# Black Scholes model (geometric Brownian) [closed]

I am asked the following question:

In the Black-Scholes model, one assumes that the stock price $S_t$ follows a geometric Brownian motion, i.e. $S_t = S_0 \exp\left(\left(\mu − \frac{\sigma^2}{2}\right)t + \sigma B_t\right), t \ge 0$ is time (in years), where $B_t$ is standard Brownian motion, $\mu$ is the drift, and $\sigma > 0$ is the volatility of the stock. If $\mu = 0.02, \sigma = 0.25$ and the stock price at time $0$ is $S_0 = 30$. Then, I am asked to determine the probability that the stock price exceeds $35$ after half a year as well as to determine the probability that the stock price is outside the interval $[30,35]$ after $2$ years.

Although we've seen the basic model, I wonder how to apply it when we consider a drift.

## closed as off-topic by Did, Namaste, marwalix, zhoraster, RohanDec 20 '16 at 6:26

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The question leads to a simple calculation: The probability that the stock price exceeds 35 after half a year is $$P(S_{0.5} \ge 35)$$. But you can easily transform a probability regarding $S_t$ to a probability regarding $B_t$ by: \begin{align*} P(S_t \ge c) &= P\left( S_0 \exp\left(\left(\mu − \frac{\sigma^2}{2}\right)t + \sigma B_t\right) \ge c\right) \\ &= P\left(B_t \ge \frac{1}{\sigma}\left(\ln\frac{c}{S_0} + \left(\frac{\sigma^2}{2} - \mu\right)t\right)\right)\end{align*}
But $B_t$ is a normal with mean $0$ and variance $t$ so you can use the distribution function for a normal.
Take $c=35$, plug in your given values and you are done…