dstefan
06-10-2008, 04:12 PM
Chapter 3, Problem 9
The sensitivity of the vega of a portfolio with respect to volatility and to the price of the underlying asset are often important to estimate, e.g., for pricing volatility swaps. These two Greeks are called volga and vanna and are defined as follows:
volga(V) ~=~ \frac{\partial (\mbox{vega}(V))}{\partial\sigma}; vanna(V) ~=~ \frac{\partial (\mbox{vega}(V))}{\partial S}
It is easy to see that
volga(V) ~=~ \frac{\partial^2 V}{\partial \sigma^2}; vanna(V) ~=~ \frac{\partial^2 V}{\partial S \partial \sigma}
The name volga is the short for ``volatility gamma".
Also, vanna can be interpreted as the rate of change of the Delta with respect to the volatility of the underlying asset, i.e.,
vanna(V) ~=~ \frac{\partial (\Delta(V))}{\partial \sigma}
(i) Compute the volga and vanna for a plain vanilla European call option on an asset paying dividends continuously at the rate q.
(ii) Use the Put-Call parity to compute the volga and vanna for a plain vanilla European put option.
Solution:
(i) Recall that
vega(C) = S e^{-q(T-t)} \sqrt{T-t}~\frac{1}{\sqrt{2 \pi}} e^{-\frac{d_1^2}{2}}; \Delta(C) = e^{-q(T-t)} N(d_1)
where
d_1 = \frac{\ln \left( \frac{S}{K} \right) ~+~ \left(r-q+\frac{\sigma^2}{2}\right)(T-t)}{\sigma \sqrt{T-t}} = \frac{\ln \left( \frac{S}{K} \right) + (r-q)(T-t)}{\sigma \sqrt{T-t}} ~+~ \frac{\sigma \sqrt{T-t}}{2}
Then,
volga(C) = \frac{\partial (\mbox{vega}(C))}{\partial\sigma} ~=~ - S e^{-q(T-t)} \sqrt{T-t} \frac{1}{2 \pi} ~d_1 e^{-\frac{d_1^2}{2}} \frac{\partial d_1}{\partial \sigma}
\mbox{vanna}(C) = \frac{\partial (\Delta(C)) }{\partial \sigma} ~=~ e^{-q(T-t)} N'(d_1) ~\frac{\partial d_1}{\partial \sigma} ~=~ e^{-q(T-t)} ~\frac{1}{\sqrt{2 \pi}} e^{-\frac{d_1^2}{2}} ~\frac{\partial d_1}{\partial \sigma}
Note that
\frac{\partial d_1}{\partial \sigma} = -~\frac{\ln \left( \frac{S}{K} \right) + (r-q)(T-t)}{\sigma^2 \sqrt{T-t}} + \frac{\sqrt{T-t}}{2} = -~\frac{\ln \left( \frac{S}{K} \right) ~+~ \left(r-q-\frac{\sigma^2}{2}\right)(T-t)}{\sigma^2 \sqrt{T-t}} = -~\frac{d_2}{\sigma}
We conclude that
volga(C) = S e^{-q(T-t)} \sqrt{T-t} ~\frac{1}{2 \pi} e^{-\frac{d_1^2}{2}} ~\frac{d_1 d_2}{\sigma}
vanna(C) = -~e^{-q(T-t)} ~\frac{1}{\sqrt{2 \pi}} e^{-\frac{d_1^2}{2}} ~\frac{d_2}{\sigma}
(ii) By differentiating the Put-Call parity P + S e^{-q(T-t)} - C = K e^{-r(T-t)} with respect to \sigma, we find that
vega(P) ~=~ \frac{\partial P}{\partial \sigma} ~=~ \frac{\partial C}{\partial \sigma} ~=~ vega(C)
Therefore,
volga(P) = \frac{\partial (\mbox{vega}(P))}{\partial\sigma} ~=~ \frac{\partial (\mbox{vega}(C))}{\partial\sigma} ~=~ \mbox{volga}(C)
vanna(P) = \frac{\partial (\mbox{vega}(P))}{\partial S} ~=~ \frac{\partial (\mbox{vega}(C))}{\partial S} ~=~ \mbox{vanna}(C)
The sensitivity of the vega of a portfolio with respect to volatility and to the price of the underlying asset are often important to estimate, e.g., for pricing volatility swaps. These two Greeks are called volga and vanna and are defined as follows:
volga(V) ~=~ \frac{\partial (\mbox{vega}(V))}{\partial\sigma}; vanna(V) ~=~ \frac{\partial (\mbox{vega}(V))}{\partial S}
It is easy to see that
volga(V) ~=~ \frac{\partial^2 V}{\partial \sigma^2}; vanna(V) ~=~ \frac{\partial^2 V}{\partial S \partial \sigma}
The name volga is the short for ``volatility gamma".
