Applied and computational mathematics

Mathematical finance

Applies probability and stochastic calculus to price options and manage financial risk.

IntuitionNo free lunches: how the price of a future promise gets pinned down

Suppose someone offers you the right, but not the obligation, to buy a share of stock next year at today's price. How much should that right cost? You cannot just guess: if the price is set too low, anyone could buy this right for nothing, wait, and pocket a certain profit if the stock happens to rise — a free lunch that real markets do not allow to persist. Mathematical finance turns this single idea, called no-arbitrage (no risk-free profit from nothing), into a precise machine for pricing every kind of financial contract.

n

-step Cox–Ross–Rubinstein binomial tree produces a

\mathrm{Bin}(n, p)

distribution of log-returns (violet bars), which converges as

n \to \infty

to the Gaussian Black–Scholes density (amber curve).

The simplest such contracts are options. A European call option gives its holder the right to buy one share at a fixed strike price $K$ on a fixed maturity date $T$ ; a European put option gives the right to sell at $K$ on $T$ . Whether the holder actually exercises the right depends only on the stock price $S_T$ at maturity, so the whole contract reduces to a single number: its payoff.

UndergraduatePayoffs, replication, and the price today

Definition: European call and put payoffs

At maturity $T$ , a call is worth $\max(S_T - K, 0)$ : exercised only if the stock is above the strike, in which case it is worth the difference. A put is worth $\max(K - S_T, 0)$ : exercised only if the stock is below the strike. Write $C$ for the price today of the call and $P$ for the price today of the put, both on the same non-dividend-paying stock, same strike $K$ , same maturity $T$ .

C_T = \max(S_T - K, 0), \qquad P_T = \max(K - S_T, 0)

These are the payoffs at maturity; the hard question is what $C$ and $P$ must be today, long before $S_T$ is known. The key trick, used throughout mathematical finance, is replication: build a portfolio of stock and risk-free bonds whose payoff at $T$ exactly matches the option's payoff in every possible scenario. Since two portfolios with identical future payoffs must have identical prices today — otherwise one could sell the expensive one, buy the cheap one, and pocket the difference risk-free — the option's price equals the cost of building its replica.

From one step to infinitely many: three pricing models for the same call option
Model	Time discretization	Underlying price dynamics	Call price
One-step binomial tree	Single period of length $T - t$	$S$ moves to $Su$ or $Sd$	$C = \dfrac{p\,C_u + (1-p)\,C_d}{1+r}$
$n$ -step binomial tree	$n$ periods of length $T/n$	Recombining lattice of up/down moves	Backward induction node by node; converges to Black–Scholes as $n \to \infty$
Black–Scholes (continuous time)	Continuous time	Geometric Brownian motion $dS_t = \mu S_t\,dt + \sigma S_t\,dW_t$	$C = S\Phi(d_1) - Ke^{-r(T-t)}\Phi(d_2)$

C = S\,\Phi(d_1) - Ke^{-r(T-t)}\,\Phi(d_2), \qquad d_{1,2} = \frac{\ln(S/K) + \left(r \pm \frac{\sigma^2}{2}\right)(T-t)}{\sigma\sqrt{T-t}}

UndergraduateTwo theorems: an exact relation, and a pricing equation

Theorem: Put–call parity

For a European call and put on the same non-dividend-paying stock, with the same strike $K$ and maturity $T$ , and a constant risk-free rate $r$ : $C - P = S - K e^{-r(T-t)}$ , where $S$ is the stock price at time $t \le T$ .

Why is it true?

A call plus enough cash to grow into $K$ by maturity, and a put plus one share of stock, are just two different ways of guaranteeing you end up holding exactly one share worth $\max(S_T, K)$ at time $T$ . If two recipes always produce the same dish, they must cost the same today.

Proof

Build two portfolios today, at time $t$ . Portfolio A holds one call option plus an amount of cash $Ke^{-r(T-t)}$ invested at the risk-free rate, so that it grows to exactly $K$ by time $T$ . Portfolio B holds one put option plus one share of the stock.

Compare their values at maturity $T$ in the two possible cases. If $S_T \ge K$ : the call is exercised and worth $S_T - K$ , and the cash has grown to $K$ , so Portfolio A is worth $(S_T - K) + K = S_T$ . The put expires worthless, and the stock is worth $S_T$ , so Portfolio B is worth $0 + S_T = S_T$ . The two portfolios agree.

