From Simulated Paths to an Uncertainty Interval
A path simulation produces a random payoff, not the exact mean payoff. Repeating the simulation with independent paths gives a sample average, and the central limit theorem describes the average’s sampling variability when the payoff has finite, nonzero variance. A confidence interval summarizes this Monte Carlo uncertainty; it does not certify that an individual interval contains the unknown mean.
For context, work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Consider an SDE \(dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t\) on \([0,T]\). The coefficient \(\mu\) is the drift and \(\sigma\) is the diffusion coefficient. For example, measurable coefficients that are globally Lipschitz in the state variable and satisfy a linear growth bound give a unique strong solution under the standard existence theorem. Its Ito integral is defined for the progressively measurable integrand \(s\mapsto\sigma(s,X_s)\) when \(\mathbb{E}\int_0^T\sigma(s,X_s)^2\,ds<\infty\). More generally, an Ito integral \(\int_0^T H_s\,dB_s\) below is understood for a progressively measurable \(H\) in \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\).
Fix a payoff rule \(G\), which may depend on the whole simulated path, and let \(Y=G(X)\) denote the resulting payoff. The discussion applies as well when \(X\) is produced by a fixed numerical scheme: then \(Y\) is the payoff output by that scheme. We estimate the mean \(m=\mathbb{E}[Y]\) of the random variable actually being simulated. In either case, assume \(\mathbb{E}[Y^2]<\infty\) and \(\operatorname{Var}(Y)=\sigma_Y^2>0\). Generate \(n\) independent copies \(Y_1,\ldots,Y_n\), each with the same distribution as \(Y\), by simulating independent paths. The time steps within one path are not \(n\) independent payoff observations.
The standard error estimates the standard deviation of the sample average, not the spread of individual payoffs. Since independent copies have variance \(\sigma_Y^2\), \(\operatorname{Var}(\overline{Y}_n)=\sigma_Y^2/n\), so the population standard deviation of the average is \(\sigma_Y/\sqrt n\). The unknown \(\sigma_Y\) is replaced by \(s_n\).
The Studentized Central-Limit Interval
Proof. We first establish consistency of the estimated variance. Write \(m=\mathbb{E}[Y_1]\). The strong law of large numbers, applied to the integrable variables \(Y_i\) and \((Y_i-m)^2\), gives \(\overline{Y}_n\to m\) and \(n^{-1}\sum_{i=1}^n(Y_i-m)^2\to\sigma_Y^2\) almost surely. Expanding around \(m\) gives \(s_n^2=\frac{n}{n-1}\left(\frac1n\sum_{i=1}^n(Y_i-m)^2-(\overline{Y}_n-m)^2\right)\). The first term in parentheses tends to \(\sigma_Y^2\), and the second tends to zero. Thus \(s_n^2\to\sigma_Y^2\) almost surely and \(s_n\to\sigma_Y>0\) almost surely. By the iid central limit theorem, \(\sqrt n(\overline{Y}_n-m)/\sigma_Y\) converges in distribution to \(N(0,1)\). Slutsky’s theorem, using \(s_n/\sigma_Y\to1\) in probability, implies that \(T_n=\sqrt n(\overline{Y}_n-m)/s_n\) converges in distribution to \(N(0,1)\) wherever \(s_n>0\).
For completeness, define \(T_n=0\) on \(\{s_n=0\}\). This convention does not change the limiting distribution: \(s_n\to\sigma_Y>0\) in probability implies \(\mathbb{P}(s_n=0)\to0\). On \(\{s_n>0\}\), the event that \(m\) belongs to the displayed interval is exactly \(\{|T_n|\le z_{1-\alpha/2}\}\). The limiting normal distribution assigns probability zero to the two boundary points, so convergence in distribution gives limiting probability \(1-\alpha\) for this event. The coverage probability on \(\{s_n=0\}\), under any interval convention, can change the total probability by at most \(\mathbb{P}(s_n=0)\), which tends to zero. This proves the assertion. The iid central limit theorem and the strong law used here are standard probability results; their general proofs are not developed in this tutorial. \(\square\)
This is an asymptotic interval: its coverage approaches \(1-\alpha\) as the number of independent paths grows. It is not a claim that the fixed, observed interval has probability \(1-\alpha\) of containing the fixed number \(m\). Rather, under repeated simulation with the same sampling procedure, the proportion of intervals containing \(m\) approaches the stated level. For a 95% interval, \(z_{0.975}\) is approximately \(1.96\).
Worked Example: A 95% Interval from Five Payoffs
Suppose five independent simulated paths produce payoffs \(0,1,2,3,4\). Compute the estimated standard error and the 95% central-limit interval.
The sample mean is \(\overline{Y}_5=(0+1+2+3+4)/5=10/5=2\). The squared deviations from 2 are \(4,1,0,1,4\), whose sum is \(10\). Hence \(s_5^2=10/(5-1)=10/4=2.5\), and \(s_5=\sqrt{2.5}\). The estimated standard error is
Using \(1.96\) as the 95% normal critical value, the half-width is \(1.96\sqrt{0.5}\approx1.3859\). The interval is therefore \([2-1.3859,\ 2+1.3859]\), or approximately \([0.614,\ 3.386]\) to three decimal places.
