Hedging an Inverse Price Exposure with a Linear Instrument
Summary
The note examines how to hedge an inverse-price product using a second instrument whose payoff changes linearly with the same underlying. It distinguishes a finite-period payoff, based on the change in the reciprocal of price, from a possible continuously valued inverse exposure. In both interpretations discussed, the inverse product’s value rises when the underlying price rises, so a short position in the linear product can offset that directional exposure.
For the continuous interpretation, the response differentiates the inverse value with respect to the underlying to identify its price sensitivity. It also raises the possibility that the stated finite-period expression is intended as an approximation to a log-price payoff, in which case volatility may matter. The question’s timing and payoff definitions are ambiguous, so the response does not establish a single hedge ratio; the correct quantity depends on the intended contract and exposure.
Key ideas
- An inverse-price exposure can increase when the underlying price rises.
- A linear instrument with the same underlying can offset that direction when held short.
- The inverse payoff’s sensitivity depends on the underlying price.
- Ambiguous timing and payoff definitions prevent a unique hedge ratio.
- A log-price interpretation may connect the finite-period expression to volatility.
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Full text
# Hedging an Inverse Product
# Hedging an Inverse Product
We have two different products that follow the same price $S(t)$ for all time $t$. The payout for product one is given by $w_1(\frac{100}{S(t)} - \frac{100}{S(t + \Delta t)})$ and the payout for product two is $w_2(S(t + \Delta t) - S(t))$. Where $w_1$ and $w_2$ are the quantities to buy or sell of product one and two respectively.
At time 0 given we buy $w_1$ of product one, to hedge against a change in price we should sell $w_2 = \frac{100w_1}{S(0)}$ units of product two? The only reason I am unsure is that the inverse product is in the inverse units of product two, so does that change how I should hedge product one with product two?
## Answer by kurtosis (score 1, accepted)
https://quant.stackexchange.com/a/57309
Discretized Time?
If we take $t=0$ and $t+\delta t=\tau$, you have payouts of: $$ \begin{align} \text{Payoff 1} &= \omega_1 \left(K_1 - \frac{100}{S_\tau}\right) \quad \text{and} \\ \text{Payoff 2} &= \omega_2 (S_\tau - K_2). \end{align} $$
Payoff 1 clearly increases with $S_\tau$ as does Payoff 2. (Imagine the underlier at 100 and then it goes up to 101: Payoff 1 would be $\frac{100}{100} - \frac{100}{101}>0$.) Therefore, to hedge Payoff 1 you would sell some of the contract offering Payoff 2.
Continuous Time?
However... I suspect the $t$ is, as you say, a time index which continues to increase. Then, Payoff 2 is clearly a "delta-1" investment. However, your discretized statement of Payoff 1 is very odd. So, expressing everything in continuous time (no more $t+\Delta t$), I suspect you mean that $$ \text{Payoff 1} = \omega_1 \frac{-100}{S_t}. $$ In that case, we can look at the derivative of the intrinsic value: $$ \frac{\partial \text{Payoff 1}}{\partial S_t} = \omega_1\frac{100}{S_t^2}. $$ In this case as well, you would sell some of the contract offering Payoff 2 to hedge holding the contract offering Payoff 1.
Log-Payout?
Finally, there is a slight chance this discretization is coming from considering a contract paying off $\log(\text{underlier})$, since the derivative of $\log(S_t)$ is $1/S_t$). If that is the case and your discretized payoffs were just a sort of Taylor Series approximation, then Payoff 1 will be related to the volatility of $S-t$.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.