Monte Carlo

Idea: A price with no formula

Monte Carlo: write the price as $E[f(X)]$, simulate many futures, average the payoffs.

• Needs only draws and a payoff, no density
• A grid of $10$ points over $6$ factors: $10^6=1000000$ cells
• Its error depends on paths, not on factors

Formula: The estimator

$\hat{\theta }=\frac{f(X_1)+\cdots +f(X_n)}{n}$

The sample mean of simulated payoffs tends to the truth as n grows.

Example: A short run by hand

A call with strike $100$. Simulated final prices: $92$, $104$, $117$, $99$, $108$. What is the estimate?

1. Payoffs: $0$, $4$, $17$, $0$, $8$
2. Sum: $0+4+17+0+8=29$
3. Estimate: $\frac{29}{5}=5.8$

Formula: Standard error

$\mathrm{SE}=\sigma /\sqrt{n}$

Use the sample s for σ, and report the price with about 2 standard errors either side.

Quick check: Size the error bar

Simulated payoffs have $s=12$ over $n=900$ paths. What is the standard error?

Common mistake: Expecting the error to fall with n

The error falls with $\sqrt{n}$, not with $n$. Doubling the paths does not halve it.

Instead Cutting error by $5$ takes $5^2=25$ times the paths: $400\times 25=10000$.

Idea: A probability is an average too

Average an indicator: the share of paths on which the event happened.

• Variance per path: $p(1-p)$
• Rare $p=0.0001$ over $10000$ paths: $10000\times 0.0001=1$ hit
• The error is as large as $p$ itself: plain runs starve

Quick check: Error of a probability

An event has $p=0.2$. What is the standard error of its estimate from $10000$ paths?

Idea: Say it out loud

State the estimator, then its error: the mean payoff with standard error $\frac{s}{\sqrt{n}}$.

• More accuracy costs the square of the factor
• Each extra digit: $10^2=100$ times the compute