{\mathbb P}[YZ].$$

If both $\mathbb Q\ll\mathbb P$ and $\mathbb P\ll\mathbb Q$, the measures are equivalent, written $\mathbb Q\sim\mathbb P$. In that case $Z>0$ $\mathbb P$-almost surely and

$$\frac{d\mathbb P}{d\mathbb Q}=\frac{1}{Z} \qquad \mathbb Q\text{-almost surely}.$$

Absolute continuity is directional; equivalence is not. That is why swapping $d\mathbb Q/d\mathbb P$ and $d\mathbb P/d\mathbb Q$ without taking the reciprocal is a serious error.

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A Finite-State Check

Take $\Omega=\{H,T\}$. Under $\mathbb P$, the coin is fair. Under $\mathbb Q$, heads has probability 0.7:

$$\mathbb P(H)=\mathbb P(T)=0.5,\qquad \mathbb Q(H)=0.7,\quad\mathbb Q(T)=0.3.$$

The density is the state-by-state likelihood ratio,

$$Z(H)=\frac{0.7}{0.5}=1.4,\qquad Z(T)=\frac{0.3}{0.5}=0.6.$$

Its expectation under $\mathbb P$ is

$$0.5(1.4)+0.5(0.6)=1.$$

A likelihood ratio is dimensionless. Both numerator and denominator measure probability mass on the same state, or probability density relative to the same base measure. This unit check catches invalid expressions that divide a probability by an asset price or multiply two densities where the chain rule requires a ratio.

For a payoff $Y(H)=10$, $Y(T)=2$,

$$\mathbb E^{\mathbb Q}[Y]=0.7(10)+0.3(2)=7.6,$$

while reweighting under $\mathbb P$ gives

$$\mathbb E^{\mathbb P}[YZ] =0.5(10)(1.4)+0.5(2)(0.6) =7.6.$$

The continuous-time formulas look more elaborate, but they do the same job.

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Chain Rule and Reverse Densities

Radon-Nikodym derivatives compose like likelihood ratios. If $\mathbb R\ll\mathbb Q\ll\mathbb P$, then

$$\frac{d\mathbb R}{d\mathbb P} =\frac{d\mathbb R}{d\mathbb Q} \frac{d\mathbb Q}{d\mathbb P} \qquad \mathbb P\text{-almost surely},$$

where a suitable version of the first factor is used. This is a product, not a sum. It lets a pricing calculation move through several convenient measures without deriving one large density from scratch.

For equivalent $\mathbb P$ and $\mathbb Q$,

$$\frac{d\mathbb P}{d\mathbb Q} =\left(\frac{d\mathbb Q}{d\mathbb P}\right)^{-1} \qquad \mathbb Q\text{-almost surely}.$$

The almost-sure measure changes sides because the reciprocal is being treated as a $\mathbb Q$-random variable. These details are invisible in a finite state space with all probabilities positive, but they matter when densities can vanish.

There is also a conditional form. For a sub-sigma-algebra $\mathcal G\subseteq\mathcal F$,

$$\left.\frac{d\mathbb Q}{d\mathbb P}\right|_{\mathcal G} =\mathbb E^{\mathbb P}\!\left[ \left.\frac{d\mathbb Q}{d\mathbb P}\right|_{\mathcal F} \,\middle|\,\mathcal G \right].$$

The density process in continuous time is this identity applied to the nested information sets $\mathcal F_t$.

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Density Processes and Bayes' Formula

Suppose $Z_T=d\mathbb Q/d\mathbb P|_{\mathcal F_T}$. The information available before $T$ is represented by the density process

$$Z_t =\left.\frac{d\mathbb Q}{d\mathbb P}\right|_{\mathcal F_t} =\mathbb E^{\mathbb P}[Z_T\mid\mathcal F_t].$$

Thus $Z_t$ is a non-negative $\mathbb P$-martingale with $Z_0=1$ when $\mathcal F_0$ is trivial. Conditional expectations obey the change-of-measure version of Bayes' formula:

$$\mathbb E^{\mathbb Q}[Y\mid\mathcal F_t] =\frac{\mathbb E^{\mathbb P}[Z_TY\mid\mathcal F_t]}{Z_t}.$$

This denominator is easy to omit. At $t=0$, where $Z_0=1$, the formula reduces to the unconditional identity.

For Brownian drift changes, a common candidate density is

$$Z_t=\exp\!\left( -\int_0^t\theta_s\,dW_s^{\mathbb P} -\frac12\int_0^t\theta_s^2\,ds \right).$$

Itô's formula shows that this is a local martingale. It is not automatically a true martingale. Novikov's condition is one sufficient check; the distinction is developed in the Girsanov explainer.

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Importance Sampling

Importance sampling is a direct computational use of the same theorem. Suppose $X\sim N(0,1)$ under the target measure $\mathbb P$, but a rare right-tail payoff is inefficient to estimate from ordinary draws. Simulate instead under $\mathbb Q$, where $X\sim N(m,1)$ with $m>0$.

The density needed to recover the $\mathbb P$-expectation is

$$\frac{d\mathbb P}{d\mathbb Q}(x) =\frac{\exp(-x^2/2)}{\exp(-(x-m)^2/2)} =\exp\!\left(-mx+\frac12m^2\right).$$

Therefore

$$\mathbb E^{\mathbb P}[h(X)] =\mathbb E^{\mathbb Q}\!\left[ h(X)\exp\!\left(-mX+\frac12m^2\right) \right].$$

For $m=2$, a simulated value $x=2$ receives weight $e^{-2}=0.135335\ldots$. The shifted simulation produces more right-tail observations, then down-weights each according to how much more likely it was under $\mathbb Q$. The estimator remains unbiased when the likelihood ratio and support conditions are correct.

