Hedge variance swapping by vanilla option(constant vega portfolio against underlying asset)

Question

One book said hedging variance swaps $$I= \sqrt{\dfrac{1}{t}\int^t_0\sigma^2(S,t)}d t$$ by vanilla option，say value $V(S,E;\sigma)$(Black-Scholes fomula) where $S$ is underlying asset, $E$ is the strike price. Then he constructed a portfolio with value $$P=\int^{\infty}_0f(E)V(S,E;\sigma)d E$$ then compute the vega of portfolio: $\textrm{Vega}_P= \dfrac{\partial P}{\partial\sigma},$ then let $$\dfrac{\partial \textrm{Vega}_P }{\partial S}=0$$ obtain $f(E) = \dfrac{k}{E^2}.$

His conclusion is variance swaps can be hedged with vanilla option, using the 'one over strike squared' rule. I can not understand:

1.What's the meaning of the representation of $P$(why take the integral) i.e how do we implement this portfolio by vanilla option, hold $F(E)$ share?

2.Why we need constant vega against $S$ i.e how to hedge?

Answer 1

1. An integral is used because the portfolio contains an infinite number of instruments. More specifically, the idea is to hold a continuous strip of $f(E)$ units of European vanilla options struck at $E$ for each $E \in [0,\infty[$. Of course this remains a theoretical concept: to build a similar portfolio in practice, one must consider a partition of the truncated strike domain (hence a finite number of options): $$ P = \sum_{i=1}^N f(E_i) V(S,E_i;\sigma) $$

2. You need a constant, non-zero, Vega against $S$ because you would like a product which is sensitive to volatility and independent of the path the asset will take, i.e. a pure volatility bet.

Have a look at this well-known deck by JP Morgan here. It may help you better understand the practical implications.