SABR Delta Risk Under Different Volatility Parameterizations

# A question about Hagan's 2002 SABR paper "Managing Smile Risk"


# A question about Hagan's 2002 SABR paper "Managing Smile Risk"












I'm reading Hagan's 2002 paper Managing Smile Risk originally published on the WILMOTT magazine, and got something confusing.

The set up: Consider a European call option on an asset $A$ with exercise date $t_{ex}$ , and strike $K$. If the holder exercises the option on $t_{ex}$, then on the settlement date t_set he receives the underlying asset A and pays the strike $K$.

Let $BS(f,K,σ_B,t_ex)$ be Black’s formula for, say, a call option. According to the SABR model, the value of a call is as equation (3.5) $$V_{call}=BS(f,K,σ_B (K,f),t_{ex})$$

Here $\sigma_B$ is the implied volatility. Also, at-the-money volatility is defined as the implied vol when $K$ is $f$: $$\sigma_{ATM} := \sigma_B(f,f)$$

The SABR model is $$d\hat{F}=α\hat{F}^β dW_1,\quad \hat F(0)=f $$ $$d\hat α = v \hat \alpha dW_2, \quad \hat \alpha(0) = \alpha$$ $$dW_1 dW_2=ρ dt$$

Then in pp95, it starts to discuss the two approaches to calculate Delta risk --

The delta risk expressed by the SABR model depends on whether one uses the parameterization $α$, $β$, $ρ$, $ν$ or $\sigma_{ATM}$, $\beta$, $\rho$, $\nu$.

- Suppose first we use the parameterization $α$, $β$, $ρ$, $ν$, so that $σ_B (K,f)≡σ_B (K,f ; α,β,ρ,ν )$. Differentiating respect to $f$ yields the Δ risk in equation (3.9) $$Δ≡\frac{∂V_{call}}{∂f}=\frac{∂BS}{∂f}+\frac{∂BS}{∂σ_B} \frac{∂σ_B(K,f;α,β,ρ,ν ))}{∂f}$$

- Now suppose we use the parameterization $σ_{AMT}$, $β$, $ρ$, $ν$. Then $α$ is a function of $σ_{ATM}$ and $f$ defined implicitly by $\sigma_{ATM} := \sigma_B(f,f)$. Differentiating (3.5) now yields the Δ risk as in Equation (3.10) $$ Δ≡\frac{∂BS}{∂f}+\frac{∂BS}{∂σ_B} \left[\frac{∂σ_B (K,f; α,β,ρ,ν)}{∂f} + \frac{∂σ_B (K,f; α,β,ρ,ν)}{∂α} \frac{∂α(σ_{ATM},f)}{∂f}\right] $$

Then in the 2nd approach, it says:

> The delta risk is now the risk to changes in $f$ with $σ_{ATM}$ held fixed. The last term is just the change in $α$ needed to keep $σ_{ATM}$ constant while $f$ changes. Clearly this last term must just cancel out the vertical component of the backbone, leaving only the sideways movement of the implied volatility curve. Note that this term is zero for $β=1$.

I'm lost here, what is the last term ?

I tried to calculate "the change in $α$ needed to keep $σ_{ATM}$ constant while $f$ changes", the change of $\sigma_{ATM}$ is

$$\delta \sigma_{ATM} = \frac{\partial\sigma_{ATM}}{\partial f} \delta f + \frac{\partial \sigma_{ATM}}{\partial \alpha} \delta \alpha$$

So to keep $\sigma_{ATM}$ constant, i.e. make the change $\delta \sigma_{ATM} = 0$, the change in $\alpha$ is

$$\delta \alpha = - \frac{ \frac{\partial \sigma_{ATM} } {\partial f} } { \frac {\partial \sigma_{ATM}}{\partial \alpha} } \delta f$$

But such amount is not in equation (3.10).

So what is the the last term indeed?