If $S_T < K$ : the call expires worthless, and the cash is still worth $K$ , so Portfolio A is worth $0 + K = K$ . The put is exercised and worth $K - S_T$ , and the stock is worth $S_T$ , so Portfolio B is worth $(K - S_T) + S_T = K$ . The two portfolios again agree, this time both equal to $K$ .

So in every possible outcome, Portfolio A and Portfolio B have exactly the same value at $T$ . If their values at $t$ differed, an arbitrageur could sell the more expensive portfolio, buy the cheaper one, invest the difference at the risk-free rate, and at $T$ collect a risk-free profit regardless of what the stock does. Since real markets do not allow such riskless profit to persist, the two portfolios must have the same price at $t$ : $C + Ke^{-r(T-t)} = P + S$ , which rearranges to $C - P = S - Ke^{-r(T-t)}$ .

Theorem: The Black–Scholes equation, via Itô's lemma

Suppose the stock price follows geometric Brownian motion $dS_t = \mu S_t\,dt + \sigma S_t\,dW_t$ , and let $V(S,t)$ be the no-arbitrage price of a derivative paying a function of $S_T$ at maturity. Then $V$ must satisfy $\dfrac{\partial V}{\partial t} + \dfrac{1}{2}\sigma^2 S^2 \dfrac{\partial^2 V}{\partial S^2} + rS\dfrac{\partial V}{\partial S} - rV = 0$ .

Why is it true?

Ordinary calculus says the change in $V(S_t,t)$ comes only from $\partial V/\partial t$ and $\partial V/\partial S$ . But $S_t$ jitters randomly, and Itô's lemma shows that this randomness feeds back into $V$ through an extra term built from the second derivative $\partial^2 V/\partial S^2$ — a correction with no analogue in ordinary calculus. Once that extra term is accounted for, a portfolio that holds the derivative and exactly the right amount of stock can be made completely riskless, and a riskless portfolio can only earn the risk-free rate $r$ .

Proof

Apply Itô's lemma to $V(S_t,t)$ , treating $V$ as a smooth function of the random process $S_t$ and of time. It states $dV = \left(\dfrac{\partial V}{\partial t} + \mu S_t \dfrac{\partial V}{\partial S} + \dfrac{1}{2}\sigma^2 S_t^2 \dfrac{\partial^2 V}{\partial S^2}\right)dt + \sigma S_t \dfrac{\partial V}{\partial S}\,dW_t$ . Unlike ordinary calculus, an extra $\frac{1}{2}\sigma^2 S_t^2 \partial^2 V/\partial S^2$ term appears, coming from the nonzero quadratic variation of Brownian motion.

Now build a hedged portfolio $\Pi = V - \Delta S_t$ , holding one derivative and short $\Delta$ shares of stock, with $\Delta$ chosen to be $\partial V/\partial S$ . Its change is $d\Pi = dV - \Delta\,dS_t$ . Substituting the expression for $dV$ and $dS_t = \mu S_t\,dt + \sigma S_t\,dW_t$ , and choosing $\Delta = \partial V/\partial S$ , the random terms $\sigma S_t \partial V/\partial S\,dW_t$ and $-\Delta \sigma S_t\,dW_t$ cancel exactly.

What remains is $d\Pi = \left(\dfrac{\partial V}{\partial t} + \dfrac{1}{2}\sigma^2 S_t^2 \dfrac{\partial^2 V}{\partial S^2}\right)dt$ , a purely deterministic (riskless) change over the instant $dt$ , with no dependence left on the stock's drift $\mu$ .

A riskless portfolio must earn exactly the risk-free rate, or an arbitrage would be possible by borrowing or lending against it. So $d\Pi = r\Pi\,dt = r(V - S_t\,\partial V/\partial S)\,dt$ . Equating the two expressions for $d\Pi$ and dividing by $dt$ gives $\dfrac{\partial V}{\partial t} + \dfrac{1}{2}\sigma^2 S^2 \dfrac{\partial^2 V}{\partial S^2} = rV - rS\dfrac{\partial V}{\partial S}$ , which rearranges into the Black–Scholes equation.

AdvancedReal-World Applications and Worked Examples

Banks use these ideas every day to price and hedge options books, insurers use them to value guarantees embedded in life and annuity products, and corporate treasurers use them to decide how much to pay for protection against currency or interest-rate swings. The binomial tree is the workhorse for hand calculation and for options with early-exercise features; the Black–Scholes formula is the workhorse for quick, closed-form estimates.

