Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 35 / 4

For an invertible continuous cumulative distribution function, monotonicity givesThus each transformed observation has the desired distribution. More generally use the quantile function $q(u)=\inf \{x:F(x)\ge u\}$; the same event identity proves inverse transform sampling even when there are atoms. Independent uniform inputs give independent transformed samples; mere uniform marginal distributions do not imply independence between different samples.

One route is inverse transform sampling: if $\Phi$ is the standard normal cumulative distribution function, then $\mu +\sigma {\Phi }^{-1}(U)$ has the normal distribution with mean $\mu$ and variance ${\sigma }^2$.

Alternatively the Box-Muller transform uses two independent uniform inputs $U_1,U_2\in (0,1)$:To verify this, set $R=\sqrt{-2\log U_1}$ and $\Theta =2\pi U_2$. The radial density is $r{e}^{-r^2/2}$ for $r>0$, and the angle is independent and uniform. Their joint density is $r{e}^{-r^2/2}/(2\pi )$. Changing to Cartesian coordinates divides by the Jacobian determinant $r$, givingThis factors into two standard normal densities, proving both normality and independence. Then $\mu +\sigma Z_j$ provides the required means and variances. Implementations must avoid the zero input to the logarithm; ideal continuous uniforms hit it with probability zero.

Let $K$ denote the strike, $\tau =T-t$, and suppose $S>0$, $K>0$, $\tau >0$ and $\sigma >0$. Under the Black-Scholes model and the risk-neutral measure, $\log S_T$ is normal withApplying risk-neutral valuation to the European call option payoff givesUse $\xi =\Phi ((u-m)/v)$, whose differential is precisely the normal density times $du$. The threshold is ${\xi }_K=\Phi ((\log K-m)/v)$, soEquivalently, rescale the tail interval to obtain $g_{\mathrm{tail}}(\eta )=e^{-r\tau }(1-{\xi }_K)\{\exp [m+v{\Phi }^{-1}({\xi }_K+(1-{\xi }_K)\eta )]-K\}$ on $0<\eta <1$.

For independent uniform inputs, the Monte Carlo estimator is ${\hat{C}}_N=N^{-1}{\sum }_{j=1}^{N}g(U_j)$. It is unbiased, converges by the strong law of large numbers, and has variance $\mathrm{Var}(g(U))/N$. The payoff has finite second moment because the lognormal stock has finite moments. Its estimated standard error is the sample standard deviation divided by $\sqrt{N}$.

Three useful methods are antithetic variates, a control variate, and stratified sampling. Their purpose is to reduce sampling variance at comparable computational cost, rather than merely to take more samples.

For antithetic variates, generate independent pairs $(U,1-U)$ and average both payoffs. Since the call integrand is increasing, the two values have nonpositive covariance. Indeed, for an independent copy $U^{\prime }$, twice that covariance is the expectation of $[g(U)-g(U^{\prime })][g(1-U)-g(1-U^{\prime })]$, which is nonpositive pointwise. With $N=2m$ evaluations, the paired average has varianceFor a control variate, use the same simulated discounted stock value $Y=e^{-r\tau }{S}_T$, whose known mean is $S$. Replace the discounted call payoff $H$ by $H-\beta (Y-S)$, preserving its mean. Minimizing its quadratic variance givesA predetermined coefficient or an independent pilot estimate preserves exact unbiasedness; fitting the coefficient from the same observations requires care.

For stratified sampling, divide $[0,1]$ into $q$ equal intervals, sample independently and uniformly within each, and average their $q$ payoffs. If $v_j$ is the conditional variance in interval $j$, this unbiased estimator has variance $q^{-2}{\sum }_j{v}_j$. The law of total variance gives $\mathrm{Var}(H)\ge q^{-1}{\sum }_j{v}_j$, so this is at most the variance $\mathrm{Var}(H)/q$ of $q$ ordinary independent samples. Stratification removes randomness in how many samples fall in each region, which is particularly valuable when conditional payoff means vary strongly.