**Simple return:** Suppose an asset price moves from $P_0$ to $P_1$ in the course of one period. The simple return $R$ over that period is the percentage change: $ R=\frac{P_1-P_0}{P_0}=\frac{P_1}{P_0}-1 $ **Log return:** The log return $r$ is defined as the natural logarithm of the price ratio $\frac{P_1}{P_0}$. $ r=\ln \left(\frac{P_1}{P_0}\right)$ To set the log return $r$ into relation with the simple return $R$ we can substitute $(1+R)$ for the price ratio. By exponentiating we can also get the invers relationship of simple return in terms of log returns. $ \begin{align} r&=\ln(1+R) \\[2pt] e^r&=1+R \\[2pt] R &= e^r-1 \end{align} $ **Advantages of Using Log Returns:** - Convenient for mathematical modeling because log returns are additive over time. In contrast, simple returns need to be multiplied sequentially (order matters). - Under the assumption that financial returns are following a [[Lognormal Distribution]], the $r$ in the exponent follows a [[Gaussian Distribution]]. ## Taylor Series Expansion By applying the natural logarithm on the price ratio, we are effectively changing our return measure. To demonstrate when these two measures are visibly different, we apply [[Taylor Series#Taylor Series Expansion|Taylor Series Expansion]] to the simple return. Taylor Series for a general function $g(x)$: $ g(x) = g(c) + g^\prime(c)(x-c) + \frac{g^{\prime\prime}(c)}{2!}(x-c)^2 + \frac{g^{\prime\prime\prime}(c)}{3!}(x-c)^3 + \dots $ Taylor Series for the simple return function $(e^r-1)$: $ \underbrace{e^r-1}_{R}=\left (1+r+\frac{r^2}{2}+\frac{r^3}{6}+ \cdots \right)-1 $ When $r$ is small, then the higher order terms like $\frac{r^2}{2}, \frac{r^3}{6}$ can be neglected. In such cases we can approximate that $R \approx r$. $ \begin{align} R= e^r-1 &\approx (1+r)-1 \\ e^r-1 &\approx r \end{align} $ ## Continuous Compounding The log return also reflects the idea of continuous compounding. **Ordinary Compounding:** When you earn a simple annual interest rate $i$, but it is compounded $n$ times per year, then your initial investment effectively grows by: $ \left(1+\frac{i}{n}\right)^n$ **Continuous Compounding:** When $n \to \infty$, the limit converges to an exponential function. We say that the compounding interval gets continuous. $ \lim_{n \to \infty} \left ( 1+ \frac{i}{n}\right)^n= e^i $ **Derivation of Continuous Compounding:** To prove that the limit tends to $e^i$, we can use [[Combining Limits#Continuous Mapping Theorem|Continuous Mapping Theorem]] by applying a function (natural logarithm) on both sides. $ \begin{align} \lim_{n \to \infty} \ln \left ( 1+ \frac{i}{n}\right)^n&= \ln(e^i) \\[2pt] \lim_{n \to \infty} n*\ln \left ( 1+ \frac{i}{n}\right)&= i \end{align} $ We can see that as $n$ grows large, the factor outside the logarithmic function increases, while fraction $\frac{i}{n}$ shrinks, pulling the $\ln$ function closer to $0$. To be more precise, when $n$ is large, we can take the following approximation: $ \ln\left(1+\frac{i}{n}\right) \approx\frac{i}{n}$ Thus the whole limit expression can be written as an approximation: $ \lim_{n \to \infty} n*\frac{i}{n} \approx i$