Variance Swap Replication and Exposure to Volatility Skew

# Why is a variance swap long skew?


# Why is a variance swap long skew?












I can appreciate the mathematical derivation, but can anyone explain this in a more intuitive sense?

I often come across the mistaken belief that due to the replicating portfolio being long more downside contracts than upside contracts, the variance swap is long skew. But this is incorrect, these weightings are just to ensure that the $vega exposure is equal both on the downside and the upside.

Is it something to do with volga, i.e. the vega of the higher vol downside contracts will increase faster (due to an increase in skew) than the vega of the lower vol contracts will decrease?

## Answer by Quantuple (score 11)

https://quant.stackexchange.com/a/27542

As I've mentioned in a comment, it would be wrong to think that entering a variance swap specifically amounts to being "long skew".

What you can say however is that, in the absence of jumps (i.e. in a pure diffusion framework, see here and here for further info), the fair variance strike $K_{var}$ at which a variance swap with notional $N$ and payoff

$$ N \times ( \sigma^2_{\text{realised}}(0,T) - K_{var} ) $$

should trade at the following par rate (or variance strike)

$$ K_{var} = \frac{2}{B(0,T)T} \left[ \int_0^{F(0,T)} \frac{P(K,T)}{K^2} dK + \int_{F(0,T)}^\infty \frac{C(K,T)}{K^2} dK \right] $$ where $T$ figures the contract's maturity date, $\sigma^2_{\text{realised}}(0,T)$ the variance of log-returns that will realise over the horizon $[0,T]$, $B(0,T)$ the discount factor, $P(K,T)$ and $C(K,T)$ European option prices with strike $K$ and maturity $T$ and $F(0,T)$ the forward price.

Thus, the price of a variance swap is simply a scaled integral of the OTMF price curve: $$ K_{var} \propto \int_0^\infty \frac{V(K,T)}{K^2} dK $$ $$ V(K,T) = \begin{cases} P(K,T) & \text{if } K < F(0,T) \\ C(K,T) & \text{otherwise} \end{cases} $$

Now, assume the following situation where, $S_0=100$, $r=q=0$ (no risk-neutral drift), $T=1$ along with 3 shapes of implied volatility smile at $T$: flat, pure skew, pure convexity. If you compute the fair variance strike $K_{var}$ under these different configurations, you will see that both negative skew and positive convexity have a positive impact and not specifically skew as you seem to indicate. See the simulations below where I've expressed the "variance price" as $\sqrt{K_{var}}\times 100$ similarly to what is done for volatility indices such as the VIX.

## Answer by will (score 11)

https://quant.stackexchange.com/a/27560

If you take Quantuple's stuff a little further, you can really see whether you're long skew. You can pretty easily see the dependence on convexity too (though it should be obvious that you're long convexity).

So first off, we need some smile parametrisation that lets us easily control convexity and skew. I just went with a made up one;

$$\mathrm{convexity} = \mathrm{C} = \left. \frac{\partial^2 \sigma}{\partial K^2} \right|_{K=F} \\ \mathrm{skew} = \mathrm{S} = \left. \frac{\partial \sigma}{\partial K} \right|_{K=F} \\ \sigma_{\mathrm{atm}} = \sigma(F) $$

which gives:

$$ \frac{1}{2} C (K-F)^2 + S(K-F) + \sigma_\mathrm{atm} $$

*Note that i understand this is not a proper smile, i'm just using it as a simple example.

Then if you look a pretty extreme range of smiles;

$$ -0.001 \leqslant \mathrm{S} \leqslant 0.001\\ 0 \leqslant \mathrm{C} \leqslant 0.0001\\ \sigma_\mathrm{atm} = 20\% $$

you get a bunch of images like Quantuple's in the other answer:

But, we can do better:

So yes, you are long skew. But only a really tiny amount - you're much longer convexity.

