# Does the fact that volatility is not constant imply existence of skew?

I had a question regarding the existence of the volatility skew. I've tried researching it a fair bit and I come across a few different explanations: 1. Market participants like buying downside puts and selling upside calls and as a function of supply and demand implied vols are skewed, 2. stock returns are not lognormally distributed (fat tails) so options that are far away from ATM are priced at a higher volatility, 3. volatility is not constant and when stocks decline vol ticks up and vice versa when the market rallies.

I understand why the first 2 contribute to skew, but I don't understand the third. Why does the fact that volatility can change imply the existence of a skew, why doesn't that imply a flat skew that simply shifts levels vertically? So what if when markets drop vol increases, why does this imply a skew? I hope this question makes sense, I've read a few explanations on skew but I still don't quite understand why the third causes it (the other 2 make sense).

• Consider a simple qualitative example. Assumptions: S&P is now 2075 and VIX is now 16% and we believe that if S&P goes to 2055 the vol will increase to 20%. Consider a put option with a strike close to 2055. Then in the future states where S&P goes to about 2055, the option will have value (it will be near ATM) and this value will be based on a volatilty of 20. This possibility is taken into account now, by the market pricing the option with a B.S. IV greater than 16. Nov 9 '15 at 16:50
• In other words the valuation of an option must consider likely future paths. The paths that are relevant to the valuation of a put with K << S0 are precisely those where local volatility will be higher than it is now, those where vol rises as S drops. Nov 9 '15 at 17:00
• So is it fair to say that the volatility of a strike, is the vol the market expects of the stock it the stock starts trading at that level? I never thought o fit that way although that would make sense. So if ATM ( say the 40 strike) is 40 vol, and the 30 strike has a 60 vol, the skew implies that if stock drops to 30 vol would rise to 60? Nov 9 '15 at 18:34

It is not the fact that volatility is time varying that creates the skew per se, but the fact that volatility is negatively correlated with the spot. That is to say, as the stock/index price declines volatility will tend on average to increase, and vice versa. Time varying volatility itself would create a more symmetric 'smile'.

Edit:

Suppose that you have a very simple case where volatility can take only two values $(\sigma_L,\sigma_H) = (0.10,0.50)$ with 50-50 probability. Also say that the drift is zero under both regimes. This is the simplest case of 'stochastic volatility,' if you want.

Then, the price of the option will just be the 50-50 weighted average of the two Black-Scholes prices $$C = 0.5\ BS(S,X,r,\sigma_L,T)+0.5\ BS(S,X,r,\sigma_H,T)$$ Equivalently, the risk neutral distribution of the log-price will be the 50-50 mixture of two Gaussians with different volatilities which exhibits fat tails.

Intuitively, when you are at the money both parts of the weighted sum contribute to the uncertainty, and the implied vol is roughly the average of the two vols, around 0.30. A Gaussian with this volatility will exhibit practically zero probability mass beyond $\pm 1.8$ which is the 6-sigma event point. However, the 50-50 mixture still has some considerable mass, since for the high volatility regime this is only a 3.5-sigma event. To mimic this mass a higher implied vol is required. This is symmetric, hence a 'smile'.

You can see that graphically below, where I have calculated the implied volatility for this example. Now suppose that we have a 'negative correlation', that is to say we still have a 50-50 mixture but now when vol takes the 'low' value 0.10 then the drift is +0.10 positive, but when vol takes the 'high' value of 0.50 then the drift is -0.10 negative.

In that case the +6-sigma event is not symmetric to the -6-sigma event. In fact, implied vol on one side will converge to the 'high' vol of 0.50 while the other side will converge to the 'low' vol of 0.10, following the drifts. Hence we are presented with a 'skew' rather than a 'smile'

Ok, this picture looks a bit odd! This is due to the extreme parameter values in the example, and reflects the 'weight' each distribution has in the mixture at each point. In any case, here is the Python code that created the pics to play with.

# parameters
sH = 0.50 # High vol regime
mH =-0.10 # Drift in high vol regime
sL = 0.10 # Low vol regime
mL =+0.10 # Drift in low vol regime
wH = 0.50 # Mixing weight

# code
import numpy as np
from scipy.stats import norm
from scipy.optimize import root
import matplotlib.pyplot as plt

N = norm(0,1).cdf
n = norm(0,1).pdf

def bs(S, X, r, sigma, T):
d1 = np.log(S/X)+(r+0.5*sigma*sigma)*T
d1 = d1/sigma/np.sqrt(T+1E-10)
d2 = d1 -sigma*np.sqrt(T)
return S*N(d1) -X*np.exp(r*T)*N(d2)
def iv(S, X, r, T, C):
return root(lambda s: bs(S, X, r, s, T)-C, 0.50*np.ones(C.shape)).x

L = 2.5
X = 100.*np.exp(np.linspace(-L, L, 101))
cH = bs(100., X, mH, sH, 1.)
cL = bs(100., X, mL, sL, 1.)
c = wH*cH+(1-wH)*cL

m =np.log1p(wH*(np.exp(mH)-1)+(1-wH)*(np.exp(mL)-1))
v = iv(100., X, m, 1., c)

fg = plt.figure()