Math 5632 Notes. Chapter 19. Sampling of a Uniform Distribution

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1 Math 5632 Notes Chapter 19 Sampling of a Uniform Distribution Uniform distribution over [0, 1]: One example, Rand() in Excel Uniform distribution over [a, b]: If U is uniform over [0, 1], then a + (b a)u is uniform over [a, b]. Sampling the Standard Normal Distribution N (0, 1) Central Limit Approximation: Let u 1, u 2,..., u 12 be twelve independent samples of the uniform distribution over [0, 1]. Then is a sample of N (0, 1). 12 z = u i 6 Inverse CDF Method: Let N (or Φ) be the CDF of the standard normal distribution. Since N is strictly increasing, N 1 is well defined. If u is sample of the uniform distribution over [0, 1], then z = N 1 (u) is a sample of N (0, 1). Sampling Stock Price under BS Under Black-Scholes, S(T ) = S(0)e (α δ σ2 /2)T +σz(t ) = S(0)e (α δ σ2 /2)T +σ T Z Z(t) is a BM and Z is standard normal. where Sample of S(T ): Let z be a sample of N (0, 1), then S T = S(0)e (α δ σ2 /2)T +σ T z is a sample of S(T ). Sample of a Path: Divide [0, T ] into n equal subintervals, each has length h = T/n, and the ith subinterval is [(i 1)h, h]. Recursively, we have the following samples of S at the end of each subinterval: S h = S 0 e (α δ σ2 /2)h+σ hz 1 S 2h = S h e (α δ σ2 /2)h+σ hz 2 S nh = S (n 1)h e (α δ σ2 /2)h+σ hz n where the z i s are the independent samples of N (0, 1). These S ih s will be used to stimulate the prices of Asian options later. Substituting S h into S 2h and simplifying, we get S 2h = S 0 e (α δ σ2 /2)2h+σ h(z 1 +z 2 )

2 Substituting this S 2h into S 3h and simplifying, and so on, we get S T = S nh = S 0 e (α δ σ2 /2)nh+σ h(z 1 + +z n) = S 0 e (α δ σ2 /2)T +σ T ( ) z1 + +zn n Example. (SOA Sample, Problem 52) The price of a stock is to be estimated using simulation. It is known that: The time-t stock price, S t, follows the lognormal distribution: ( ) St ln N (( α σ 2 /2 ) t, σ 2 t ) S 0 S 0 = 50, α = 0.15, and σ = 0.3. The following are three uniform (0, 1) random numbers: , , Use each of these three numbers to simulate a time-2 stock price. Calculate the mean of the three simulated prices. Solution. The corresponding standard normal samplings are N 1 (0.9830) = , N 1 (0.0384) = , N 1 (0.7794) = α σ 2 /2 = 0.105, σ t+0.3 tz = 0.09 S t = S 0 e sampled stock prices and their average are S 2 = 50e z. The 50e ( ) = , 50e ( ) = , 50e ( ) = , Monte Carlo Valuation = Let V (S, T ) be the time T payoff of a derivative which depends on the final stock price S(T ), like an European call option, or a path of the stock price, like an Asian option. Let S 1, S 2,..., S n be n random samples of S under risk neutral probability. Then the Monte Carlo estimate of the time 0 value of the derivative is the discounted average value of the sampled payoffs: V = 1 n e rt n V (S i, T ). (Why do we use risk neutral probability to sample the stock price?) Example. Under Black-Scholes, the volatility of a stock is 0.3, and the interest rate is The current price of the stock is 10. Using 40 samples, estimate the price of a 1-year European call option on the stock with strike price 11.

