# Why do the prices of deep in-the-money options increase with volatility in the Black Scholes framework? [closed]

I can understand that volatility increases the value of an option when a stock is out/at the money. Then more volatility means the stock's distribution gets more upside without suffering a greater probability of ending out of the money.

But imagine that a stock was far in the money. If I have a distribution that has a low volatility and thus greater chance of ending in the money, if I was risk averse, I might prefer that to an option that has more upside but more downside as well. Using the same thought process, If I was risk neutral, I might be indifferent to each and thus they may have the same value to me.

Does this mean then that increasing volatility disproportionately increase the upside of the option relative to the increase in the probability of ending out of the money? Is there an intuitive explanation or a way to show this effect mathematically?

• Because of limited liability and convexity. Yes, very good and very bad events get more likely as volatility rises. But the option’s losses are limited, it thus benefits overall from increases in volatility. Two caveats: some barrier and compound options actually have negative vegas. Also, we’re talking of partial equilibrium (idiosyncratic) effects. If aggregate (or systematic) volatility rises, prices of vanilla (ITM) option can decrease! Feb 14 at 17:45
• I like your comment on convexity. Are you referring to the convexity of the density of the terminal GBM stick price? Do you think the asymmetrical shape of the lognormal distribution plays a part? Feb 14 at 21:11
• Thank you! I was referring to the convexity of the option payoff and the option value Feb 14 at 21:12
• So we think of our option payoff as the truncated first moment of the terminal distribution (from the strike to infinity). Why does this integral gain from greater volatility? Because, if it is in the money it is also making outcomes near the mean less probable and thus they have a lower payoff Feb 14 at 21:16
• Hence the argument about limited liability in my first comment Feb 14 at 21:16

First note that the sensitivity of price to implied vol must be qualitatively symmetric for out-of-the-money and in-the-money options, since by put-call parity an ITM call behaves the same as an OTM put.

Second, for options that are far from at-the-money (whether they are deep ITM or deep OTM) the price does not particularly increase with volatility! You can see this by plotting the vega (derivative of price wrt implied volatility) of an option against moneyness, where moneyness is defined in terms of forward price $$F$$, strike $$K$$, implied volatility $$\sigma$$ and time to expiry $$\tau$$

$$m= \frac{\log(K/F)}{\sigma \sqrt{\tau}}$$

Plotting vega against moneyness gives a chart like this You can see that for deeply ITM or OTM options (e.g. $$|m| > 3$$) the vega is practically zero, so the option does not materially increase in price as implied volatility increases.

• Hi Chris, I understand that the effect of increasing volatility is minimal for in-the-money options, what I would like to know is why it's positive. Specifically for in the money options. Why should raising volatility make my option more valuable if I am now being exposed to greater odds of ending out of the money? Feb 15 at 9:56
• Because the option payoff is convex (the point @Kevin made in their first comment). A simple example may help. Say the strike is 100, the current price is 115 and the stock is known to either expire at 105 or 125, both with 50% probability. Then the price should be about 0.5 x 25 + 0.5 x 5 = 15 and the chance of expiring OTM is 0%. Now assume volatility doubles so the stock will expire either at 95 or 135, again with 50% probability. The new price is 0.5 x 35 + 0.5 x 0 = 17.5 and the chance of expiring OTM is now 50%. The extra upside value more than compensates for the additional downside. Feb 15 at 10:46
• I see. Are there any mathematical proofs for how volatility increases the value of an option? Feb 15 at 13:51
• What do you mean by a mathematical proof? You could simply look at the sensitivity of Black-Scholes price to the implied volatility parameter (equal to $S \times \phi(d_1) \times \sqrt(T)$) and note that it is the product of three positive terms, which shows that volatility increases the value of an option in this model. Feb 15 at 15:59