Hull states that option prices increase with an increase in volatility.

I think that statement could be false in a specific scenario: when we are considering a deeply in-the-money European put option.

Since we are deeply in-the-money, the price of the underlying would be close to zero. Since the price of the underlying can't be negative, the effect of volatility would be asymetric: it would be more likely for the share price to recover than to fall anymore, simply because there isn't a lot of scope for a fall to happen from an already near-zero share price.

So a higher volatility is more likely to lead to a recovery of the share price, reducing the payoff, in turn reducing the price of the European put.

Is my reasoning wrong? Thanks in advance!


If you hold an option, you're always vega long, i.e. if volatility increases, your position increases as well - regardless of moneyness and the option type (put or call). Note firstly that by the model-free put-call parity, put and call options have the same vega (i.e. changes in volatility affect put and call prices in an identical way).

Let now $K\gg S_t$, then your put option is deep ITM but a corresponding call option would be deep OTM and what about the logic ''the call has nothing to lose and can only win, so increasing volatility should increase the call price'' but that would then also imply increasing put prices.

In the Black-Scholes model, \begin{align*} \mathrm{Vega} &= S_te^{-qT}\varphi(d_1)\sqrt{T-t} \\ &= Ke^{-r(T-t)}\varphi(d_2)\sqrt{T-t} \end{align*} which is always positive. Here, $\varphi$ denotes the probability distribution function of a normally distributed random variable.

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    $\begingroup$ (1) I do realize call prices actually increase as volatility increases. (2) I do realize the put-call parity implies an identical vega for puts and calls. (3) But I would like to know how the reasoning in the question details is internally inconsistent. Thank you for a wonderful answer! $\endgroup$ – Dhruv Gupta Aug 2 '19 at 5:03
  • $\begingroup$ Number one: If you’re question was right, then OTM call prices decrease when volatility increases which is not true. Furthermore, volatility is (in the BSM model) proportional to the stock price, i.e. if $S_t$ is low, then (even if $\sigma$ is high), the stock price will not much that much. So there is no real „probability of recovery“. Quite the opposite, it is possible that the stock price drops further which would increase the value of your put. Finally, your reasoning explains why deep ITM puts may have a positive theta and a negative time value. $\endgroup$ – Kevin Aug 2 '19 at 6:33

Maybe it will help your intuition if you think in terms of log-moneyness $\ln S/K$ instead of $S/K$. Let's look at a `deep' in the money put $K=100, S=10$. That sounds really deep in the money, but the value of log-moneyness for this situation is only $-2.3$, which is not that much if you consider the possible range of $\ln S/K$ is $(-\infty,\infty)$. So when you say there isn't much scope for the share price to fall further that isn't really true. Log-moneyness is a symmetric measure of moneyness given that $S$ is log-normal, and is in a sense a more correct measure of moneyness.


I just want to add a simple piece to this reasoning, that is very intuitive and not excessively mathematic, since the mathematic explanation has already been given in the other answers (I like to base my mathematical understanding on logic intuitive reasoning).

Just consider what a put option is: it is a contract to sell at the strike K and buy at the final price P of the underlying. Consider your percentage return on a put, which is K/P-1. When P is close to 0 but it is not yet 0, your percentage return is high but still discrete because at maturity you will have to pay P. Assume now that P is exactly 0 (which in practice means that the company will never recover and your contract will be paying for sure at maturity T, or maybe will be terminated even sooner if in real practice that are clauses in certificates allowing to terminate the contract in case of special events like defaults or restructuring). This means that your return will be K/0 which is infinite. Clearly, the percentage difference between K/P1 with a low P1 and K/P2 with P2=0 is still big even when P1 is already low (it is the difference between a high real number and infinite).. this may help you in 2 words with no mathematics justify the fact that the Vega remains positive.

the opposite would be true for example in the unlikely case of a barrier put option that has a clause according to which the option pays K-P at maturity, but if P=0 (Knock-out barrier is touched) then the option pays nothing (I.e. an option that protects the issuer from an extreme scenario on the underlying), then yes, for some low P, the Vega would become negative.


We can also think in a more discrete way.

Lets take for example a put option with $5 strike.

If the underlying asset has 50% chance of valuing 2 and 50% of valuing 4, the price of the asset would be 3 and the value of put would be (3*0,5 + 1*0,5) =2

Now, assume an increase in volatility and asset can value 1 or 5 with same probabilities... Asset price continue to be 3 and put values 2,5 (4*0,5 + 1*0,5)

Now to the extreme, asset can value 0 or 6 with same probability. Asset still 3 but due to higher volatility, put values 3 (6*0,5 + 0*0,5).

So, you are capturing a positive vega effect.

Now, assume a new scenario in which underlying asset values either 0 or 7 with 50% probability. We can see a volatility increase, but put still values 3 (6*0,5 + 0*0,5). Still, it does not mean you are not long vol. Since the underlying asset can now value 0 or 7, it should be priced at 3,5 and thus an increase in price in the asset should decrease the value of your put. The fact that it didn't, is because although you are short delta, you are still long vol and it offset the delta effect.

The "intuition" of short vega when deep in the money on puts comes from the idea that prices have more space and probability to go up then down, in which case the price of the underlying should be higher than it is, which is a delta effect.


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