Measure theory in option pricing

The twelve lessons before this one built measure theory and probability from the ground up and never mentioned a price. "Probability spaces and Kolmogorov's axioms" turned a measure of total mass 1 into a model of chance, "Random variables and their distributions" described what a model produces, and "Conditional probability and conditional expectation" showed how to update it. "Risk-neutral probabilities" showed that an option is priced by an average taken with a probability that is not the real-world one. This lesson puts the two sides together on one small market, a call option on a stock that moves up or down for three days. Every symbol in the price formula turns out to be an object from this track, so the lesson goes through them one at a time, adds the pieces of measure theory that pricing needs and the earlier lessons did not give, the filtration, absolute continuity and the Radon–Nikodym density, and ends with the limits that connect the three-day market to continuous time.

A three-day market

A stock starts at $S_0 = 100$. Each day its price is multiplied by $1.1$ (an up day, $u$) or by $0.9$ (a down day, $d$). Interest is zero, so cash keeps its value; with a positive rate the formula below only gains a constant factor $e^{-rT}$, which changes none of the measure theory. A European call with strike $K = 100$ expires at the end of day 3 and pays $(S_3 - K)^+ = \max(S_3 - 100, 0)$. Its price, as "Risk-neutral probabilities" derives it, is

Three days of up and down moves give $2^3 = 8$ price paths. Three up days end at $S_3 = 100 \cdot 1.1^3 = 133.1$; two up days and one down day, in any order, end at $108.9$; one up day ends at $89.1$; three down days end at $72.9$. Only the first two kinds pay: $33.1$ on one path and $8.9$ on three.

The figure shows the eight paths. Select a symbol of the formula to see what it is on this market, switch between the risk-neutral measure $\mathbb{Q}$ and the real-world measure $\mathbb{P}$, and move the strike. In its starting position the price reads $7.475$, the call pays with probability $0.5$ and the expected stock price is $100$. Switch to $\mathbb{P}$ and the same readouts become $9.142$, $0.575$ and $103.03$; the rest of the lesson explains why these numbers differ and which of them is the price.

Outcomes and events

The first symbol hidden in the formula is the sample space $\Omega$. "Probability spaces and Kolmogorov's axioms" described outcomes with two dice: an outcome is a complete description of what happened, there a pair of numbers. On this market an outcome has to settle everything the model can ask, so it is a whole path, not one day's move: $\Omega = \{uuu, uud, udu, udd, duu, dud, ddu, ddd\}$. All the randomness of the model is the question of which $\omega \in \Omega$ happens. Everything else, the prices $S_1, S_2, S_3$ and the payoff, is a function of $\omega$: once the path is known, so is every price on it.

An event is a set of paths. The call pays on $\{S_3 > 100\} = \{uuu, uud, udu, duu\}$, four paths; the first day is up on $\{uuu, uud, udu, udd\}$. On a finite $\Omega$ nothing stops us from calling every set of paths an event, so the σ-algebra is $\mathcal{F} = 2^{\Omega}$ with $2^8 = 256$ events. The axioms of "σ-algebras and Borel sets" start to matter only when $\Omega$ is infinite. A model with a price at every instant has uncountably many paths, and then, as Vitali's set showed, some collections of paths cannot receive a probability at all.

A probability measure is a measure of total mass 1. "Measuring the size of sets" described measures on a finite set as weighted point masses, and that is exactly what $\mathbb{P}$ is here: eight point masses, one on each path, so that $\mathbb{P}(A) = \sum_{\omega \in A} \mathbb{P}(\{\omega\})$ for every event $A$.

