Stochastic Calculus

Brownian motion, Ito’s rule and the ideas behind Black-Scholes.

1. 01Brownian motion at two timesEasyAbout 3 min

   W is a standard Brownian motion with W_0 = 0. What are the mean and the variance of W_4? What is the covariance of W_2 and W_5? What is the probability that W_4 is above 2?

   Hint

   W_t is normal with mean 0 and variance t. For the covariance, write W_5 as W_2 plus an independent increment.

   Answer

   W_4 has mean 0 and variance 4. The covariance of W_2 and W_5 is 2. The probability that W_4 exceeds 2 is about 0.159.

   Solution

   By definition W_4 is normal with mean 0 and variance 4, so its standard deviation is 2.

   W_5 = W_2 + (W_5 - W_2), and the increment is independent of W_2. So Cov(W_2, W_5) = Var(W_2) = 2. In general the covariance is the smaller of the two times.

   W_4 > 2 means W_4 / 2 > 1 for a standard normal, which has probability 1 - Φ(1), about 0.159.

   Mark as done
2. 02Ito on the log of a priceMediumAbout 6 min

   A price follows dS = μS dt + σS dW with μ = 0.08 and σ = 0.20 per year, and starts at S_0 = 100. Use Ito’s rule to find the dynamics of ln S. What are the median and the mean of the price after 10 years?

   Hint

   For f(S) = ln S you need f′ = 1/S and f″ = -1/S^2, and (dS)^2 = σ^2 S^2 dt.

   Answer

   d(ln S) = 0.06 dt + 0.20 dW. After 10 years the median price is about 182.21 and the mean is about 222.55.

   Solution

   Ito’s rule gives d(ln S) = dS/S - (1/2)(dS)^2/S^2 = μ dt + σ dW - (1/2)σ^2 dt.

   So ln S has the constant drift μ - σ^2/2 = 0.08 - 0.02 = 0.06 and the volatility 0.20.

   ln S_10 is normal with mean ln 100 + 0.6, so the median of S_10 is 100 e^0.6, about 182.21.

   The mean is S_0 e^(μT) = 100 e^0.8, about 222.55. The mean sits above the median because the lognormal has a long right tail.

   The gap of σ^2/2 per year is the volatility drag of module 10 of the course.

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3. 03The hedging argumentHardAbout 15 min

   A stock follows dS = μS dt + σS dW, and money in the bank grows at the rate r. Derive the partial differential equation for the price V(S, t) of an option by holding the option and a short position of Δ shares. Then check it with numbers for a call with S = 100, K = 100, r = 0.03, σ = 0.20 and one year left.

   Hint

   Apply Ito’s rule to V. Choose Δ so that the dW terms cancel. A portfolio without risk has to earn the bank rate.

   Answer

   The equation is V_t + (1/2)σ^2 S^2 V_SS + rS V_S - rV = 0. For the call the closed form price is about 9.41, and its time decay, gamma and delta satisfy the equation up to rounding.

   Solution

   Ito’s rule gives dV = (V_t + μS V_S + (1/2)σ^2 S^2 V_SS) dt + σS V_S dW.

   The portfolio Π = V - ΔS changes by dΠ = dV - Δ dS. With Δ = V_S the dW terms and the μ terms cancel, which leaves dΠ = (V_t + (1/2)σ^2 S^2 V_SS) dt.

   This change has no randomness, so it must equal the bank return rΠ dt = r(V - S V_S) dt, or else there is an arbitrage.

   Setting the two equal gives V_t + (1/2)σ^2 S^2 V_SS + rS V_S - rV = 0. The drift μ has dropped out.

   For the call, the closed form gives V ≈ 9.413, V_S ≈ 0.599, V_SS ≈ 0.0193 and V_t ≈ -5.380. Then -5.380 + 0.5 × 0.04 × 10,000 × 0.0193 + 3 × 0.599 - 0.03 × 9.413 is 0 up to rounding.

   from math import erf, exp, log, sqrt
   
   def call(S, K=100, r=0.03, s=0.2, T=1.0):
       N = lambda x: 0.5 * (1 + erf(x / sqrt(2)))
       d1 = (log(S / K) + (r + s * s / 2) * T) / (s * sqrt(T))
       return S * N(d1) - K * exp(-r * T) * N(d1 - s * sqrt(T))
   
   S, h, r, s = 100.0, 0.01, 0.03, 0.2
   V = call(S)
   V_S = (call(S + h) - call(S - h)) / (2 * h)
   V_SS = (call(S + h) - 2 * V + call(S - h)) / h ** 2
   V_t = -(call(S, T=1 + 1e-5) - call(S, T=1 - 1e-5)) / 2e-5
   print(round(V, 3), round(V_t + 0.5 * s * s * S * S * V_SS + r * S * V_S - r * V, 6))

   Mark as done

Compare your own answer with the worked solution.