Also, vanna can be interpreted as the rate of change of the Delta with respect to the volatility of the underlying asset, i.e.,
vanna(V) ~=~ \frac{\partial (\Delta(V))}{\partial \sigma}
(i) Compute the volga and vanna for a plain vanilla European call option on an asset paying dividends continuously at the rate q.
(ii) Use the Put-Call parity to compute the volga and vanna for a plain vanilla European put option.
Solution:
(i) Recall that
vega(C) = S e^{-q(T-t)} \sqrt{T-t}~\frac{1}{\sqrt{2 \pi}} e^{-\frac{d_1^2}{2}}; \Delta(C) = e^{-q(T-t)} N(d_1)
where
d_1 = \frac{\ln \left( \frac{S}{K} \right) ~+~ \left(r-q+\frac{\sigma^2}{2}\right)(T-t)}{\sigma \sqrt{T-t}} = \frac{\ln \left( \frac{S}{K} \right) + (r-q)(T-t)}{\sigma \sqrt{T-t}} ~+~ \frac{\sigma \sqrt{T-t}}{2}
Then,
volga(C) = \frac{\partial (\mbox{vega}(C))}{\partial\sigma} ~=~ - S e^{-q(T-t)} \sqrt{T-t} \frac{1}{2 \pi} ~d_1 e^{-\frac{d_1^2}{2}} \frac{\partial d_1}{\partial \sigma}
\mbox{vanna}(C) = \frac{\partial (\Delta(C)) }{\partial \sigma} ~=~ e^{-q(T-t)} N'(d_1) ~\frac{\partial d_1}{\partial \sigma} ~=~ e^{-q(T-t)} ~\frac{1}{\sqrt{2 \pi}} e^{-\frac{d_1^2}{2}} ~\frac{\partial d_1}{\partial \sigma}
Note that
\frac{\partial d_1}{\partial \sigma} = -~\frac{\ln \left( \frac{S}{K} \right) + (r-q)(T-t)}{\sigma^2 \sqrt{T-t}} + \frac{\sqrt{T-t}}{2} = -~\frac{\ln \left( \frac{S}{K} \right) ~+~ \left(r-q-\frac{\sigma^2}{2}\right)(T-t)}{\sigma^2 \sqrt{T-t}} = -~\frac{d_2}{\sigma}
We conclude that
volga(C) = S e^{-q(T-t)} \sqrt{T-t} ~\frac{1}{2 \pi} e^{-\frac{d_1^2}{2}} ~\frac{d_1 d_2}{\sigma}
vanna(C) = -~e^{-q(T-t)} ~\frac{1}{\sqrt{2 \pi}} e^{-\frac{d_1^2}{2}} ~\frac{d_2}{\sigma}
(ii) By differentiating the Put-Call parity P + S e^{-q(T-t)} - C = K e^{-r(T-t)} with respect to \sigma, we find that
vega(P) ~=~ \frac{\partial P}{\partial \sigma} ~=~ \frac{\partial C}{\partial \sigma} ~=~ vega(C)
Therefore,
volga(P) = \frac{\partial (\mbox{vega}(P))}{\partial\sigma} ~=~ \frac{\partial (\mbox{vega}(C))}{\partial\sigma} ~=~ \mbox{volga}(C)
vanna(P) = \frac{\partial (\mbox{vega}(P))}{\partial S} ~=~ \frac{\partial (\mbox{vega}(C))}{\partial S} ~=~ \mbox{vanna}(C)