Information and the filtration

A trader does not see $\omega$ at once; the path is revealed one day at a time. What is known after day $t$ is the σ-algebra $\mathcal{F}_t$ of the events that are decided by then. Before trading starts nothing is known, so $\mathcal{F}_0 = \{\emptyset, \Omega\}$. After day 1 the trader knows whether the first move was up, so $\mathcal{F}_1$ is generated by the two blocks

As "σ-algebras and Borel sets" explained for finite σ-algebras, these two blocks are the atoms of $\mathcal{F}_1$ and the σ-algebra holds all unions of atoms: $2^2 = 4$ events, namely $\emptyset$, $A_u$, $A_d$ and $\Omega$. After day 2 there are four atoms ($uu\cdot$, $ud\cdot$, $du\cdot$, $dd\cdot$) and $2^4 = 16$ events; after day 3 every path is an atom of its own and $\mathcal{F}_3 = 2^\Omega$. A chain of σ-algebras

in which each one contains the previous one is a filtration: information only grows, because a question answered today stays answered tomorrow.

"Measurable functions" showed that on a partition a function is measurable exactly when it is constant on every atom. So $S_1$ is $\mathcal{F}_1$-measurable, being $110$ on $A_u$ and $90$ on $A_d$, while $S_3$ is not: it takes three values, $133.1$, $108.9$ and $89.1$, on $A_u$. A process whose value at each time is measurable with respect to that time's σ-algebra is called adapted; "Filtrations, adapted processes and stopping times" studies such processes.

Two measures on the same events

Two probability measures matter on this market, and both live on the same pair $(\Omega, \mathcal{F})$. Under the real-world measure $\mathbb{P}$ the stock goes up each day with probability $p = 0.55$. The risk-neutral measure $\mathbb{Q}$ uses the up probability that makes today's price equal to the expected price tomorrow, as derived in "Risk-neutral probabilities": after an up day the price is $110$, after a down day $90$, and $0.5 \cdot 110 + 0.5 \cdot 90 = 100 = S_0$. In general

Under either measure the three days are independent, so the probability of a path is a product of one-day probabilities, the product measure of "Independence and product measures". A path with $k$ up days has probability $p^k (1 - p)^{3-k}$ under $\mathbb{P}$ and $q^k (1 - q)^{3-k}$ under $\mathbb{Q}$:

• three up days: $\mathbb{P} = 0.166375$, $\mathbb{Q} = 0.125$;
• two up days (three paths, each): $\mathbb{P} = 0.136125$, $\mathbb{Q} = 0.125$;
• one up day (three paths, each): $\mathbb{P} = 0.111375$, $\mathbb{Q} = 0.125$;
• no up day: $\mathbb{P} = 0.091125$, $\mathbb{Q} = 0.125$.

The two measures weigh the same events differently. The first day is up with probability $\mathbb{P}(A_u) = 0.55$ and $\mathbb{Q}(A_u) = 0.5$; the call pays with probability $\mathbb{P}(S_3 > 100) = 0.57475$ and $\mathbb{Q}(S_3 > 100) = 0.5$.

Absolute continuity and the Radon–Nikodym density

There is a familiar way to turn one measure into another by reweighting. The rod of "Measuring the size of sets" had length 1 and density $\rho(x) = 2x$, so its mass was $\mu(A) = \int_A 2x \, dx$: the measure $\mu$ is length with every point weighted by $\rho$. The segment $[0, 1/2]$ has length $1/2$, but its mass is $\int_0^{1/2} 2x \, dx = 1/4$.

Reweighting has one unavoidable consequence. A set that length cannot see, a null set of "Lebesgue measure and null sets", carries no mass after reweighting either, because $\int_A \rho \, d\lambda = 0$ whenever $\lambda(A) = 0$. So a measure $\nu$ can be a reweighting of $\mu$ only if $\mu(A) = 0$ implies $\nu(A) = 0$. This property has a name: $\nu$ is absolutely continuous with respect to $\mu$, written $\nu \ll \mu$. Two measures are equivalent when each is absolutely continuous with respect to the other, that is, when they have exactly the same null sets.

Two non-examples show what can go wrong. A point mass $\delta_0$ is not absolutely continuous with respect to length, since $\lambda(\{0\}) = 0$ but $\delta_0(\{0\}) = 1$. The uniform measure on $[0, 1/2]$, which gives a set $A$ the mass $2\lambda(A \cap [0, 1/2])$, is absolutely continuous with respect to length on $[0, 1]$, but the two are not equivalent: the segment $(1/2, 1]$ has length $1/2$ and mass 0.

