SABR Goes Normal: Deriving the Equivalent Local Volatility Density

# Balland - SABR goes normal


# Balland - SABR goes normal












To summarise this very long post : please help me understand the undetailed proof of the quoted paper. I am not comfortable using a result I do not fully understand.

I am reading Balland & Tran SABR goes normal paper and struggle to understand the computations from the first section (equivalent SABR local volatility).

His notations for a general LSV model with lognormal stochastic volatility is $$ \begin{cases} d S_t = \sigma_t \varphi \left(S_t\right) dW_t^1 \\ d \sigma_t = \gamma \sigma_t dW_t^2 \\ \langle W^1, W^2\rangle_t = \rho t \end{cases} $$ Introducing the process $J$ as $$ J_t \equiv J\left(S_t, \sigma_t\right), \quad J \left(x, y\right) := \frac{1}{y} \int_K^x{\frac{\mathrm{d}u}{\varphi \left(u\right)}} $$ We compute partial derivatives \begin{align} \partial_x J & = \frac{1}{y \varphi \left(x\right)} \\ \partial_{xx}^2 J & = - \frac{\varphi' \left(x\right)}{y \varphi \left(x\right)^2} \\ \partial_y J & = - \frac{J}{y} \\ \partial_{yy}^2 J & = \frac{2 J}{y^2} \\ \partial_{xy}^2 J & = - \frac{1}{y^2 \varphi \left(x\right)} \end{align} to apply Itō formula. \begin{align} dJ_t & = \frac{d S_t}{\sigma_t \varphi \left(S_t\right)} - J_t \frac{d \sigma_t}{\sigma_t} + \frac{1}{2} \left[\frac{2 J_t}{\sigma_t^2} d \langle \sigma \rangle_t - \frac{\varphi' \left(S_t\right)}{\sigma_t \varphi \left(S_t\right)^2} d \langle S \rangle_t - 2 \frac{d \langle S, \sigma \rangle_t}{\sigma_t^2 \varphi \left(S_t\right)} \right] \\ & = dW_t^1 - \gamma J_t dW_t^2 + \left[\gamma^2 J_t - \rho \gamma - \frac{1}{2} \sigma_t \varphi' \left(S_t\right) \right] dt \\ & = q \left(J_t\right)^{\frac{1}{2}} d \bar{W}_t + \left[\gamma^2 J_t - \rho \gamma - \frac{1}{2} \sigma_t \varphi' \left(S_t\right) \right] dt \end{align}

Noticing that $\delta \left(J_t\right) \equiv \sigma_t \, \delta \left(S_t - K\right)$, we get $$ \varphi \left(K\right) \mathbb{E} \left[\sigma_t^2 \delta \left(S_t - K\right) \right] = \varphi \left(K\right) \mathbb{E} \left[\sigma_t \delta \left(J_t \right)\right] = \varphi \left(K\right) \sigma_0 \, \mathbb{E} \left[ e^{\gamma W_t -\frac{1}{2} \gamma^2 t} \delta \left(J_t\right)\right] $$

Let us define the probability $\hat{\mathbb{Q}}$ by $$ \frac{\mathrm{d} \hat{\mathbb{Q}}}{\mathrm{d} \mathbb{Q}} = e^{\gamma W_t^2 - \frac{1}{2} \gamma^2 t} $$ By Girsanov's theorem, $$ \begin{cases} \hat{W}^1 := W^1 - \rho \gamma \cdot \\ \hat{W}^2 := W^2 - \gamma \cdot \end{cases} $$ are $\hat{\mathbb{Q}}$-Wiener processes with marginal covariance process $\Sigma$. The $\hat{\mathbb{Q}}$-dynamics of $J$ is $$ d J_t = d \hat{W}_t^1 - J_t \gamma_t \, d\hat{W}_t^2 - \frac{1}{2} \sigma_t \varphi' \left(S_t\right) dt $$ Since a linear combination of Gaussian random variables (Wiener process increments) is also Gaussian, we can rewrite that dynamics using a single $\hat{\mathbb{Q}}$-Wiener process $\hat{U}$ $$ d J_t = \sqrt{q \left(J_t\right)} \, d\hat{U}_t - \frac{1}{2} \sigma_t \varphi' \left(S_t\right) dt; \quad q : j \mapsto \left(1 - 2 \rho \gamma j + \gamma^2 j^2 \right) $$

