Estimating a European Call Price by Monte Carlo

From Terminal Prices to a Call-Price Estimate

A European call pays \((S_T-K)^+=\max(S_T-K,0)\) at its maturity \(T\). Under the risk-neutral measure, its time-zero price is the expected discounted payoff. Monte Carlo estimation replaces that expectation with the average of finitely many simulated payoffs. The estimate is random: its standard error describes the sampling variability of the average, not the size of the error in any one particular run.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\), and let \(\mathbb{Q}\) be the risk-neutral measure, equivalent to \(\mathbb{P}\) on the time interval under consideration. Under \(\mathbb{Q}\), let \(B_t\) be Brownian motion and suppose the stock follows geometric Brownian motion with constant interest rate \(r\) and volatility \(\sigma>0\):

Here \(rS_t\) is the drift term and \(\sigma S_t\) is the diffusion coefficient. The stochastic integral is an Ito integral: its integrand is adapted and progressively measurable and belongs to \(\mathcal{L}^2\) on each finite time interval, meaning \(\mathbb{E}^{\mathbb{Q}}\int_0^T H_s^2\,ds<\infty\). In particular, the integral with integrand \(\sigma S_s\) is well-defined, since the geometric-Brownian solution has finite second moments on finite intervals.

The terminal-price formula follows by applying Ito's formula to the logarithm of \(S_t\), or equivalently by checking the explicit solution in the SDE. In that calculation, the correction from \(dB_t\,dB_t=dt\) is essential; the exponent contains \(-\tfrac12\sigma^2T\), not zero. Direct terminal simulation uses one normal draw per price and does not require simulating all the intermediate points of a Brownian path.

Why the Estimator Works

Proof. Set \(m=\mathbb{E}^{\mathbb{Q}}[Y_1]\) and \(v=\operatorname{Var}^{\mathbb{Q}}(Y_1)\). These quantities are finite: \(0\le (S_T-K)^+\le S_T\), and the lognormal terminal price has finite second moment. Independence gives \(\mathbb{E}^{\mathbb{Q}}[\widehat C_n]=\frac1n\sum_i\mathbb{E}^{\mathbb{Q}}[Y_i]=m=C\). Also, the covariance of distinct \(Y_i,Y_j\) is zero, so expanding the variance of the sum gives \(\operatorname{Var}^{\mathbb{Q}}(\widehat C_n)=n^{-2}\sum_i v=v/n\). For the sample variance, write \(\overline Y-m=\frac1n\sum_i(Y_i-m)\). Expanding squares gives \(\sum_i(Y_i-\overline Y)^2=\sum_i(Y_i-m)^2-n(\overline Y-m)^2\). Taking expectations yields \(nv-n(v/n)=(n-1)v\). Dividing by \(n-1\) shows \(\mathbb{E}^{\mathbb{Q}}[s_Y^2]=v\), and hence \(\mathbb{E}^{\mathbb{Q}}[s_Y^2/n]=v/n\). \(\square\)

The theorem establishes the estimator’s mean and variance; it does not assert that a particular estimate must be close to the price. The reported standard error estimates the standard deviation of repeated estimates of the same size. In a large sample, an approximate interval is often reported as \(\widehat C_n\pm1.96\,\widehat{\operatorname{SE}}\). This is a normal approximation, not an exact finite-sample guarantee; it is most useful when the sample is sufficiently large and the payoff distribution is not too skewed.

The benchmark for comparison is the Black–Scholes call formula, a standard result for this constant-parameter model: \(C_{\mathrm{BS}}=S_0\Phi(d_1)-Ke^{-rT}\Phi(d_2)\), where \(d_1=(\log(S_0/K)+(r+\tfrac12\sigma^2)T)/(\sigma\sqrt T)\) and \(d_2=d_1-\sigma\sqrt T\), for \(K>0\). The Monte Carlo result should fluctuate around this benchmark, with typical sampling fluctuations decreasing in proportion to \(1/\sqrt n\). This comparison is a diagnostic, not a proof that a particular simulation is correctly implemented.

Worked Calculations

Implementation Checks and Common Mistakes

In a simulation, generate independent standard normal draws and use the exact terminal transition. For every draw, compute the exponential, apply the positive-part operation, discount the payoff, and then calculate the mean and sample standard deviation of the discounted payoffs. The order matters: the standard error is computed from the \(Y_i\), so both the payoff transformation and discounting must occur before calculating it.

Common mistake: using the physical drift in a risk-neutral price. A simulation under the physical measure \(\mathbb{P}\) may use a different drift, such as an expected stock return \(\mu\). Substituting \(\mu\) for \(r\) in the risk-neutral terminal-price formula generally estimates a different expectation. For the no-dividend model here, pricing uses the risk-neutral measure \(\mathbb{Q}\), under which the drift is \(r\), and discounts by \(e^{-rT}\).

Common mistake: reporting the payoff standard deviation as the standard error. The standard deviation \(s_Y\) measures the spread of individual discounted payoffs. The estimated standard error of their mean is \(s_Y/\sqrt n\). Confusing the two overstates the estimated uncertainty in the mean by a factor of \(\sqrt n\).

A useful implementation check is to compare the Monte Carlo estimate with \(C_{\mathrm{BS}}\) for a case where the benchmark is available. Also check basic constraints: each simulated payoff is nonnegative, and the call price cannot exceed the discounted stock value \(S_0\) in this model. A mismatch may reflect ordinary sampling variation, a coding error, an incorrect discount factor, or use of the wrong drift; the benchmark comparison alone does not identify which one.