The Radon–Nikodym theorem says that the necessary condition is also sufficient. Let $\mu$ and $\nu$ be σ-finite, which means that the space is a countable union of sets of finite measure, as $\mathbb{R}$ is the union of the intervals $[n, n+1]$; every probability measure qualifies. If $\nu \ll \mu$, there is a measurable function $Z \ge 0$, called the density or Radon–Nikodym derivative and written $\frac{d\nu}{d\mu}$, such that

The second formula follows from the first by the steps of "The Lebesgue integral": indicators, then simple functions, then monotone limits. The density is unique up to a $\mu$-null set, so two densities agree almost everywhere. The proof is not reproduced here.

On a finite space the density is simply a ratio of masses, $Z(\omega) = \nu(\{\omega\})/\mu(\{\omega\})$. On the market, $Z = \frac{d\mathbb{Q}}{d\mathbb{P}}$ divides the $\mathbb{Q}$-probability of each path by its $\mathbb{P}$-probability:

• $Z = 0.7513$ on $uuu$, since $0.125 / 0.166375$;
• $Z = 0.9183$ on each path with two up days;
• $Z = 1.1223$ on each path with one up day;
• $Z = 1.3717$ on $ddd$.

$Z$ is below 1 where $\mathbb{P}$ puts more weight than $\mathbb{Q}$ and above 1 where it puts less, so averaging it with the weights of $\mathbb{P}$ gives exactly the total mass of $\mathbb{Q}$: $\mathbb{E}_{\mathbb{P}}[Z] = \mathbb{Q}(\Omega) = 1$. The density also respects information. Restricted to $\mathcal{F}_1$ it is the ratio of the block probabilities, $0.5 / 0.55 = 0.9091$ on $A_u$ and $0.5 / 0.45 = 1.1111$ on $A_d$.

Equivalent measures share their null sets, and so they share their almost sure statements: an event of $\mathbb{P}$-probability 1 has $\mathbb{Q}$-probability 1. On the tree both measures charge every path, so nothing is lost by switching. This is the property that makes a change of measure safe in continuous time. "Change of measure and Girsanov's theorem" builds an equivalent measure $\mathbb{Q}$ under which a Brownian motion with drift becomes driftless. Continuity of the paths, nowhere differentiability and the value of the quadratic variation, collected in "Properties of Brownian motion", are almost sure properties of the paths, so they hold under $\mathbb{Q}$ exactly as under $\mathbb{P}$. A change of measure can move the drift but cannot touch the volatility. For the same reason a pricing measure has to be equivalent to the real-world one: an event that cannot happen must not carry value, and an event that can happen must not be ignored.

Equivalence is what lets the density run both ways: if $\mathbb{Q} \approx \mathbb{P}$, then $\frac{d\mathbb{P}}{d\mathbb{Q}} = 1/Z$. On the market this is the reciprocal ratio, $1/0.7513 = 1.331$ on $uuu$, which is $0.166375 / 0.125$.

The same theorem gives conditional expectation its existence. For $X \ge 0$ and a σ-algebra $\mathcal{G}$, the set function $G \mapsto \int_G X \, d\mathbb{P}$ on $\mathcal{G}$ is absolutely continuous with respect to $\mathbb{P}$, because a $\mathbb{P}$-null set has integral 0. Its Radon–Nikodym density is a $\mathcal{G}$-measurable function, and that function is $\mathbb{E}[X \mid \mathcal{G}]$, defined in "Conditional probability and conditional expectation" by $\int_G \mathbb{E}[X \mid \mathcal{G}] \, d\mathbb{P} = \int_G X \, d\mathbb{P}$.

