Why a Sum of Correlated Geometric Brownian Stocks Is Not Geometric Brownian Motion

# Correlated stock prices and geometric Brownian motion


# Correlated stock prices and geometric Brownian motion












I have two uncorrelated stocks which follow geometric Brownian motion, as follows

$$\begin{aligned} dS_a &= \mu_aS_adt + \sigma_aS_adW\\ dS_b &= \mu_bS_bdt + \sigma_bS_b dW \end{aligned}$$

Does a portfolio of these stocks also follow geometric Brownian motion?

I have determined that

$$dS_a + dS_b = (\mu_aS_a + \mu_bS_b)dt + (\sigma_aS_a + \sigma_bS_b)dW$$

which does not follow geometric Brownian motion. I'm stuck now on what happens if the $S_a$ and $S_b$ are correlated? How does this change?

## Answer by Sesame (score 1)

https://quant.stackexchange.com/a/44547

You posed a quiet vague question but I will try to reply to it. Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space. We denote $X$ the portfolio of two stocks which follow geometric brownian motions, i.e. for all $t \in \mathbb{R}^+$, \begin{align*} X_t = S^a_t + S^b_t \end{align*} where $S^a_t$ and $S^b_t$ have the following $SDE$: \begin{align*} dS^{x}_t = \mu^x_tdt + \sigma^x_tdW^x_t \end{align*} where $x = \lbrace{a,b\rbrace}$, $\mu^a \neq \mu^b$ and $\sigma^a \neq \sigma^b$. Suppose that $W_t^a$ and $W_t^b$ are two correlated brownian motions, i.e. $d<W^a, W^b>_t = \rho dt$. By applying the Ito formula to the function $\phi(x,y) = x+y$, we have : \begin{align*} dX_t &= dS_t^b + dS_t^a \\ &= (S_t^a\mu^a_t + S_t^b\mu^b_t)dt + S_t^a\sigma^a_tdW^a_t + S_t^b\sigma^b_tdW^b_t \end{align*} As you can see, we supposed that the portfolio is a linear function to the stocks (i.e. sum of the stocks). Thus, the correlation will not change anything as the second derivative of the function $\phi$ is zero. So the portfolio is not a geometric brownian motion.

Now, if we want to be more general we can suppose that the portfolio is a function of the stocks and smooth enough to apply the Ito formula (we can first suppose $\mathcal{C}^2(\mathbb{R}_+^2,\mathbb{R}_+)$. We have then: \begin{align*} dX_t = d\phi(S_t^a, S_t^b) = \partial_x\phi dS_t^a + \partial_y\phi dS_t^b + \frac12\left[\partial_{xx}\phi <S^a>_t + \partial_{yy}\phi <S^b>_t + 2\partial_{xy}\phi <S^a,S^b>_t\right] \end{align*} Then we can try to find $\phi$ such that $X$ is a geometric BM.