# Relation between IV and SD

In option pricing, volatility naturally appeared through the Black-Scholes (BS) model where it was a coefficient for the linear diffusion term $\sigma S\,\mathrm dW_t$, and as such represented the time-scaled Standard Deviation (SD) of log-returns: $$\mathrm{Var}[\log S_t] = \mathrm{Var}[\sigma W_t] = (\sigma \sqrt t)^2.$$ In particular, if $T$ is maturity, the SD of the log-return on the initial investment $S_0$ is given by $\sigma \sqrt T$. Due to imprecision of the BS model, one deals with the BS Implied Volatility (IV) which fits the market prices of vanilla options with BS formula. The IV surface can thus have a pretty general shape, and we can compute the implied distribution of $S_T$ given the market prices of the option. I thus wonder, whether the SD of $S_T$ given by implied distribution can be computed in terms of IV surface, and whether it can happen that for some IV surfaces the SD is infinite.

Edit: just to clarify my question. Let us assume that the expiry $T$, interest rate $r$ and current underlying level $S$ are fixed. Denote by $\sigma(K)$ the IV smile at time $T$. If $C(K,\sigma)$ denotes the price of a European call with a strike $K$, volatility $\sigma$ and other parameters given above, then the implied distribution of $S_T$ has density $$f(K) = \frac{\mathrm d^2}{\mathrm d K^2}C(K,\sigma(K))$$ and hence the SD of $\log S_T$ is finite iff $$\int_0^\infty (\log x)^2f(x)\mathrm dx <\infty.$$ My question is whether there is a choice of $\sigma(K)$ which turns the latter integral to be infinite, but yet satisfies all necessary non-arbitrage conditions (non-negative vertical spreads etc.)

• Please write Implied Vol(IV) and Standard Deviation(SD) at least once in the question. Having too many acronyms is quite off putting.
– ash
Nov 18 '14 at 12:12

In my mind volatility (SD) of a stock and implied volatility (IV) are two quite different things:

• volatility is usually measured backwards looking. The common methods (empirical, GARCH, ..) look into the past. Measuring the risk of owning the stock in the future is often based on these backwards looking observations. We try to measure risk in the real world here.

• (BS-) implied volatility is some number that matches traded option prices with the pair moneyness/time to maturity. It is basically forward looking. You have different numbers for different levels of moneyness and time to maturity. Here we try to find implied parameters that live in the risk-neutral world

So to answer your question: In my mind the risk-neutral world and the "real" world can not be connected that easily. Concerning the implied distribution: there is research based on this. Basically you derive the risk-netrual distribution of the stock directly from traded option prices - not from implied vol. You can start your research here.

Concerning limit behaviour of geometric BM models: If maturity tends to zero, the option price is positive and the option is out of the money, then $\sigma$ tends to infinity. This fact is one of the critical points of this modelling approach. This is overcome by using linear BM models (such as the Bachelier model) or models with jumps.

• Thanks, I do understand the difference in how IV and SD/realized volatility are treated by practitioners. I also know how to get risk-neutral distribution through the option prices, which we can get through the IV surface as an example. I wonder, whether we can give such a surface that an SD of stock log return would be infinite. Nov 18 '14 at 11:48
– Ric
Nov 18 '14 at 12:57
• By $\sigma$ here you mean the IV or the SD (I'm interested in the latter as indicated in the OP)? It is clear that IV of the out of the money strikes with positive option prices tends to infinity, not sure whether the SD of log return in implied distribution does this. Nevertheless, in any case even if it tends to infinity, it is still likely it takes finite values. I wondered whether for some IV surfaces case implied distributions of log return with infinite SD, not just arbitrarily large SD. Nov 18 '14 at 13:32
• I understand now as you edited the question - but I can not really answer this.
– Ric
Nov 18 '14 at 14:11
• I see, thanks for the effort anyways - I also liked the example that you provided in the last paragraph regarding the expiration approaching. Can you still clarify, whether you meant high values of IV or of SD there? Nov 18 '14 at 14:17

Well, the implied volatility surface is given by (the Dupire equation)

$$\sigma^2(K, T) = 2 \frac{ \frac{\partial C}{\partial T} }{K \frac{\partial^2 C}{\partial K^2}}$$

so at least theoretically if either $K$ or $\frac{\partial^2 C}{\partial K^2}$ becomes very small the implied volatility can tend towards infinity. Whether or not you can find quoted call options with these properties is another question :)

Hope I did understand your question correctly.

• Not exactly, here you say whether IV goes negative. I rather wondered, whether we can construct an IV surface $\sigma(K,T)$ in such a way, that the corresponding implied distribution of the stock return has infinite variance. Nov 18 '14 at 11:42
• The Dupire equation gives the local volatility, and not the implied volatility. Each one can be recovered from the other, but they are not the same thing, except in particular cases, e.g. if both are constant. May 13 '18 at 19:14