Payoffs, distributions and the price

$S_3$ is a function on $\Omega$: $S_3(uuu) = 133.1$, $S_3(uud) = 108.9$, and so on. Every question of the form $S_3 \le x$ is an event, so $S_3$ is a random variable in the sense of "Random variables and their distributions". The payoff $X = (S_3 - K)^+$ is a continuous function of $S_3$, and "Measurable functions" showed that a continuous function of a measurable function is measurable, so $X$ is a random variable too.

A random variable carries the probability on $\Omega$ over to the real line; that image is its distribution. Under $\mathbb{Q}$ the payoff is $33.1$ with probability $1/8$, $8.9$ with probability $3/8$ and $0$ with probability $1/2$, an atom at 0 collecting the four paths that end at or below the strike. Under $\mathbb{P}$ the same three values have probabilities $0.166375$, $0.408375$ and $0.42525$. The values are the same, the weights are not.

The expectation averages the payoff over the outcomes. $X$ is a simple function, $33.1 \cdot \mathbf{1}_{\{uuu\}} + 8.9 \cdot \mathbf{1}_{\{uud, udu, duu\}}$, and "The Lebesgue integral" defines the integral of a simple function as the sum of each value times the measure of the set where it is taken:

The same integral under $\mathbb{P}$ gives $\mathbb{E}_{\mathbb{P}}[X] = 9.14155$, the payoff a real-world investor expects. It is not the price, which is $V_0 = 7.475$: a portfolio of stock and cash that replicates the call costs $7.475$ whatever $p$ is. What singles out $\mathbb{Q}$ is that it makes the stock price a martingale, $\mathbb{E}_{\mathbb{Q}}[S_3] = 100 = S_0$, while $\mathbb{E}_{\mathbb{P}}[S_3] = 103.0301$. The density links the two averages, $\mathbb{E}_{\mathbb{Q}}[X] = \mathbb{E}_{\mathbb{P}}[Z X]$.

The price does not depend on $p$. Changing the real-world probability changes $\mathbb{E}_{\mathbb{P}}[X]$ but not $V_0$, because $V_0$ is the cost of a replicating portfolio and $\mathbb{Q}$ is the measure that reproduces that cost.

Call prices as tail integrals

The price can also be computed by layers. "The Lebesgue integral" cut the area under a function into horizontal layers, and "Expectation, variance and inequalities" turned this into the tail formula $\mathbb{E}[X] = \int_0^\infty \mathbb{P}(X > t) \, dt$. For the call payoff, $X > t$ exactly when $S_3 > K + t$, so substituting $s = K + t$ gives

On the tree, $s \mapsto \mathbb{Q}(S_3 > s)$ is a staircase. For $s$ between $100$ and $108.9$ four paths end above $s$, so it equals $0.5$; between $108.9$ and $133.1$ only $uuu$ does, so it equals $0.125$; beyond $133.1$ it is 0. The area under the staircase to the right of the strike is

the same price, summed by layers instead of by paths. The price of a call is an area under the survival curve of the terminal price, and raising the strike cuts the left part of that area away.

Distributions from call prices

Raise the strike by 1 and the area loses a strip: $C(K) - C(K+1) = \int_K^{K+1} \mathbb{Q}(S_3 > s) \, ds$. When the staircase is flat on the strip, its area is simply its height. Between $100$ and $101$ it is flat at $0.5$, and indeed $C(101) = 0.5 \cdot 7.9 + 0.125 \cdot 24.2 = 6.975$, so

Two call prices have revealed a probability. Let the strike step shrink to zero. Then the derivative is $\frac{\partial C}{\partial K} = -\mathbb{Q}(S_3 > K)$, which gives the distribution function of "Random variables and their distributions", $F_{\mathbb{Q}}(K) = \mathbb{Q}(S_3 \le K) = 1 + \frac{\partial C}{\partial K}$. Differentiating once more, $\frac{\partial^2 C}{\partial K^2}$ is the density of $S_3$ under $\mathbb{Q}$ whenever a density exists.

