Interpreting Price Elasticity of Return and Leverage
Summary
The note derives the price elasticity of an investment’s return when return is defined relative to initial wealth. With initial wealth normalized to one, return is price minus one, so elasticity is price divided by price minus one. The apparent mismatch with leverage comes from treating elasticity as a fixed leverage multiple: it instead measures the local percentage response of return to a percentage change in price at a particular price level.
An example compares equal one-unit price increases at different starting prices. The absolute return gain is the same, but its percentage relative to the existing return shrinks as the price rises. Thus the calculated elasticity falls as the investment appreciates. The explanation clarifies the distinction between a local sensitivity measure and leverage as commonly used for exposure relative to capital; it does not develop a general leveraged portfolio model or discuss financing costs and losses.
Key ideas
- Return elasticity measures the local percentage sensitivity of return to price.
- With initial wealth set to one, the elasticity is price divided by price minus one.
- Equal absolute price changes can represent different percentage changes in return depending on the starting price.
- The elasticity declines as price rises, so it should not be read as a fixed leverage multiple.
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# How to reason about leverage in terms of elasticity
# How to reason about leverage in terms of elasticity
Return of an investment for a given period is by definition: $$r = \frac{P}{W_0} - 1$$ where $P$ is the price of the investment at the end of the period, and $W_0$ is the initial investment. I want to understand how much the return changes in terms of percentage, with respect to that of the price. This naturally is the definition of elasticity.
Elasticity can be calculated as: $$\epsilon = \frac{\partial{r}}{\partial{P}} \cdot \frac{P}{r} = \frac{P}{P-W_0}$$
Now if we have a leverage of 2x, we should have a elasticity of 2. However, I do not seem to be able to connect these two concepts. If $W_0$ is normalized to 1, then the formula becomes: $$\epsilon = \frac{P}{P-1}$$
When $\epsilon = 2$, this implies that P = 2. However, this says that only when $P$ is twice that initial investment $W_0$ that the leverage is 2. When price is higher, the leverage decreases.
Where did I get the logic wrong?
## Answer by Marco Breitig (score 1, accepted)
https://quant.stackexchange.com/a/11213
Your math is right. If we normalize $W_{0}=1$, we have a return of $$ r\left(P\right) = P-1 $$ and a price-elasticity of return of $$ \epsilon\left(P\right) = \frac{P}{P-1} \mathrm{.} $$ If your price $P$ goes from its initial value $W_{0}=1$ to $P=2$, you make a return of $r\left(2\right) = 1$. If your investment has a final price of $P=3$, your return would be $r\left(3\right) = 2$. So when you go from a price of $P=2$ to a price of $P=3$, your return goes from $r\left(2\right) = 1$ to $r\left(3\right) = 2$. For a price of $P=4$ your return wold be $r\left(4\right) = 3$. So if you go from a price $P=3$ to a price of $P=4$ (note that the price rises again $1$ unit), your return only increases from $r\left(3\right) = 2$ to $r\left(4\right) = 3$. The first time your return doubled, the second time you only made $50\%$ more return.
That is what the equation says. You look at relative the local change of return due to price changes (the derivative). This is always constant. But it makes a huge difference for your gain in return if your price goes from $1$ to $2$ or if it goes, say, from $101$ to $101$.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.