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?


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

  • $\begingroup$ I get $0.133526$, do you agree with me? By the way, do you know a good resource that explains the idea behind this calculation on a basic level? Thanks for your time $\endgroup$ – BillyJean Sep 9 '17 at 21:15
  • $\begingroup$ @BillyJean No I think you made an error (remember variance vs std deviation). What part? The drift part, the figuring out the mean and variance part or the computing $P(Z>z)$ part? Sorry don't have a good reference on hand. The last part will be covered in any probability theory or statistics book. The second part is applying the basic properties (or the definition depending how you look at it) of brownian motion... any reference that covers Brownian motion should tell you the distribution at time $t.$ For the first part perhaps look for some lecture notes on stochastic processes for finance. $\endgroup$ – spaceisdarkgreen Sep 9 '17 at 22:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.