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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?

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1 Answer 1

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 $101$.

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