First, I can't find a purely "financial" explanation for this.
Also the only mathematical explanation I've found so far was using the large deviations theory, which is quite complex.
Is there a rather simple mathematical explanation ?
Thanks !
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1$\begingroup$ Are you talking about the smile observed on the market or the smile produced by a model? $\endgroup$– AFKFeb 22, 2015 at 15:45
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$\begingroup$ The implied volatily in the Black Scholes model. $\endgroup$– DarkFeb 22, 2015 at 20:46
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$\begingroup$ It was obvious that you implied the volatility by applying the inverse of the Black Scholes function to call prices. The question is which call prices? Are they market prices or are they prices produced by a model? $\endgroup$– AFKFeb 22, 2015 at 21:52
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$\begingroup$ Market prices. But I think I have found an explanation : the call price seen as a function of the volatility is bounded between $S_t - K*B(t,T)$ and $S_t$. (where $S_t$ is the spot price and $B(t,T)$ the price at time tof zero-coupon of maturity T.) When T increases, $B(t,T)$ decreases to 0. Hence the flattening. $\endgroup$– DarkFeb 24, 2015 at 8:22
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7 Answers
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The central limit theorem guarantees, under fairly general assumptions, that the sum of returns becomes more normally distributed as the number of returns grows (technically, defining a return as $\mathrm{log}(S_{t+\Delta t}/S_t)$, $\sum_i ^n \mathrm{log}(S_{t+\Delta t i}/S_{t+\Delta t (i-1)} \to \mathcal{N}(\cdot,\cdot)$ as $ n \to \infty $). Thus, as $T$ gets larger, the Black Scholes assumption of normally distributed log returns becomes more and more valid. This is exemplified by the flattening implied volatility smile.
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1$\begingroup$ I don't see how this could work: In a BS world you have infinitessimally many price paths aggregated at every time step - this is why you have, due to the CLT, a Gaussian distribution everywhere. I don't see how you could get a Gaussian that is even more Gaussian. Now, looking at the smile (and at real return data) we see that they are obviously not Gaussian. Why? Who knows, most probably because the CLT can't be used because of autocorrelation and ill-defined variance. When the CLT doesn't hold it cannot hold more at later timesteps. So either way I don't see how this could work. $\endgroup$– vonjdFeb 24, 2015 at 13:23
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2$\begingroup$ The CLT still holds with some autocorrelation as long as the variance is finite. My statement is not that 1 day or 1 year returns are Gaussian, but that as the return period gets longer (for example, 100 year returns, if we had that data) would become more Gaussian. This can be demonstrated by bootstrapping clearly non-normal returns over long time periods. Consider a simple binomial tree model. Each step is clearly not Gaussian (it has only two possible outcomes!) but the returns converge over time to Gaussian even if we don't let dt go to zero, but instead let T go to infinity. $\endgroup$– user9403Feb 24, 2015 at 13:46
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2$\begingroup$ The CLT explanation is intuitive. However, the CLT is not the reason for the flattening of the smile. All martingale models will produce a flattening smile according to Rogers and Teranchi (see statslab.cam.ac.uk/~mike/papers/parallel-shifts.pdf): "Indeed, suppose that the log stock price is a spectrally negative α-stable Levy process with α < 2, as proposed by Carr and Wu. Then the implied volatility surface flattens at long maturties despite the fact that the distribution of the standardized returns do not tend to the normal distribution!" $\endgroup$ Aug 21, 2018 at 22:15
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1$\begingroup$ @vonjd Because of the smile, the process is not gaussian. (see for example local vol, by gyongy lemma, local vol is equivalent in distribution with a GBM with smile). But if you add these non-gaussian returns over a long time period, it will become more closer to gaussian. $\endgroup$ Mar 5, 2020 at 16:35
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If skew is too high, then you can have call/put spread arbitrage. An easy way to see put spread arbitrage would be to price a digital put when using skew.