3 U Z S(T) V(T) V(0) Monte.Carlo Black6Scholes , , , , , , , , , , , , , , , , , , , , Figure 1: Monte Carlo Example 1

4 Solution. The stock price is S(T ) = 10e 0.005T +0.3 Z(T ). Figure 1 shows a simulation using Excel. The Monte Carlo evaluation is , and the Black-Scholes formula gives Example. Asian option. The Monte Carlo estimate V of V (0) is an unbiased estimate. Let V i = V (S i, T ) be the sampled payoffs of V (T ). Then E V i = E (V (T )). we have ( ) E (V ) = E 1 n n e rt V i = 1 n n e rt E (V i ) = e rt E (V (T )) = V (0) Therefore, V is an unbiased estimate of V (0). Accuracy Let X 1, X 2,..., X n be n independent samples of X. The variance σx 2 of X and the variance σn 2 of the average of the X i s have the following relation ( ) σn 2 X1 + + X n = Var = nσ2 X = σ2 X n n 2 n, and so their standard deviations satisfy σ n = σ X / n. Therefore, the larger the n is, the more reliable/accurate the sampled average is. Usually we do not know what σ X is, but we can use the sampled variance to approximate it. Efficiency Improvement Control Variate Method Example. A = A + (G G) where A and G are the Monte Carlo estimates of A and G respectively, G is the BS value, and A is the improved estimate. In general, suppose that the correlation between two asset prices X and Y is large, and Y can be computed by a pricing formula. Let X and Y be the Monte Carlo estimates of X and Y respectively, and let X = X + β(y Y ). We want to determine β such that Var(X ) is minimal. Since Var(X ) = Var[X + β(y Y )] = Var[X βy ] = Var(X) + β 2 Var(Y ) 2βCov(X, Y ) (1) is a quadratic polynomial in β, the minimum value of Var(X ) occurs at 1 β = Cov(X, Y ). Var(Y ) 1 If y = ax 2 + bx + c and a > 0, then y has a minimum value c b2 4a when x = b 2a.

5 Exercise. Substitute the expression of β above into (1) to show that the minimal Var(X ) is given by Var(X Cov(X, Y )2 ) = Var(X) Var(Y ) = Var(X) ( 1 ρ 2), where ρ is the correlation coefficient of the estimated prices X and Y. Example. (SOA Sample, Problem 75) You are using Monte Carlo simulation to estimate the price of an option X, for which there is no pricing formula. To reduce the variance of the estimate, you use the control variate method with another option Y, which has a pricing formula. You are given: 1. The naive Monte Carlo estimate of the price of X has standard deviation The same Monte Carlo trials are used to estimate the price of Y. 3. The correlation coefficient between the estimated prices of X and Y is 0.8. Calculate the minimum variance of the estimated price of X, with Y being the control variate. Solution. Let X = X + β(y Y ). Since the minimum value of Var(X ) is Var(X ) = Var(X) + β 2 Var(Y ) 2βCov(X, Y ) Cov(X, Y )2 Var(X) Var(Y ) = Var(X) ( 1 ρ 2) = = 0.9. Question. How about X and Y are negatively correlated? Stratified Sampling Example. (SOA Sample Problem 57) Michael uses the following method to simulate 8 standard normal random variates: 1. Simulate 8 uniform (0, 1) random numbers U 1, U 2,..., U Apply the stratified sampling method to the random numbers so that U i and U i+4 are transformed to random numbers V i and V i+4 that are uniformly distributed over the interval ( i 1, i 4 4), i = 1, 2, 3, 4. In each of the four quartiles, a smaller value of U results in a smaller value of V. 3. Compute 8 standard normal random variates by Z i = N 1 (V i ), where N 1 is the inverse of the cumulative standard normal distribution function. Michael draws the following 8 uniform (0, 1) random numbers: i U i

6 Find the difference between the largest and the smallest simulated normal random variates. Solution. We have i = 1, V 1 = 0 + U = 0.122, V 5 = 0 + U = i = 2, V 2 = U = , V 6 = U = i = 3, V 3 = U = , V 7 = U = i = 4, V 4 = U = , V 8 = U = Z 4 = max{z i } = N 1 (V 4 ) = and Z 5 = min{z i } = N 1 (V 5 ) = The largest difference is then Z 4 Z 5 =