Expected Return in Geometric Brownian Motion and Continuous Compounding
Summary
The document clarifies how a finite-horizon expected return relates to the instantaneous drift in a geometric Brownian motion. For a stock following a GBM without dividends, the expected simple return over a year is obtained from the expected price ratio, which equals the exponential of the drift minus one. Thus, a quoted annual simple expected return must be converted to the continuously compounded rate before identifying the GBM parameter.
The response also defines expected return over a short interval as the conditional expectation of the proportional price change; in the infinitesimal limit, this is the drift multiplied by the time increment. It notes that a stated return depends on its horizon and compounding convention. The example assumes no dividends and a constant-drift, constant-volatility GBM, so its interpretation does not automatically cover other return definitions or models.
Key ideas
- In a no-dividend GBM, the annual expected simple return is the exponential of the annual drift minus one.
- The instantaneous expected proportional price change over a small interval is approximately the drift times that interval.
- A quoted return needs a specified horizon and compounding convention.
- The relationship described assumes a constant-parameter GBM without dividends.
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# What Exactly is Expected Return
# What Exactly is Expected Return
Consider the following plot, courtesy of this page:
Regarding the $y$-axis, how does this "expected return" relate to the "instantaneous expected return" in a geometric Brownian motion (GBM)?
E.g., assume each stock price follows $dS(t) = \mu S(t) dt + \sigma S(t) dW(t)$, and so $S(t) = S(0)\exp\left(\left(\mu - \frac{\sigma^2}{2}\right)t + \sigma \sqrt{t} Z\right)$ where $Z \sim \mathcal{N}(0,1)$. Then I would calculate the (annual) expected return as $$ \mathrm{E}\left[\frac{S(1)}{S(0)} - 1\right] = \mathrm{E}\left[\exp\left(\mu - \frac{\sigma^2}{2} + \sigma Z\right)\right] - 1 = \exp\left(\mu - \frac{\sigma^2}{2} + \frac{\sigma^2}{2}\right) - 1 = e^\mu - 1, $$ where the second equality is from the moment-generating function of a normal random variable.
Take Portfolio A in the plot and suppose it's just a single stock, driven by the GBM above with instantaneous rate of return $\mu$. Portfolio A has an "expected return" of $8\%$. So, which of the following (if any) do we mean?
- $e^\mu - 1 = 8\%$
- $\mu = 8\%$
## Answer by fni (score 1)
https://quant.stackexchange.com/a/18981
Suppose we have no dividends like in Black-Scholes-Merton and in your example. Expected return between time $t$ and $t+\Delta t$ is defined as $$ \mathbb{E}_t\left[R_{t+\Delta t}\right]\equiv\mathbb{E}_t\left[\frac{S_{t+\Delta t} - S_t}{S_t}\right] = \mathbb{E}_t\left[\frac{\Delta S_t}{S_t}\right] $$ You can see that, as $\Delta t \to dt$, $\mathbb{E}_t\left[\frac{\Delta S_t}{S_t}\right] \to \mathbb{E}_t\left[\frac{dS_t}{S_t}\right] = \mu dt$.
Regarding your question, $\mu$ is the instantaneous rate of return so $e^\mu - 1 = 8\%$. Alternatively, you know that rate of returns should be defined for a certain timeframe (e.g. 4% per semester, 8% per year etc…) and with some compounding method (e.g. yearly compounding, quarterly compounding etc…). $\mu$ is the annual nominal expected interest rate obtained by continuous compounding.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.