Check: the interval is centered at the computed mean \(2\), and its total width is \(2(1.3859)\approx2.7718\). The displayed standard error is smaller than the sample standard deviation by the required factor \(\sqrt5\): \(\sqrt{2.5}/\sqrt5=\sqrt{0.5}\).
Worked Path Payoffs and Interval Width
Worked Example: A Payoff Sample with Repeated Values
Four independent paths yield payoffs \(0,0,1,1\). Calculate the estimated standard error and the 95% interval. This example also shows why a finite-sample zero sample variance cannot be dismissed as impossible.
The mean is \(\overline{Y}_4=(0+0+1+1)/4=1/2\). Each deviation from \(1/2\) has square \(1/4\), so the sum of squared deviations is \(4(1/4)=1\). Thus \(s_4^2=1/(4-1)=1/3\), and the estimated standard error is \(\sqrt{(1/3)/4}=\sqrt{1/12}=1/\sqrt{12}\approx0.2887\). The half-width is \(1.96/\sqrt{12}\approx0.5658\), giving the interval \([0.5-0.5658,\ 0.5+0.5658]\), or approximately \([-0.066,\ 1.066]\).
Check: the interval’s half-width is \(1.96\) times the computed standard error, and its midpoint is \(0.5\). Since these payoffs are either 0 or 1, the negative lower endpoint is not a possible payoff; it reflects the normal approximation, not a claim that payoffs outside \([0,1]\) can occur. If four Bernoulli payoffs were all equal, then \(s_4=0\); under success probability \(1/2\), that event has probability \(2(1/2)^4=1/8>0\). The asymptotic theorem handles this by the vanishing probability of zero sample variance as \(n\) grows, not by asserting it is impossible at finite \(n\).
Worked Example: How Sample Size Changes the Width
Suppose a pilot simulation suggests a payoff standard deviation of \(2\). Using \(1.96\) for a 95% interval, compare the estimated widths for \(n=100\) and \(n=400\), holding the standard-deviation estimate fixed.
For \(n=100\), the estimated standard error is \(2/\sqrt{100}=2/10=0.2\). The half-width is \(1.96(0.2)=0.392\), and the full width is \(2(0.392)=0.784\). For \(n=400\), the estimated standard error is \(2/\sqrt{400}=2/20=0.1\). The half-width is \(1.96(0.1)=0.196\), and the full width is \(2(0.196)=0.392\).
Check: multiplying the number of paths by four multiplies its square root by two. With the same estimated payoff spread, the standard error and interval width are therefore halved: \(0.784/2=0.392\). Actual sample standard deviations need not be exactly equal across two runs, so this comparison describes the formula’s scaling, not a guarantee about observed widths.
Common Errors and Practical Diagnostics
Using the wrong standard-error formula. A common incorrect calculation divides the sample standard deviation by \(n\), rather than by \(\sqrt n\). For the five payoffs \(0,1,2,3,4\), this error gives \(\sqrt{2.5}/5\approx0.3162\), whereas the correct standard error is \(\sqrt{2.5}/\sqrt5=\sqrt{0.5}\approx0.7071\). The incorrect value understates uncertainty because the average of \(n\) independent observations has standard deviation proportional to \(n^{-1/2}\), not \(n^{-1}\).
Confusing payoff spread and estimator precision. The sample standard deviation \(s_n\) measures the observed spread of individual payoffs. The standard error \(s_n/\sqrt n\) estimates the spread of the sample average across repeated simulations. More paths can reduce the latter without reducing the former.
Checking the wrong convergence. Useful basic diagnostics include reporting \(n\), \(\overline{Y}_n\), \(s_n\), the standard error, and the interval endpoints; verifying that the reported half-width equals \(z_{1-\alpha/2}s_n/\sqrt n\); and comparing estimates from independent batches. Batch estimates should vary on a scale consistent with their own standard errors, although agreement between batches is a diagnostic, not a proof of correctness. If paths are not independent, the iid standard-error formula and theorem above do not automatically apply.
The interval quantifies sampling uncertainty around the mean of the payoff distribution being simulated. In particular, if a numerical SDE scheme is used, the interval alone does not show that this simulated mean equals the exact SDE payoff mean. Keeping the target mean explicit prevents a confidence interval from being interpreted more broadly than its assumptions allow.
Check Your Understanding
Use the definitions and theorem above to answer the following questions.
- Why does the standard error of an average of \(n\) independent payoffs scale as \(n^{-1/2}\), rather than \(n^{-1}\)?
- For the payoffs \(1,2,2,3\), calculate the sample mean, sample variance, and estimated standard error.
- On which event is the equivalence between interval coverage and the studentized-statistic event stated directly? Why is that qualification needed?
- If the number of independent paths is multiplied by nine while the estimated payoff standard deviation stays fixed, by what factor does the estimated standard error change?
- What mean does a confidence interval from a fixed numerical SDE scheme estimate, and what does the interval alone not establish?