The standard error of a simulated option price, computed exactly, and what it says about the paths you need and the runs you already have.
Divide the standard deviation of the discounted payoff by the price accuracy you need, square it, and that is the number of paths. For a one-year at-the-money call worth 11.4541, the payoff's standard deviation is 17.43, so a standard error of 0.01 takes about 3 million paths and a 95 per cent confidence interval of ±0.01 about 11.7 million. A million paths gives a standard error of 0.0174, not the third decimal place many people assume.
Worked in full in Quantitative Finance by Julian R. Sterling, with every figure reproduced in a free workbook.See the book on Amazon →
The option is Kestrel 100, the European call that runs through Quantitative Finance, simulated under the same lognormal model as Black-Scholes so that the exact answer is known. All inputs are illustrative.
| Input | Value |
|---|---|
| Spot and strike | 100 |
| Maturity, years | 1 |
| Risk-free rate, continuously compounded | 4% |
| Volatility | 24% |
| Black-Scholes price, the target | 11.4541 |
A Monte Carlo price is the average of n independent discounted payoffs. By the central limit theorem its error shrinks with the square root of n.
Standard error = σpayoff ÷ √n
Paths needed = (z × σpayoff ÷ tolerance)2, with z = 1 for one standard error, 1.96 for 95 per cent confidence
Excel, on a column of simulated discounted payoffs: =STDEV.S(B2:B100001)/SQRT(COUNT(B2:B100001))
The square root is the whole story. Ten times the accuracy costs a hundred times the paths.
In practice you estimate σpayoff from a pilot run. For a European call under the lognormal model it has a closed form, which makes it a clean benchmark. With X = max(ST − K, 0):
E[X2] = S2e(2r+σ²)TN(d1 + σ√T) − 2K × S erTN(d1) + K2N(d2) = 471.24
Discounted second moment: e−2rT × 471.24 = 435.01
Variance = 435.01 − 11.45412 = 435.01 − 131.20 = 303.82
σpayoff = √303.82 = 17.43
The standard deviation is 1.52 times the price itself. That is typical of an option: most paths pay nothing, a few pay a great deal, and the average of a lopsided distribution converges slowly.
| Paths | Standard error | 95% half-width | Half-width as share of price |
|---|---|---|---|
| 1,000 | 0.5512 | 1.0803 | 9.43% |
| 10,000 | 0.1743 | 0.3416 | 2.98% |
| 100,000 | 0.0551 | 0.1080 | 0.94% |
| 1,000,000 | 0.0174 | 0.0342 | 0.30% |
| Tolerance | For one standard error | For 95% confidence |
|---|---|---|
| 0.100 | 30,382 | 116,714 |
| 0.050 | 121,526 | 466,856 |
| 0.010 | 3,038,159 | 11,671,392 |
| 0.005 | 12,152,636 | 46,685,567 |
The book reports four simulation runs of this option. Each is a price and a sample standard deviation; dividing one by √n and measuring the distance from 11.4541 in standard errors is the test the book insists on.
| Paths | Price | Sample sd | Standard error | Difference | In standard errors |
|---|---|---|---|---|---|
| 1,000 | 10.4523 | 16.702 | 0.5282 | −1.0018 | −1.90 |
| 10,000 | 11.2841 | 17.286 | 0.1729 | −0.1699 | −0.98 |
| 100,000 | 11.3448 | 17.388 | 0.0550 | −0.1093 | −1.99 |
| 1,000,000 | 11.4244 | 17.408 | 0.0174 | −0.0297 | −1.71 |
Every run is within two standard errors, so none is wrong on its own. The million-path run is 0.0297 below the closed form, which looks like model disagreement but is 1.71 standard errors of noise. Notice too that the sample standard deviations climb towards the exact 17.43 as n grows: here the smaller runs understated the tail, and with it their own error.
Paths are not the only lever. Two standard techniques lower σpayoff itself, and the gain enters squared.
A sixfold saving from a few lines of code is worth more than a faster machine.
The first is reporting a simulated price with no standard error, which makes it impossible to say whether a gap to another model is error or noise. The second is setting n by habit. "A million paths" sounds exhaustive, but here it fixes the price only to about ±0.03 at 95 per cent confidence. The third is comparing models on a single run: the 0.0297 spread between the closed form, the lattice and the simulation on this option is mostly the noise of the million-path run. The closed form it is measured against is worked in the Black-Scholes worked example.
Out-of-the-money options are worse: fewer paths pay anything, the payoff's standard deviation is larger relative to the price, and the paths needed for the same relative accuracy rise sharply.
The standard error of each run, live beside the reported prices, is in the free workbook for this case.
Take the sample standard deviation of the discounted payoffs and divide by the square root of the number of paths. For an at-the-money one-year call with a payoff standard deviation of 17.43, 100,000 paths give a standard error of 0.0551 and a 95 per cent interval of plus or minus 0.1080 around the estimate.
It depends on the accuracy you need. For a one-year at-the-money call worth 11.4541 with a payoff standard deviation of 17.43, a million paths give a standard error of 0.0174, so the 95 per cent interval is about plus or minus 0.0342. For a 0.01 standard error you need about 3 million paths.
On this option, pairing each draw with its mirror image cuts the variance of a pair to 86.32 against 303.82 for a single path, a gain of 1.76 times per draw. A control variate on the discounted stock price does better: with a correlation of 0.9191 it removes 84.5 per cent of the variance.
Chapter 5 of Quantitative Finance makes the standard error the quantity that decides whether a simulated price can be trusted. The book takes the same case from first principles to the decision, chapter by chapter, and every figure it prints is a live formula in the free companion workbooks.
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