Merton problem: can the stock price keep rising?

Question

I read that the stock price, $S(t)$ of the famous Merton model is given by the following differential equation $dS(t) = µS(t)dt + σS(t)dB(t).$

I gather that this is geometric Brownian motion. A path of a geometric Brownian motion keeps growing as in the Wikipedia figure.

My question are:

Does stock prices too keep growing with a positive trend? (I do not see this is obvious)

If not what is the rational for modeling stock price by geometric Brownian motion?

Can this stock price go negative?

Thanks

Answer 1

Using Ito's Lemma we can find the exact solution of this SDE

$S(t) = S(0)\exp[(\mu-\frac{1}{2}\sigma^2)t]\exp(\sigma B(t)) $

where $B(t)$ is a normally distributed random variable with mean $0$ and variance $t$.

The expected value of $S(t)$ is

$E[S(t)] = S(0) \exp (\mu t)$.

Hence, if the drift is positive, the expected value of the stock price will grow exponentially with time, but the random path will fluctuate around this trend with variance that grows approximately linearly in time:

$var[S(t)] = S^2(0) \exp (2\mu t)[\exp(\sigma^2t)-1)] = O(\sigma^2t)$

From the solution, you can see that the stock price can never be negative -- one reason why GBM is popular for modelling stock prices.

In practice, more elaborate models that account for random jumps, stochastic volatility, etc. are often used.

Answer 2

Note that you can re-write the equation as $$dS(t)/S(t) = µ dt + σ dB(t)$$ and then, $dS(t)/S(t) = d(\log S(t))$. So $\log(S(t))$ follows a random walk, but $S(t) = \exp(\log(S(t)))$ is always positive.

The "trend" term $\mu$ can be either positive or negative; declining stock prices do happen.

This geometric BM model is widely criticized, however, as an over-simplification. In particular, $\sigma$ does not stay constant.