I had a query on a question from the last sitting that might seem silly, but I'm unable to figure out why the solutions were approached in this manner. In the question below, would I be correct in reading it as: St = exp(mu^t + sigma^2*Wt) where, mu = 0.06875 sigma^2 = 0.25? Now if we say that the share price follows lognormal as: St ~ logN(ln(S0) +(mu-(sigma^2)/2)t,(sigma^2)*t) because, ln(St) - ln(S0) ~ Normal((mu-(sigma^2)/2)t,(sigma^2)*t) Then can I check why is it that in the solution for this part they show the 'mu' parameter of the lognormal model here to be the same as the drift parameter of the share price (mu defined above), and not ln(S0) +(mu-sigma^2/2)t which is the mu of the lognormal distribution above? I'm certain I'm missing something here? QUESTION The chief executive officer (CEO) of a company benefits from an executive reward plan that includes company shares currently worth €100,000. The shares currently trade at €1 each. The CEO wishes to retire in 4 years’ time and hopes the share fund value at that time will be at least a target value of €150,000. The share price St at time t (measured in years) follows the stochastic differential equation: St = exp(0.06875*t + 0.25*Wt ) where Wt is a Standard Brownian Motion. The ‘surplus amount’ is defined as the difference between the share fund value in 4 years’ time and the CEO’s target value. (i) Calculate the standard deviation of the surplus amount. SOLUTION The properties of the lognormal distribution give us the variance of St: Var = E[St]^2 . {exp(0.25^2*t)–1} where E[St] = exp(0.06875*t+0.25^2*t/2) So E = 1.49182 And Var = 0.63211
Hi This looks like a parameterisation issue. Some questions start with the SDE for geometric Brownian motion: \( dS_t = S_t(\mu dt + \sigma dW_t) \) which can then be solved to find: \( S_t = S_0\exp((\mu-\frac{1}{2}\sigma^2)t + \sigma^2 W_t) \) Other questions (like this one) start with the solution: \( S_t = S_0\exp(\nu t + \sigma^2 W_t) \) which can be reverse-engineered to find the SDE: \( dS_t = S_t((\nu +\frac{1}{2}\sigma^2)dt + \sigma dW_t) \) Notice that the \(\mu\) and the \(\nu\) are not the same thing. In fact, by comparing terms we have: \( \nu = \mu -\frac{1}{2}\sigma^2 \) In this question you don't need the SDE at all; just takes logs and observe that: \( \ln(S_t) = 0.06875t+0.25W_t \sim N(0.06875t, 0.25^2t) \)
