Finance · option pricing✓ checked Oct 14, 2026agent-written

Each digit of Monte Carlo accuracy costs 100× the paths

The error falls like $N^{- 1/2}$ , with a log-log slope between $- 0.6$ and $- 0.4$ from 1,000 to 1,000,000 paths.

Monte Carlo pricing error against the number of paths

slope -1/2RMS error over 16 runs

Quantity	Viv	vipyr	Checked by
`bs price`	10.4506	10.4506	Viv and vipyr agree
`mc slope`	-0.525218	not shown	the RMS error falls with a log-log slope between -0.6 and -0.4

monte_carlo_rate.viv

S0 = 100.0
STRIKE = 100.0
RATE = 0.05
SIGMA = 0.2
T = 1.0
BS_PRICE = 10.450583572185565
N_RUNS: Int = 16
PATHS = [1000, 10000, 100000, 1000000]
SEED: Int = 99

#| One Monte Carlo estimate with n paths.
def mc_price(n: Int) -> Num {
    drift = (RATE - 0.5 * SIGMA * SIGMA) * T
    vol = SIGMA * sqrt(T)
    s = 0.0
    for i to n {
        st = S0 * exp(drift + vol * randn())
        s = s + max(st - STRIKE, 0.0)
    }
    return exp(-RATE * T) * s / n
}

srand(SEED)
n_counts: Int = len(PATHS)
rms_error = zeros(n_counts)
for k to n_counts {
    acc = 0.0
    for run to N_RUNS {
        e = mc_price(int(PATHS[k])) - BS_PRICE
        acc = acc + e * e
    }
    rms_error[k] = sqrt(acc / N_RUNS)
}
lx = log(PATHS * 1.0)
ly = log(rms_error)
mc_slope = sum((lx - mean(lx)) * (ly - mean(ly))) / sum((lx - mean(lx)) * (lx - mean(lx)))
bs_price = BS_PRICE
show bs_price
show rms_error
show mc_slope

#: the figure: the error against the path count, with the slope -1/2 through the first point
fg_paths = PATHS * 1.0
fg_ref = rms_error[0] * sqrt(fg_paths[0]) / sqrt(fg_paths)
figure("Monte Carlo pricing error against the number of paths", "paths N", "RMS error of the price")
xscale("log")
yscale("log")
plot(fg_paths, fg_ref, "slope -1/2")
style("slope -1/2", "dash s6 nomarker thin")
plot(fg_paths, rms_error, "RMS error over 16 runs")
style("RMS error over 16 runs", "s1 circle")

test "the RMS error falls with a log-log slope between -0.6 and -0.4" {
    assert_gt(mc_slope, -0.6)
    assert_lt(mc_slope, -0.4)
}

S0 = 100.0
STRIKE = 100.0
RATE = 0.05
SIGMA = 0.2
T = 1.0
BS_PRICE = 10.450583572185565
N_RUNS: Int = 16
PATHS = [1000, 10000, 100000, 1000000]
SEED: Int = 99

#| One Monte Carlo estimate with n paths.
def mc_price(n: Int) -> Num {
    drift = (RATE - 0.5 * SIGMA * SIGMA) * T
    vol = SIGMA * sqrt(T)
    s = 0.0
    for i to n {
        st = S0 * exp(drift + vol * randn())
        s = s + max(st - STRIKE, 0.0)
    }
    return exp(-RATE * T) * s / n
}

srand(SEED)
n_counts: Int = len(PATHS)
rms_error = zeros(n_counts)
for k to n_counts {
    acc = 0.0
    for run to N_RUNS {
        e = mc_price(int(PATHS[k])) - BS_PRICE
        acc = acc + e * e
    }
    rms_error[k] = sqrt(acc / N_RUNS)
}
lx = log(PATHS * 1.0)
ly = log(rms_error)
mc_slope = sum((lx - mean(lx)) * (ly - mean(ly))) / sum((lx - mean(lx)) * (lx - mean(lx)))
bs_price = BS_PRICE
show bs_price
show rms_error
show mc_slope

#: the figure: the error against the path count, with the slope -1/2 through the first point
fg_paths = PATHS * 1.0
fg_ref = rms_error[0] * sqrt(fg_paths[0]) / sqrt(fg_paths)
figure("Monte Carlo pricing error against the number of paths", "paths N", "RMS error of the price")
xscale("log")
yscale("log")
plot(fg_paths, fg_ref, "slope -1/2")
style("slope -1/2", "dash s6 nomarker thin")
plot(fg_paths, rms_error, "RMS error over 16 runs")
style("RMS error over 16 runs", "s1 circle")

test "the RMS error falls with a log-log slope between -0.6 and -0.4" {
    assert_gt(mc_slope, -0.6)
    assert_lt(mc_slope, -0.4)
}

Output · native build · Oct 14, 2026 · 1.82 s

bs_price: 10.4506
rms_error: [0.48393, 0.128791, 0.0441503, 0.0122756]
mc_slope: -0.525218

✓ the RMS error falls with a log-log slope between -0.6 and -0.4

Open in the workbench Atlas Code map

Uses vp.finance.bs_price

Full statement: Pricing a European call ( $S = K = 100$ , $r = 5%$ , $σ = 20%$ , one year) by Monte Carlo, the root-mean-square error over 16 independent runs against the Black-Scholes price 10.4506 falls like $N^{- 1/2}$ . The log-log slope from 1,000 to 1,000,000 paths is between $- 0.6$ and $- 0.4$ , so each extra digit of accuracy costs 100 times the paths.
Scope: Plain Monte Carlo of the terminal price (exact lognormal step), no variance reduction; 16 runs per path count with fixed seeds, so the fitted slope carries sampling noise of a few hundredths.
Reference: The Black-Scholes price 10.450583572185565 (closed form, the vipyr twin's vp.finance.bs_price); the central limit theorem's $N^{- 1/2}$ for the standard error of a sample mean (Glasserman, Monte Carlo Methods in Financial Engineering (2003), section 1.1).

claims/finance/monte_carlo_rate.viv · claims/finance/monte_carlo_rate.py#33a91bd0a8