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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Sign up to join this communityFirst, 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 !
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.
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!
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.
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 $T \to \infty$.
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
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.
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.