On the tree the price $C(K)$ is piecewise linear in $K$ and its slope is $-\mathbb{Q}(S_3 > K)$: $-1$ below $72.9$, $-0.875$ on $(72.9, 89.1)$, $-0.5$ on $(89.1, 108.9)$, $-0.125$ on $(108.9, 133.1)$ and $0$ above. At each possible price the slope jumps by the probability of that price. At $108.9$ it jumps from $-0.5$ to $-0.125$, by $0.375 = \mathbb{Q}(S_3 = 108.9)$. A butterfly of three call prices measures the jump: $C(K-h) - 2C(K) + C(K+h)$ divided by $h$. At $K = 108.9$ and $h = 1$ the prices are $3.525$, $3.025$ and $2.9$, and $3.525 - 2 \cdot 3.025 + 2.9 = 0.375$.

Call prices at all strikes therefore determine the whole distribution of the terminal price under $\mathbb{Q}$, and only under $\mathbb{Q}$: they say nothing about the real-world distribution. "Static replication: Breeden–Litzenberger and Carr–Madan" turns this into a method that reads the distribution from option prices and prices other payoffs from it.

Quantiles and Value at Risk

A risk manager asks a different question from the pricer: not what the call is worth, but how much can be lost. Take one share bought at 100. The loss is $L = 100 - S_3$, and risk is about what will really happen, so the measure is $\mathbb{P}$. The loss takes the values $-33.1$ (a gain), $-8.9$, $10.9$ and $27.1$ with probabilities $0.166375$, $0.408375$, $0.334125$ and $0.091125$, so its distribution function jumps to $0.166375$, $0.57475$, $0.908875$ and $1$ at those values.

The quantile function of "Random variables and their distributions" is $F^{-1}(u) = \inf\{x : F(x) \ge u\}$. Value at Risk at level $u$ is exactly this quantile of the loss:

the smallest loss that is exceeded with probability at most $1 - u$. At level $0.5$ it is $-8.9$: the median outcome is a gain. At level $0.9$, $F_L(-8.9) = 0.57475$ is below $0.9$ and $F_L(10.9) = 0.908875$ is above it, so the value at risk is $10.9$. At level $0.95$ it is $27.1$.

The same position under $\mathbb{Q}$ has a value at risk of $27.1$ at level $0.9$, against $10.9$ under $\mathbb{P}$. The two measures answer different questions: prices come from $\mathbb{Q}$, risk figures from $\mathbb{P}$, and a number computed under the wrong one answers a question nobody asked. "Value at Risk and expected shortfall" develops the risk measures built on the quantile.

Tomorrow's value as a conditional expectation

After day 1 the call has a new value that depends on what happened, so it is a random variable, not a number. "Conditional probability and conditional expectation" defined the conditional expectation given a partition as the average of $X$ over each atom. On $A_u$ the four remaining paths end at $133.1$, $108.9$, $108.9$ and $89.1$, each with conditional probability $\mathbb{Q}(\{\omega\}) / \mathbb{Q}(A_u) = 0.125 / 0.5 = 1/4$, so the average payoff is $(33.1 + 2 \cdot 8.9 + 0)/4 = 12.725$. On $A_d$ it is $(8.9 + 0 + 0 + 0)/4 = 2.225$:

$V_1$ is $\mathcal{F}_1$-measurable, being constant on each atom, and averaging it once more gives back the price: $\mathbb{E}_{\mathbb{Q}}[V_1] = 0.5 \cdot 12.725 + 0.5 \cdot 2.225 = 7.475 = V_0$. This is the tower property.

The same can be done one day at a time. After day 2 the four atoms give $V_2 = 21$ after $uu$ (the average of $33.1$ and $8.9$), $4.45$ after $ud$ and after $du$, and $0$ after $dd$. Each day-1 value is the average of its two day-2 children, $0.5 \cdot 21 + 0.5 \cdot 4.45 = 12.725$ and $0.5 \cdot 4.45 + 0.5 \cdot 0 = 2.225$, and the price is the average of those. Repeated averaging from the last day back to the first is exactly the backward induction of "Multi-step trees and backward induction", and the tower property is why it gives the same price as the single average over eight paths.

