Hi The solution to the SDE is: \(S_t = S_0 e^{0.275t + 0.5B_t}\) Therefore: \(ln(S_t/S_0) = 0.275t + 0.5B_t\) From the definition of standard Brownian motion, \(B_t \sim N(0,t)\), therefore \(0.5B_t \sim N(0,0.25t)\) and: \(ln(S_t/S_0) \sim N( 0.275t ,0.25t)\) Steve
Hi Steve Can you please explain how you get to 0.5Bt~N(0,0.25t) from Bt~N(0,t)? Can you also please explain how you get the mu parameter for the distribution of St? Thanks
Since Bt~N(0,t) then 0.5Bt will be normally distributed, we just need to know its mean and variance: E[0.5Bt]=0.5*E[Bt]=0 Var(0.5Bt) = 0.5^2 * Var(Bt) = 0.25t Therefore 0.5Bt~N(0,0.25t). Hope that helps.
Thanks Steve that is really helpful. I knew that I was missing something simple... Can you explain how, in this question, we get to S_t ~ logN(logS_0 + 0.275t, 0.25t)? I understand the variance part as you have explained above. Thanks
Hi If you're happy with \(ln(S_t/S_0) \sim N( 0.275t ,0.25t)\) from above, then we just need to note that: \(ln(S_t/S_0) = ln(S_t) - ln(S_0)\) Therefore we have: \(ln(S_t) - ln(S_0) \sim N( 0.275t ,0.25t)\) which implies that: \(ln(S_t) \sim N(ln(S_0)+ 0.275t ,0.25t)\) Let me know if that's not what you meant.
