Implied Volatility and the Tails of the Risk-Neutral Distribution

# Relation between IV and SD


# 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.)

## Answer by Richi Wa (score 3)

https://quant.stackexchange.com/a/15515

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.

## Answer by DoubleTrouble (score -1)

https://quant.stackexchange.com/a/15514

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.