LessonThis is a lesson restated in new words. Any figure in it is an example from the source, not this desk's record.
Monte Carlo prices an option by averaging discounted payoffs. That average is the same no-arbitrage value as the Black-Scholes hedging argument when a formula exists, because Feynman-Kac ties the hedging equation to a risk-neutral expectation. When the equation has no closed form, the simulation is often the practical way to compute the expectation, provided each path is a draw from the model you chose.
Why it matters
Black-Scholes reaches its equation by hedging away the underlying's randomness and requiring a riskless portfolio to earn the riskless rate. The fundamental theorem gives a different-looking object: the discounted risk-neutral expectation of the payoff. If the option value solves the equation, the lecture says it must equal that expectation. A formula is preferable when you have one, because each new parameter set is then an evaluation, not a new experiment.
Many payoffs have no analytical solution. Finite differences can still attack the equation. For a complicated payoff, simulating the expectation can be the easier numerical route. The law of large numbers is why a long average can stand in for the integral you cannot solve.
The idea in plain words
A discrete expectation is a probability-weighted sum. A fair die gives 3.5: each face, times one-sixth, added up. If the sum is awkward or infinite, repeated rolls still approach 3.5 as the count grows. A continuous expectation is an integral of the variable against its density. For a standard normal the integrand is an odd function over the whole line, so the integral is zero. Draws from that same normal pull the sample mean to zero as well.
A stochastic process is a random variable indexed by time. Brownian motion has independent increments, so the value at a fixed time T is normal, with variance equal to that time. In the lecture's picture, 100 simulated terminals already begin to fill the theoretical histogram. The same logic prices an option only if you draw the terminal law, or the whole path when the payoff needs it, from the model you intend. Write the risk-neutral expectation as an integral against the density and you often cannot finish it. The long-run average is the approximation.
The recipe has a fixed shape. Choose a model for the underlying: arithmetic or geometric Brownian motion, an additive or multiplicative process, Heston, even a rough specification. Discretize, for example with Euler-Maruyama, using a short time step and a Brownian increment, and simulate to maturity. Compute that path's payoff, whether the contract is a barrier, an Asian, or something else. Discount to the present, store the number, and repeat.
How to use it
The lecture applies the recipe to a rainbow option, a best-of call on two underlyings. At T you take the better of the two spots and pass it through the call payoff. Both underlyings are given Heston dynamics, with a correlation structure if you want one, under a risk-neutral parameter set. Heston has many free parameters, technically infinitely many choices of risk-neutral measure. The lecture's selection is to calibrate them to the current implied-volatility surface, then discretize with Euler-Maruyama and simulate to T.
What usually goes wrong
Simulating the wrong dynamics produces a confident average of the wrong price. The die example only converges because every roll is the same die. An option simulation only converges to the model's no-arbitrage value when the paths are draws from that model, including the risk-neutral drift and the discounting.
Using Monte Carlo out of habit when a formula exists is wasteful, not incorrect: both routes should agree. Using it on a payoff that depends on the path, while storing only the terminal spot, drops information the contract needs. Calibration is part of the recipe for Heston, not an optional chart. Without a parameter set tied to liquid instruments, the average is an average under a measure you did not justify.
Questions people ask
Why can an average replace the integral?
For a fixed distribution, the sample mean settles on the theoretical expectation. The die settles on 3.5 and the standard normal settles on zero. The option case is the same idea with a harder expectation.
When is simulation the better tool?
When the hedging equation or the expectation has no closed form, and the payoff is easier to evaluate on a path than on a grid. Finite differences remain available for the equation itself.
What model did the rainbow example use?
Heston dynamics on both underlyings, calibrated to the volatility surface, then an Euler-Maruyama path and a best-of call payoff, discounted and averaged.