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?

  • 1
    $\begingroup$ not a very clear question... are you asking about the relationship of implieds to delta moves.. look up sticky delta and sticky strike.. $\endgroup$
    – cdcaveman
    Feb 18, 2013 at 5:18
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    $\begingroup$ @jessica, your question is indicating that you are not affiliated with this industry and thus other boards should be serving you better. You can easily google what the impact of changes in implied vol is on option prices. Your question contains logical errors (a: there is no iVol for a stock, only iVol of an option which is linked to the expected future volatility of the underlying asset's price returns, b: price does not always decline when iVol increases, c: a long put position benefits from generally increasing iVol and short delta exposure when prices of the underlying decrease. $\endgroup$
    – Matt Wolf
    Feb 18, 2013 at 6:03
  • $\begingroup$ "c3: a long put position makes money on both, increasing iVol (if it indeed inceases) and your short delta exposure, there is no offset." hmm I see. Though isn't it more often than not that iVol and the delta position for a put are going in opposite directions? $\endgroup$
    – jessica
    Feb 18, 2013 at 6:08
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    $\begingroup$ @jessica, iVol is not necessarily positively correlated with the direction of the underlying. That is why I said "generally" (and that does not even apply to certain assets, for example iVol of options on agricultural futures ticks "generally" up when the futures price increases). Re delta, you need to become clear what your delta exposure is: Long put = short delta, short call = short delta,...you then know exactly what your delta impact will be from rising or falling underlying asset prices. $\endgroup$
    – Matt Wolf
    Feb 18, 2013 at 6:24
  • $\begingroup$ It's worth noting that the conditions to make such a trade profitable were very visible during the GameStop short squeeze in January of 2021 and the second squeeze in February of 2021. Simply by moving upwards in px, IVs increased substantially, and both sides of the options chain gained in value. The opposite was true as the underlying fell $\endgroup$ Jun 4, 2022 at 19:56

1 Answer 1


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.

  • $\begingroup$ Sigma is not the sign for Vega or implied volatility.. its the sign for standard deviation $\endgroup$
    – cdcaveman
    Feb 20, 2013 at 8:54
  • $\begingroup$ @cdcaveman I use $\nu$ for Vega and $\sigma$ for the volatility inputted in the pricing model. $\endgroup$
    – SRKX
    Feb 20, 2013 at 11:28
  • $\begingroup$ @cdcaveman Um, you do realize that volatility and standard deviation are synonymous in Black Scholes, right? $\endgroup$ Feb 20, 2013 at 12:24
  • $\begingroup$ >In the case of the put, you know that Δ<0 (i.e. the price of the put V decreases as the price of the underlying S increases) and you hence need to buy Δ of S to make your resulting portfolio have a Δ=0 @SRKX: Sorry but I think you're wrong. If the prices goes up & the put price decrease, then you should sell stock. $\endgroup$
    – dns
    Nov 7, 2014 at 18:25

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