Here's python code for this if you want to mess around with it.

```
import numpy as np

def CND(X):
  a1,a2,a3,a4,a5 = 0.31938153, -0.356563782, 1.781477937, -1.821255978, 1.330274429
  L = np.abs(X)
  K = 1.0 / (1.0 + 0.2316419 * L)
  w = 1.0 - 1.0 / np.sqrt(2*np.pi)*np.exp(-L*L/2.) * (a1*K + a2*K*K + a3*np.power(K,3) + a4*np.power(K,4) + a5*np.power(K,5))
  if X<0:
    w = 1.0-w
  return w

def BlackSholes(cp,S,X,T,r,v):
  d1 = (np.log(S/X)+(r+v*v/2.)*T)/(v*np.sqrt(T))
  d2 = d1-v*np.sqrt(T)
  if cp=='c':
    return S*CND(d1)-X*np.exp(-r*T)*CND(d2)
  else:
    return X*np.exp(-r*T)*CND(-d2)-S*CND(-d1)

def C(S,X,T,r,v):
    return BlackSholes("c", S, X, T, r, v)
def P(S,X,T,r,v):
    return BlackSholes("p", S, X, T, r, v)

def B(r,t):
    return np.exp(-r*t)

def vol(k, vol_atm, convexity, skew, atm=100, max_vol=1):
    v = 0.5*convexity*k**2 + (skew - convexity*atm)*k + vol_atm + 0.5*convexity*atm**2 - skew*atm
    return max(1e-5,min(v, max_vol))

import scipy.integrate as integrate
import scipy.special as special
def var_swap(S,T,r,atm_vol, convexity, skew):
    F = S/B(r,T)
    return np.sqrt((2 / (T*B(r,T))) * (integrate.quad(lambda k: P(S, k, T, r, vol(k, atm_vol, convexity, skew, atm=F)) * k**-2, 0, F)[0] + integrate.quad(lambda k: C(S, k, T, r, vol(k, atm_vol, convexity, skew, atm=F)) * k**-2, F, F*5)[0]))

r = 0.0
T = 1.0
S = 100.0
F = S/B(r,T)
print F
atm_vol = 0.2
convexity = 0.0001
skew = 0.001

ks = [k for k in range(1, int(F*2))]

n_scenarios = 20
skews = np.linspace(-skew, skew, n_scenarios)
convexities = np.linspace(0, convexity, n_scenarios)

plot_smiles = False

if plot_smiles:

  import colorsys
  blues = [colorsys.hsv_to_rgb(h, 1, 1) for h in np.linspace(0.5, 0.65, n_scenarios)]
  reds = [colorsys.hsv_to_rgb(h, 1, 1) for h in np.linspace(0.0, 0.15, n_scenarios)]

  from matplotlib import pyplot
  fig = pyplot.figure()
  ax_smiles = fig.add_subplot(1,1,1)
  ax_opts = ax_smiles.twinx()

  for i, (convexity, skew) in enumerate(zip(convexities, skews)):
    vols = [vol(k, atm_vol, convexity, skew, atm=F) for k in ks]
    opts = [BlackSholes("p" if k < F else "c", S, k, T, r, vol(k, atm_vol, convexity, skew, atm=F)) * k**-2 for k in ks]

    ax_smiles.plot(ks, vols, color=blues[i])
    ax_opts.plot(ks, opts, color=reds[i])

  pyplot.show()

else:

  CC = np.linspace(0, convexity,n_scenarios)
  SS = np.linspace(-skew, skew,n_scenarios)

  CC, SS = np.meshgrid(CC, SS)

  VV = np.empty(CC.shape)
  for i in range(CC.shape[0]):
    for j in range(CC.shape[1]):
      VV[i,j] = var_swap(S, T, r, atm_vol, CC[i,j], SS[i,j])

  from mpl_toolkits.mplot3d import Axes3D
  from matplotlib import cm
  from matplotlib.ticker import LinearLocator, FormatStrFormatter
  import matplotlib.pyplot as plt

  fig = plt.figure()
  ax = fig.gca(projection='3d')
  surf = ax.plot_surface(CC, SS, VV, rstride=1, cstride=1, cmap=cm.jet, linewidth=0, antialiased=True)

  ax.set_xlabel("Convexity")
  ax.set_ylabel("Skew")
  ax.set_zlabel("Var. Swap par rate")
  ax.set_ylim(ax.get_ylim()[::-1])

  plt.show()
```

There is a caveat here though - this is the theoretical value of a variance swap. The market does not trade these according to the theoretical value, there is a difference which I have not seen a way of accounting for yet. The solution to this is that you store a table of varswap par rate spreads which can be interpolated and applied to varswaps at the corresponding start and end dates.

This spread does not come from stochastic vol, it seems to me to be some sort of insurance against unpleasantly large costs when something happens. The alternative is trading corridor variance swaps (i.e. variance only accrues when the index is inside a corridor), to limit the likelihood of this downside.

You can easily get skew exposure with trades like the above though, if variance is only accruing when the underlying is above/below a certain level, then you will be long/short skew when the underlying is near the barrier - because when you're only looking at one side of a point, skew and convexity have similar effects.

## Answer by user34971 (score 3)

https://quant.stackexchange.com/a/40996

In addition to the answers already given, another way to look at it in the context of a stochastic volatility model is as follows:

The skew is heavily influenced by the correlation between the spot and the volatility. However, pricing a derivative on volatility does not depend on the correlation parameter. Hence a variance swap is not long (or short) skew / independent of the correlation parameter, as is a volatility swap and other pure volatility derivatives.

It (variance strike and vol derivatives in general) does depend on the convexity though (which is determined by the vol of vol).

## Answer by Odyssey (score 0)

https://quant.stackexchange.com/a/81832

apologize for digging out this old post, also apologize if my question might sounds stupid

- so the final conclusion is that: long var swap indeed long skew? In Will's excellent analysis and plot, it shows long skew, however it's for a specific example , would in general long Var swap also long skew?

- as original poster asked, intuitively why long Var Swap is long skew? In the static replication, var swap is replicated by 1/k^2(otm put +otm call), is it because more skew means: otm put is more expensive though otm call is cheaper, but due to the weight 1/k^2, then benefit more from more expensive otm put than the lost of cheaper otm call?

if anyone can help to explain? thank you very much!