The direction here is deliberately the reverse of the common pricing formula. We simulate under $\mathbb Q$ but want a $\mathbb P$-expectation, so the weight is $d\mathbb P/d\mathbb Q$. Using $d\mathbb Q/d\mathbb P$ would square the intended tilt rather than undo it.

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Change of Numeraire

A numeraire is a strictly positive traded asset used as the unit of account. If $M$ and $N$ are valid numeraires with associated measures $\mathbb Q^M$ and $\mathbb Q^N$, then the density process for moving from the $M$-measure to the $N$-measure is

$$\left. \frac{d\mathbb Q^N}{d\mathbb Q^M} \right|_{\mathcal F_t} =\frac{N_t/M_t}{N_0/M_0} =\frac{N_tM_0}{M_tN_0}.$$

The order is worth checking. Under $\mathbb Q^M$, $N_t/M_t$ is a martingale, so the right-hand side is non-negative and has expectation one. The Nanyang Technological University notes derive the forward-measure case.

If $X_T$ is a claim, its time-zero price can be written under either numeraire:

$$X_0 =M_0\mathbb E^{\mathbb Q^M}\!\left[\frac{X_T}{M_T}\right] =N_0\mathbb E^{\mathbb Q^N}\!\left[\frac{X_T}{N_T}\right].$$

For the money-market account $B_t$ and a zero-coupon bond $P(t,T)$, take $M=B$ and $N_t=P(t,T)$. Since $P(T,T)=1$,

$$\left. \frac{d\mathbb Q^T}{d\mathbb Q^B} \right|_{\mathcal F_T} =\frac{B_0}{B_TP(0,T)}.$$

Under the $T$-forward measure, asset prices divided by $P(t,T)$ are martingales. The phrase "forward measure" is specific to a bond numeraire; a general $N$ defines an $N$-numeraire measure.

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A Stock-Numeraire Example

In Black-Scholes with no dividends, constant $r=5\%$, $\sigma=20\%$, $S_0=100$ and $T=1$, use the bank account $B_t=e^{rt}$ as the original numeraire and the stock as the new one. The terminal density is

$$\left. \frac{d\mathbb Q^S}{d\mathbb Q^B} \right|_{\mathcal F_T} =\frac{S_TB_0}{B_TS_0}.$$

On a path with $W_T^{\mathbb Q^B}=0.5$,

$$S_T=100e^{(0.05-0.5\times0.20^2)+0.20(0.5)} =113.8828.$$

Therefore

$$\frac{S_T}{S_0e^{0.05}} =1.083287.$$

The stock measure gives this above-average stock path about 8.33% more likelihood than the bank-account measure. Across all paths the density still averages to one. Under $\mathbb Q^S$, ratios such as $X_t/S_t$ are martingales for suitable traded assets $X$. Sharpe ratios are not martingales merely because the stock is the numeraire.

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What Can Go Wrong

Wrong orientation. To convert a $\mathbb P$-expectation into a $\mathbb Q$-expectation, use $d\mathbb Q/d\mathbb P$. The reciprocal belongs to the reverse conversion.

A strict local martingale. A positive stochastic exponential can have expectation below one. Then it is not a probability density on the stated horizon.

Loss of equivalence. Absolute continuity allows $Z=0$ on a set with positive $\mathbb P$-probability. Finance often needs equivalence so that the pricing measure does not declare a physically possible event impossible.

An invalid numeraire. A process must be strictly positive and tradeable for the standard numeraire theorem. A convenient state variable or a portfolio that can hit zero does not qualify automatically.

Confusing pricing with forecasting. The density used to construct an equivalent martingale measure is selected for no-arbitrage valuation. It is not an estimate of real-world probabilities.

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Frequently Asked Questions

Is a Radon-Nikodym derivative a probability density?

It is a density of one measure relative to another. An ordinary probability density $f_X$ is the special case $d\mathbb P_X/d\lambda$, where $\lambda$ is Lebesgue measure.

When does the derivative exist?

For probability measures, $d\mathbb Q/d\mathbb P$ exists when $\mathbb Q\ll\mathbb P$. Equivalence is stronger than existence and makes the reverse derivative available too.

Why must its expectation equal one?

Because $\mathbb Q(\Omega)=1$. Applying the defining identity to the whole sample space gives $1=\mathbb E^{\mathbb P}[d\mathbb Q/d\mathbb P]$.

What is the difference between $Z_T$ and $Z_t$?

$Z_T$ defines the measure on terminal information. The density process $Z_t=\mathbb E^{\mathbb P}[Z_T\mid\mathcal F_t]$ is its likelihood ratio using only information known by time $t$.

How does this connect to Feynman-Kac?

The derivative changes which path law supplies an expectation; Feynman-Kac connects an expectation under a chosen diffusion law to a linear PDE. They are separate steps in the pricing ladder.

Is the Radon-Nikodym derivative unique?

Only up to $\mathbb P$-almost-sure equality. Changing its value on a $\mathbb P$-null set does not change any $\mathbb Q$-probability.