Probability of geometric Brownian motion I want to solve the following question: Let $X(t)$ be the price of JetCo stock at time $t$ years from the present.  Assume that $X(t)$ is a geometric Brownian motion with zero drift and volatility $\sigma = 0.4/yr$.  If the current price of JetCo stock is 8.00 USD, what is the probability that the price will be at least 8.40 USD six months from now.
Here is my attempt: Since $X(t)$ is a geometric Brownian motion, $\log(X(t))$ is a regular Brownian motion with zero drift and $\sigma = 0.4 / yr$. We want to find the probability that $\log(X(1/2)) \geq \log(8.40)$ given that $\log(X(0))= \log(8.00)$.
What can I do from here?
 A: You are on the right track so far, but you have the drift wrong. The drift of $\log(X(t))$ is, from Ito's lemma, $-\frac{1}{2}\sigma^2.$
After that you use the fact that since it's a brownian motion starting from $\log(8.0)$ the distribution is $$\log(X(t)) \sim N\left(\log(8)-\frac{1}{2}\sigma^2t,\sigma^2t\right).$$ So let $Z$ be normal with mean $\log(8)-\frac{1}{2}\sigma^2\left(\frac{1}{2}\right)$ and variance $\sigma^2\left(\frac{1}{2}\right).$ To finish the problem, you need to compute $$ P(Z>\log(8.40)).$$
A: Since $X(t)$ is a geometric Brownian motion, we recall that $\log(X(t))$ is a regular Brownian motion with zero drift and $\sigma=0.4/\text{yr}$. As we want to know the probability that $\log(X(1/2))\geq \log(8.40)$ given that $\log(X(0))\geq\log(8.00)$, this means that
\begin{align*}
\log(X(1/2)-\log(X(0))\geq \log(8.40)-\log(8.00)&=\log\left(\frac{8.40}{8.00}\right)\\
&=\log(1.05)\\
&=0.0488\\
\end{align*}
In this case,
\begin{align*}
Z&=(\log(X(1/2))-\log(X(0))-(0)(1/2))/(0.4\sqrt{0.5})\\
&=(\log(X(1/2))-\log(X(0))/0.283\\
\end{align*}
is a standard normal random variable. Therefore, we just calculate the probability
\begin{align*}
\mathbb{P}(\log(X(1/2))-\log(X(0))>0.0488)&=\mathbb{P}\left(\frac{\log(X(1/2))-\log(X(0))}{0.283}>\frac{0.0488}{0.283}\right)\\
&=\mathbb{P}(Z>0.172)\\
&=1-\mathbb{P}(Z\leq0.172)\\
&=1-\Phi(0.172)\\
&=1-0.568\\
&\approx0.432\\
\end{align*}
where $\Phi(x)=\frac{1}{\sqrt{2\pi}}\int^x_{-\infty}e^{-\frac{1}{2}t^2}dt$.
