Why a Driftless GBM Has a Unit Expected Price Ratio

# Modelling driftless stock price with geometric Brownian motion


# Modelling driftless stock price with geometric Brownian motion












I wish to understand some basic fact about the (primitive) simulation of stock prices with geometric Brownian motion.

If $S(t)$ is the stock price at time $t$, and the stock price follows geometric Brownian motion distribution, then it should satisfy $$dS(t) = S(t)\left(\mu dt + \sigma d B(t)\right)$$ where $B(t)$ is a standard linear Brownian motion, and $\mu$ is sometimes called drift. Solving this for $S(t)$ gives $$S(t) = S(0)\cdot \exp\left(\sigma B(t) + \left(\mu-\frac{\sigma^2}{2}\right)t\right)$$

Assuming there is no drift (there is no trend, perhaps), we obtain the following:$$S(t) = S(0)\cdot \exp\left(\sigma B(t) -\frac{\sigma^2}{2}t\right)$$ hence $$\ln\frac{S(1)}{S(0)} \sim \mathcal{N}(0,\sigma^2)-\frac{\sigma^2}{2}$$ and in particular \begin{eqnarray*} \mathbb{E}\left(\ln\frac{S(1)}{S(0)}\right) &=& \int_{-\infty}^\infty x\cdot\mathbb{P}\left(\mathcal{N}(0,\sigma^2) = x +\frac{\sigma^2}{2}\right) dx\\ &=& \int_{-\infty}^\infty x \cdot \frac{1}{\sigma\sqrt{2\pi}}\cdot\exp\left(-\frac{\left(x+\frac{\sigma^2}{2}\right)^2}{2\sigma^2}\right) dx \end{eqnarray*}

It looks like $$\mathbb{E}\left(\ln\frac{S(1)}{S(0)}\right) = -\frac{\sigma^2}{2}$$ which is what I would have expected. However, the more interesting expectation I couldn't compute: \begin{eqnarray*} \mathbb{E}\left(\frac{S_1}{S_0}\right) &=& \int_{-\infty}^\infty x\cdot \mathbb{P}\left(\exp\left(\mathcal{N}(0,\sigma^2)-\frac{\sigma^2}{2}\right)=x\right)dx\\ &=& \int_{-\infty}^\infty x\cdot\mathbb{P}\left(\mathcal{N}(0,\sigma^2)=\ln{x}+\frac{\sigma^2}{2}\right)dx\\ &=& \int_{-\infty}^\infty \frac{x}{\sigma\sqrt{2\pi}}\cdot \exp\left(-\frac{\left(\ln{x} + \frac{\sigma^2}{2}\right)^2}{2\sigma^2}\right)dx = ? \end{eqnarray*}

I would expect that expectation to be 1, as there is no drift. However, I can't do the integration. Is that true to assume that the result of the integral is 1?

## Answer by ajc3 (score 4, accepted)

https://quant.stackexchange.com/a/11534

To solve the expectation directly, you need to remember that a density function is not the same as the probability of the event.

We have, $\frac{S_1}{S_0} \sim \ln \mathcal{N} \left(-\frac{\sigma^2}{2},\sigma\right)$, therefore,

\begin{eqnarray} \mathbb{E}\left(\frac{S_1}{S_0}\right) &=& \int_{-\infty}^\infty x\, f_{\frac{S_1}{S_0}}(x)dx\\ &=&\int_{-\infty}^\infty x \, \frac{1}{x\sqrt{2\pi}\sigma}\exp\left[-\frac{\left(\ln x+\frac{\sigma^2}{2}\right)^2}{2\sigma^2}\right]dx. \end{eqnarray} The usual substitution method aims to manipulate the expression into looking like a standard normal. To do this we use $z=\frac{\ln x + \frac{\sigma^2}{2}}{\sigma}-\sigma$ and we get the substitutions $dx=x \sigma dz$ and $x=\exp(\sigma z + \frac{\sigma^2}{2})$.

\begin{eqnarray} \mathbb{E}\left(\frac{S_1}{S_0}\right) &=& \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi} \sigma} \sigma \exp\left(\sigma z + \frac{\sigma^2}{2}\right) \exp\left( \frac{-z^2 - 2\sigma z -\sigma^2}{2} \right)dz\\ &=&\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{z^2}{2}\right) dz\\ &=&1. \end{eqnarray}

## Answer by wsw (score 1)

https://quant.stackexchange.com/a/11527

If $S_t = S_0 e^{(\mu-\sigma^2/2)t + \sigma W_t}$, we can compute $$\mathbb{E}^Q\left[S_T\middle\vert \mathcal{F}_0\right] = S_0 e^{r T} = \text{forward price of } S_T \text { at time } 0. $$ To show the details, $\mathbb{E}^Q\left[S_T\middle\vert \mathcal{F}_0\right] = S_0 e^{(r-\sigma^2/2) T} \mathbb{E}^Q\left[e^{\sigma W_T} \middle\vert \mathcal{F}_0\right]$. To compute $\mathbb{E}^Q\left[e^{\sigma W_T} \middle\vert \mathcal{F}_0 \right]$, note that, for $X \sim \mathcal{N}(\mu,\sigma^2)$, $\mathbb{E}\left[e^X \right]=e^{\mu+\sigma^2/2}$.

In your case, since $r=0$ and $T = 1$, $$\mathbb{E}^Q\left[\frac{S_1}{S_0}\middle\vert \mathcal{F}_0\right] = 1. $$