Example: Pricing a call with a one-step binomial tree

A stock trades today at $S_0 = 100$ . Over the next year it will either rise to $S_u = 110$ or fall to $S_d = 90$ . The risk-free rate is $r = 0.05$ per year (simple, one period). Find the price of a European call with strike $K = 100$ and maturity one year.

Solution

First find the risk-neutral probability $p$ of the up move — the probability under which the discounted stock price is a martingale, not the real-world probability. It solves $S_0 = \dfrac{p\,S_u + (1-p)\,S_d}{1+r}$ , giving $p = \dfrac{(1+r)S_0 - S_d}{S_u - S_d} = \dfrac{1.05\times 100 - 90}{110 - 90} = \dfrac{15}{20} = 0.75$ .

Next compute the call's payoff in each branch: if the stock rises, $C_u = \max(110 - 100, 0) = 10$ ; if it falls, $C_d = \max(90 - 100, 0) = 0$ .

Finally discount the risk-neutral expected payoff at the risk-free rate: $C_0 = \dfrac{p\,C_u + (1-p)\,C_d}{1+r} = \dfrac{0.75\times 10 + 0.25\times 0}{1.05} = \dfrac{7.5}{1.05} \approx 7.14$ .

So the call is worth about $7.14$ today — notice that the real-world probability of the stock going up never entered the calculation, only $p$ , $r$ , and the two possible payoffs.

Example: Checking put–call parity to find a missing price

A stock trades at $S = 50$ . A European call with strike $K = 48$ and maturity $T = 0.5$ years is quoted at $C = 6.5$ . The risk-free rate is $r = 0.04$ . Use put–call parity to find the price $P$ of the European put with the same strike and maturity.

Solution

Start from put–call parity, $C - P = S - Ke^{-r(T-t)}$ , and solve for $P$ : $P = C - S + Ke^{-r(T-t)}$ .

Compute the discount factor first: $Ke^{-rT} = 48 \times e^{-0.04\times 0.5} = 48 \times e^{-0.02} \approx 48 \times 0.9802 \approx 47.05$ .

Substitute the numbers: $P \approx 6.5 - 50 + 47.05 = 3.55$ .

So the put should trade at about $3.55$ . If it traded noticeably above or below this value while $C$ , $S$ , and $r$ stayed fixed, an arbitrageur could combine the mispriced put with the call, the stock, and borrowing or lending to lock in a riskless profit — which is exactly why parity holds so tightly in liquid option markets.

Research frontier as of 2026

As of 2026, research in mathematical finance has moved well past the constant-volatility world of the original Black–Scholes model. Rough volatility models (building on the observation, developed by Gatheral, Jaisson, and Rosenbaum around 2018, that realized volatility behaves like a very rough — low Hölder-regularity — stochastic process rather than a smooth one) now fit option-price surfaces far better than classical stochastic-volatility models, at the cost of harder-to-simulate dynamics. Deep hedging and other machine-learning approaches train neural networks directly on simulated or historical paths to hedge and price derivatives in markets with transaction costs and other frictions where closed-form formulas do not exist, replacing the analytic PDE approach with learned, data-driven hedging strategies. A third active direction is model-free and robust finance: bounding option prices using only observed market prices of other options and minimal assumptions about the underlying process, so that conclusions do not depend on trusting a specific model like geometric Brownian motion. Open questions include how to calibrate rough volatility models efficiently, how to give rigorous guarantees for learned hedging strategies, and how far model-free bounds can be tightened without extra assumptions.

Using put–call parity with $S = 100$ , $K = 100$ , $r = 0.05$ , $T - t = 1$ , and a call price $C = 10$ , what is the put price $P$ ? (Use $e^{-0.05} \approx 0.9512$ .)

In a one-step binomial model with up factor $u = 1.2$ , down factor $d = 0.8$ , and risk-free rate $r = 0.1$ per period, what is the risk-neutral probability $p$ of the up move?

Under the Black–Scholes model, if the true (real-world) expected return $\mu$ of the stock increases while its volatility $\sigma$ stays fixed, the price of a call option on it:

Which of the following is NOT an assumption of the classical Black–Scholes model?

References

Fischer Black, Myron Scholes (1973). The Pricing of Options and Corporate Liabilities · DOI:10.1086/260062
John C. Cox, Stephen A. Ross, Mark Rubinstein (1979). Option Pricing: A Simplified Approach · DOI:10.1016/0304-405X(79)90015-1
Steven E. Shreve (2004). Stochastic Calculus for Finance II: Continuous-Time Models