When using skew, the price of a digital put is:
$$DP=N(-d_2)+\frac{d\sigma}{dK}\frac{\partial V}{\partial \sigma}$$
where the price is the black scholes price of the digital put plus skew times vega of a vanilla options (puts/calls have the same vega).
However, as time to expiry gets longer, vega increases roughly with $\sqrt T$.
$$\frac{\partial V}{\partial \sigma}=e^{-rT}F\sqrt{T}n(d1)$$
If the skew is very steeply negative with a long time to expiry, that skew correction factor could take the price of a digital negative when skew is negative (as it would be for SPX puts). Probably can also prove with a little bit of work that the implied risk neutral probability density function can go negative if the skew is too high - i.e. butterfly arbitrage. If I have time later, I might work that out, but I have to get back to work!
answered Mar 24, 2017 at 14:11
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1$\begingroup$ Interesting view on the question. Notice that in the presence of a smile $\sigma(K,t)$ the risk-neutral pdf $q_{S_t}(K) = d\Bbb{Q}(S_t \leq K)/dK$ reads $$ q_{S_t}(K) = \phi(d_2) \left[ \frac{1}{K \sigma \sqrt{t}} + \frac{2d_1}{\sigma}\frac{\partial \sigma}{\partial K} + \frac{d_1d_2K\sqrt{t}}{\sigma} \left( \frac{\partial \sigma}{\partial K} \right)^2 + K\sqrt{t} \frac{\partial^2 \sigma}{\partial K^2} \right] $$ $\endgroup$ Mar 24, 2017 at 16:05
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1$\begingroup$ @Quantuple - worth a look. Thx for that. Probably equivalent to the formula Gatheral put in his parsimonious svi paper though he did it in log moneyness space. Either way, even if there is no explicit arbitrage, there would probably be some pretty compelling trades to put on with a skew for a 3 year option being the same as the skew of a 3 week option. Probably the skew in total variance space can be quite similar - but with annualized implied vols, the units suggest strongly that skew has no reason to be the same. It is all about total variance really in the end. $\endgroup$ Mar 24, 2017 at 21:02
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The volatility smile is seen when Black Scholes' model assumptions are broken.
When you see a flattening, the assumption breakage eases up a little (if the assumptions held you would se a pretty flat line!).
Usually smiles are are due to the possibility of price change momentum.
I.e. a price move in one period causes subsequent period moves in the same direction.
Think of a market crash. A large drop precedes a surge to the exit.
BS assumes returns are IID, however.
During short periods you might see such momentum, which lessens over longer horizons.
There may be a horizon where you could see mean reversion.
E.g. a large drop would be the precursor to 'corrections' back in the other direction to a longer term mean.
To sum up, what you are see is a change in autocorrelation over differing time periods.
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A wee bit late to the party but still worth posting an answer I think, especially since this question appears to be the only one asking why the IV flattens as $T \rightarrow \infty$. Furthermore, I am not entirely sure about the answer that states the flattening is due to the central limit theorem.
So, first of all recall the Black-Scholes formula for a put option: $$ P^{BS} (K, I(K)) = K N(-d_2(K)) - S N(-d_1(K)) $$ where $$ d_2 (K) = -\frac{\log K}{I(K) \sqrt T} - \frac{I(K) \sqrt T}{2} $$ $$ d_1 (K) = -\frac{\log K}{I(K) \sqrt T} + \frac{I(K) \sqrt T}{2} $$ and the implied volatility is defined such that the BS price matches the put market price $$ P^{BS} (K, I(K)) = P(K) $$ From now on I'll hide the explicit dependence of $P,I, d_2,d_1$ on $K$ to save space.
