Long Put Vega Exposure and Delta Hedging
Summary
The document explains how implied volatility affects option value independently of the underlying price. In the Black–Scholes framework, both puts and calls have positive vega, so a long put can gain value when implied volatility rises even if the underlying remains unchanged. A rise in volatility does not itself imply a fall in the underlying, however, and an upward move can reduce a put’s value through its negative delta.
To isolate volatility exposure, the answer describes delta hedging: combine the put with an underlying position that offsets its delta. This creates a delta-neutral position whose short-term value is more responsive to volatility changes. The hedge is only local; larger underlying moves introduce gamma effects and other risks, so the position may need further adjustment. The explanation is conceptual and model-based, with no empirical evidence or discussion of trading costs and hedge frequency.
Key ideas
- Both calls and puts have positive vega in the Black–Scholes framework.
- A long put can benefit from rising implied volatility even when the underlying price is unchanged.
- A rising underlying price can offset the vega gain through the put’s negative delta.
- Delta hedging uses an underlying position to reduce first-order exposure to price moves.
- Delta neutrality is local because larger moves bring gamma and other effects into play.
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Full text
# Can a long put trade be profitable through Vega even if the underlying moves upwards?
# Can a long put trade be profitable through Vega even if the underlying moves upwards?
Generally speaking, I know when implied vol increases, option prices increase for calls.
However, does the same occur for puts?
If I am expecting implied volatility to increase for an option on an underlying asset (let's say a stock) and I believe the price will decline as implied vol rises, would the best strategy be to buy a put, as opposed to buying a call (forget strangle/straddles for the moment being)?
Is it possible for a long put position to be profitable if the gain on vega, due to increase in implied vol on either the upside or downside, is larger enough to offset the short delta position?
## Answer by SRKX (score 3, accepted)
https://quant.stackexchange.com/a/7326
First, notice that the two greeks you mentioned in your question are simply the partial derivatives of the value of the option $V$ with respect to two different variables $S$ (the price of the underlying) and $\sigma$ (the volatility of the underlying):
$$\Delta = \frac{\partial V}{\partial S} \quad \text{and} \quad \nu=\frac{\partial V}{\partial \sigma}$$
As mentioned in the comments, volatility is not per-say a sign of declining prices, but, at least under the Black-Scholes model, $\nu$ is positive for both puts and calls meaning that the value of both types of options is expected to rise if $\sigma$ goes up.
So, from a mathematical point of view, if only $\sigma$ goes up (i.e. $S$ stays the same) then yes the trade would be profitable even by buying a put option. However, it is fair to say that this situation is not very realistic and that in real-world you would be exposed to wild moves in $S$ which would affect the price of the option through the $\Delta$.
As a result, what you would like to do is to is to perform delta hedging which consists in offsetting your exposure to $S$ by buying or selling an amount of the underlying $S$ corresponding to $\Delta$. In the case of the put, you know that $\Delta<0$ (i.e. the price of the put $V$ decreases as the price of the underlying $S$ increases) and you hence need to buy $\Delta$ of $S$ to make your resulting portfolio have a $\Delta=0$: it is delta-neutral.
Once you've done that, you've removed your exposure to changes in $S$ and you could consider that your put position is mainly determine by the remaining $\nu$. Some funds such as the Amundi Volatility Funds do exactly that.
This reasoning though is really perfectly working only for small changes in $S$, as bigger changes would also be sensitive to other greeks such as $\Gamma = \frac{\partial^2V}{\partial S^2}$. You can also hedge against this kind of move using a similar reasoning.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.