How Variance and Volatility Swaps Differ in Volatility-of-Volatility Exposure

# Is variance swap long volatility of volatility?


# Is variance swap long volatility of volatility?












In JPM's note on variance swaps, on page 29, they say "... a long variance swap is also long volatility of volatility".

In Bennett's book Trading Volatility, on page 115, he says "... a variance swap has no vol of vol risk".

I thought Bennett is correct, that variance swaps do not have vol of vol risk, because variance swaps can be replicated by delta-hedged static vanillas, and implied vols are fixed on the strike day and delta hedges only involve the underlying. So I would usually say this instead: "variance swaps do not have vol of vol exposure, and vol swaps are short vol of vol."

I am aware of the log contract weights of $1/K^2$ etc., so it is perfectly fine to say that variance swap has skew convexity. Is it possible that the authors from JPM confused themselves with skew convexity and vol of vol? They are very different concepts to me, as I would usually think of vol of vol as the vol of variance parameter in the Heston or Bergomi models.

Anyone can clarify?

ref:

- http://quantlabs.net/academy/download/free_quant_instituitional_books_/[JP%20Morgan]%20Variance%20Swaps.pdf

- https://www.trading-volatility.com/Trading-Volatility.pdf

## Answer by Frido (score 5, accepted)

https://quant.stackexchange.com/a/74791

My two cents:

Let's agree that a derivative is long an underlying if the payoff of the derivative increases with the price of the underlying $S$.

Then buying a variance swap is going long the volatility of $S$, and it is also long the variance of $S$. It is a convex payoff of the volatility of $S$ and linear in the variance of $S$.

Similarly, buying a volatility swap is also going long the volatility of $S$ and going long the variance of $S$. A volatility swap's payoff is concave in variance of $S$ and linear in volatility of $S$.

That's all, nothing about vol of vol.

An option on a varswap or volswap or realised volatility or realised variance would be long 'vol of vol' at any time before expiry, where just as there are different notions of vol there would be different definitions of 'vol of vol'.

An instrument that would be long expected / forward start risk-neutral 'vol of vol' before and at maturity is trading the spread between the VIX future and the forward starting volswap of the same tenors. Because at maturity of the VIX future the payoff is the difference between the VIX (square root of variance swap) and the spot start volswap, which is greater than zero by Jensen's inequality.

EDIT

Following dm63's and Newquant's answers below, for which +1 for both, an edit to try to settle this good question from the OP.

Let $\bar\sigma$ be the annualised volatility over the interval $[0,T]$. Then $$ X_t := E_t [ \bar\sigma^2 ] $$ is the varswap price, and $$ Y_t := E_t \sqrt{\bar\sigma^2} $$ is the volswap price.

Now suppose that we take the varswap as the base instrument. Then relative to the varswap the volswap has vol of vol (to be precise to the vol of the varswap) exposure since $$ Y_t = E_t \sqrt{X_T^2} $$ However, the varswap, as base instrument, does not have exposure to the vol of vol, just like the SPX spot price has no vega.

If we take the volswap as the base instrument, then the volswap has no vol of vol exposure, but the varswap does (to the vol of the volswap), since $$ X_t = E_t [Y_T^2] $$

Regardless of the base/reference instrument , the difference between the two has vol of vol exposure.

## Answer by dm63 (score 5)

https://quant.stackexchange.com/a/74808

What about the following argument: a variance swap can be replicated with a portfolio of vanilla options, nearly all of which are out of the money (OTM) . But it is well known that OTM options are long vol of vol, so the variance swap is long vol of vol.

On the other hand, a variance swap is flat vol of vol versus its replication portfolio, so it all depends what you are holding constant when you ask the question.

## Answer by Newquant (score 4)

https://quant.stackexchange.com/a/74805

Since the variance swap is linear in variance. Its local volatility exposure is 2σ, with second derivative = 2. If one was to hedge this local volatility exposure using options or a volatility swap, the resulting payoff would be convex in volatility. The spread between the variance swap strike and volatility swap strike must be positive. Otherwise the variance swap can be arbitraged for free volatility of volatility exposure.