Notice also that $$ \lim_{T \rightarrow \infty} d_2 = -\infty, \quad \lim_{T \rightarrow \infty} d_1 = \infty $$
The change of the put price with respect to strike is $$ \frac{\partial P}{\partial K} = N(-d_2) + K \sqrt T N'(d_2) \frac{\partial I}{\partial K} $$ By no arbitrage, $$ 0 < \frac{\partial P}{\partial K} < 1 $$
Using the upper bound $\frac{\partial P}{\partial K} < 1$ it follows that $$ \frac{\partial I}{\partial \log K} < \frac{1}{\sqrt T} \frac{ 1 - N(-d_2)}{N'(-d_2)} $$ Instead of using the lower bound $\frac{\partial P}{\partial K} >0$, a better lower bound is given by $\frac{\partial P}{\partial K} > \frac{P}{K}$ which leads to $$ \frac{\partial I}{\partial \log K} > \frac{1}{\sqrt T} \frac{ N(d_1) - 1}{N'(d_1)} $$
Now there is something called the Mills ratio; $$ \frac{1 - N(x)}{N'(x)} < \frac{1}{x} \quad \forall x>0 $$ Since $\lim_{T \rightarrow \infty} d_2 = -\infty$, $\exists T_2$ such that $\forall T > T_2 \; d_2 < 0$. Similarly as $\lim_{T \rightarrow \infty} d_1 = \infty$, $\exists T_1$ such that $\forall T > T_1 \; d_1 > 0$. Thus when looking at $T \rightarrow \infty$ it is fine to take $d_2 < 0$ and $d_1 > 0$.
Thus for large enough $T$ $$ -\frac{1}{\sqrt T}\frac{1}{d_1} < \frac{\partial I}{\partial \log K} < -\frac{1}{\sqrt T} \frac{1}{d_2} $$ Finally then $$ 0 \leq \lim_{T \rightarrow \infty}\; \frac{\partial I}{\partial \log K} \leq 0 $$ which implies $$ \lim_{T \rightarrow \infty} \frac{\partial I}{\partial \log K} = 0 $$ It can also be shown from the above inequality that for large $T$, $\frac{\partial I}{\partial \log K} = O(\frac{1}{T})$
Although not so simple, I think this is the simplest possible maths way to show the skew flattens as $T \to \infty$.
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1$\begingroup$ This implies that the reason for the flattening of the skew is that otherwise there would be arbitrage for large enough T. $\endgroup$– dm63Jan 21 at 18:32
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$\begingroup$ @dm63 Yes, that's a good way to look at it as for large enough $T$ we have $-\frac{1}{\sqrt T}\frac{1}{d_1} < \frac{\partial I}{\partial \log K} < -\frac{1}{\sqrt T} \frac{1}{d_2}$ $\endgroup$– FridoJan 21 at 21:22
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Basically what we are talking about here is the volatility of implied volatility. This decreases with expiration - why? Because not only volatility but also the volatility of volatility is mean reverting. Meaning: Short-term implied volatilities are more volatile and the further you look into the future the flatter the smile becomes as mean reversion wins out.
Have also a look at this excellent presentation by quant legend Emanuel Derman:
http://finmath.stanford.edu/seminars/documents/Stanford.Smile.Derman.pdf
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Because as the number of data points grows, the sum of returns would become more and more normal. As that becomes normal, smile becomes flatter and flatter. Why does the sum of returns become normal as number of data points increase? Because some periods where the volatility was high would be offset by other periods when the volatility was low. You would have enough data points to get a nice normal curve for the returns distribution. Why does a normal curve give a flat vol smile? That's the assumption of the Black Scholes model "vol is constant". So it fits back in.
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everyone here is giving arbitrary answers based on various different models.
I think the real reason is that once you get past a certain time horizon, the long term vol doesn't really mean anything - if you try and trade it you're going to get a price so wide that nuances from different models don't really mean anything.
If you want to trade something where the price mainly comes from skew, or kurtosis, then the price it not so much going to be based on the level of that, bur rather the current price to enter into a hedge that is close enough at the longest reasonable maturity + expected costs to role the hedge once they become an option.