The values $V_0, V_1, V_2, V_3$ form a martingale under $\mathbb{Q}$, the subject of "Conditional expectation and martingales": the best prediction today of tomorrow's value is today's value. The stock is one too, since $\mathbb{E}_{\mathbb{Q}}[S_3 \mid \mathcal{F}_1]$ equals $110$ on $A_u$ and $90$ on $A_d$, which is $S_1$.

Limits behind continuous-time prices

Split the three days into more and more steps, with the up and down factors shrinking to match, and the tree price converges to the price of "The Black–Scholes formula". On the way the sample space changes its nature. With $n$ steps it has $2^n$ paths; in the limit an outcome is a whole continuous path, and there are uncountably many of them. Sums over paths become integrals against a measure on paths, and the question of which sets of paths can carry a probability is the one "σ-algebras and Borel sets" and "Lebesgue measure and null sets" answered for the interval.

What converges is the distribution of the terminal price. A sum of $n$ small independent up and down moves is the setting of the central limit theorem, which "Modes of convergence and the law of large numbers" states as convergence in distribution. That alone does not make expectations converge, because a call payoff is unbounded; uniform integrability is needed too, which that lesson links to convergence of means through Vitali's theorem. In the standard scaling the second moment of the terminal price stays bounded, which is enough, and the tree price converges to the Black–Scholes price.

A bounded payoff needs less. A put pays $(K - S_T)^+$, which never exceeds $K$. If the payoffs converge, the constant $K$ is an integrable bound for all of them, and the dominated convergence theorem of "Convergence theorems" lets the limit pass through the expectation. On the three-day market the put with $K = 100$ pays at most $100$ and costs $0.125 \cdot (3 \cdot 10.9 + 27.1) = 7.475$, the same as the call, because $\mathbb{E}_{\mathbb{Q}}[S_3] = K$.

Monte Carlo pricing replaces the exact sum by an average of simulated payoffs. Draw $N$ paths under $\mathbb{Q}$ and average their payoffs. By the strong law of large numbers the average converges to $7.475$ almost surely, and by the central limit theorem its error is roughly normal with standard deviation $\sigma / \sqrt{N}$, where $\sigma = 10.5252$ is the standard deviation of the payoff under $\mathbb{Q}$. With $N = 10{,}000$ paths the standard error is $0.1053$, so about 95% of the runs land within $\pm 0.2063$ of the price. "Monte Carlo option pricing" turns this into a practical method.

Summary

• An outcome is a whole path, events are sets of paths, and information is a filtration: the σ-algebras $\mathcal{F}_0 \subset \mathcal{F}_1 \subset \mathcal{F}_2 \subset \mathcal{F}_3$ grow day by day, and the price $S_t$ is measurable with respect to $\mathcal{F}_t$.
• $\nu \ll \mu$ means that $\mu$-null sets are $\nu$-null; the Radon–Nikodym theorem then writes $\nu$ as $\mu$ reweighted by a density $Z = \frac{d\nu}{d\mu}$. Equivalent measures share their null sets and almost sure statements, which is why $\mathbb{Q}$ can replace $\mathbb{P}$.
• The price is an integral under $\mathbb{Q}$ ($7.475$), not under $\mathbb{P}$ ($9.14155$); the two are linked by $\mathbb{E}_{\mathbb{Q}}[X] = \mathbb{E}_{\mathbb{P}}[Z X]$ with $Z = 0.7513, 0.9183, 1.1223, 1.3717$ on the paths.
• The layer-cake formula writes a call price as the area $\int_K^\infty \mathbb{Q}(S_T > s) \, ds$, so differences and second differences of call prices over strikes give the distribution of the terminal price under $\mathbb{Q}$.
• Value at Risk is a quantile of the loss under $\mathbb{P}$, and the value after day 1 is a conditional expectation under $\mathbb{Q}$, a block average whose own average is the price.
• Continuous-time prices are limits: convergence in distribution with uniform integrability, dominated convergence and the laws of large numbers justify the passage from trees to Black–Scholes and to Monte Carlo.