Probability of geometric Brownian motion

Question

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?

Answer 1

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 $$ P(Z>\log(8.40)).$$

Answer 2

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 $\Phi(x)=\frac{1}{\sqrt{2\pi}}\int^x_{-\infty}e^{-\frac{1}{2}t^2}dt$.