I got a question and corresponding solution, but have some difficulties in understand the lognormal distribution part of it, so I really appreciate your advice:
Question: assume zero interest rate and a stock with current price at 1 dollar that pays no dividend. When the price hits level H(H>0) for the first time you can exercise the option and received 1 dollar. What is this option worth to you today?
Solution: since the stock price follows a geometric Brownian motion under risk-neutral measure $dS = rSdt+σSdW(t)$. Since r=0, $dS = σSdW(t)$, so $d(lnS)=-0.5*σ^2dt+σdW(t)$ When t=0, we have $S_0=1,ln(S_0)=0$. Notice that S is a martingale under the risk-neutral measure, but $lnS$ has a negative drift.
Here is my first doubt: if I understand correctly, the reason why $dlnS$ has a negative drift while $dS$ does not have is because: dlnS is continuously compounded rate of stock price, due to continuously compounded feature, it takes into account volatility (or standard deviation), so its real drift should be subtracted by this volatility component, I'm wondering if my understanding is correct?
(continued from the solution) The reason is that $lnS$ follows a normal distribution, but $S$ itself follows a lognormal distribution, which is positively skewed. As $T$ approaches positive infinity, although the expected value of $S_T$ is 1, the probability that $S_T>=1$ actually approaches 0. Here is my second doubt: how do we know the probability that $S_T>=1$ actually approaches 0?