Similarly, $$ \frac{\mathbb{E} \left[\delta \left(S_t - K\right) \right]}{\varphi \left(K\right)} = \frac{\mathbb{E} \left[\frac{\delta \left(J_t\right)}{\sigma_t} \right]}{\varphi \left(K\right)} = \frac{\mathbb{E} \left[\delta\left(J_t\right) e^{- \gamma W_t^2 + \frac{1}{2} \gamma^2 t}\right]}{\sigma_0 \varphi \left(K\right)} $$ We now define the probability $\tilde{\mathbb{Q}}$ by $$ \frac{\mathrm{d} \tilde{\mathbb{Q}}}{\mathrm{d} \mathbb{Q}} = e^{- \gamma W_t^2 - \frac{1}{2} \gamma^2 t} $$ to get $$ \frac{\mathbb{E} \left[\delta \left(S_t - K\right) \right]}{\varphi \left(K\right)} = \boxed{\frac{\mathbb{E}^{\tilde{\mathbb{Q}}} \left[\delta\left(J_t\right) \right] e^{\gamma^2t}}{\sigma_0 \varphi \left(K\right)}} $$

By Girsanov's theorem, $$ \begin{cases} \tilde{W}^1 := W^1 + \rho \gamma \cdot \\ \tilde{W}^2 := W^2 + \gamma \cdot \end{cases} $$ are $\tilde{\mathbb{Q}}$-Wiener processes with marginal covariance process $\Sigma$. The $\tilde{\mathbb{Q}}$-dynamics of $J$ is \begin{align*} d J_t & = d \tilde{W}_t^1 - J_t \gamma_t \, d\tilde{W}_t^2 + \left[2 J_t \gamma^2 - \frac{1}{2} \sigma_t \varphi' \left(S_t\right) - 2 \rho \gamma\right] dt \\ & = \sqrt{q \left(J_t\right)} \, d\tilde{U}_t + \left[q' \left(J_t\right) - \frac{1}{2} \sigma_t \varphi' \left(S_t\right)\right] dt \end{align*}

Let us define the $\tilde{\mathbb{Q}}$-martingale $\bar{\rho}$ by its dynamics $$ d \bar{\rho}_t = - \frac{q' \left(J_t\right)}{\sqrt{q \left(J_t\right)}} \bar{\rho} d \tilde{U}_t $$ and initial value $\bar{\rho}_0 = 1$. We can then define another probability $\bar{\mathbb{Q}}$ by $\frac{\mathrm{d}\bar{\mathbb{Q}}}{\mathrm{d} \tilde{\mathbb{Q}}} = \bar{\rho}_t$, under which $J$ has the same dynamics as under $\hat{\mathbb{Q}}$, up to the drift of $\sigma$ : $$ d J_t = \sqrt{q \left(J_t\right)} \, d\bar{U}_t - \frac{1}{2} \sigma_t \varphi' \left(S_t\right) dt $$

He then introduces the process $X_t = \frac{q \left(J_0\right)}{q \left(J_t\right)} e^{\gamma^2 t}$ and notices that $X_t = \bar{\rho}_t e^{\frac{1}{2} \int_0^t{\varphi' \left(S_s\right) \sigma_s \frac{q' \left(J_s\right)}{q \left(J_s\right)}\mathrm{d}s}}$. Until then, OK.

First interrogation

Now, first thing he does I do not get at all is the following expression they give for the SABR density.

$$ D = \frac{\mathbb{E}^{\tilde{\mathbb{Q}}} \left[\frac{\color{blue}{q \left(J_0\right)}}{\color{red}{q \left(J_t\right)}}\delta\left(J_t\right) \right] e^{\gamma^2t}}{\color{blue}{q \left(J_0\right)} \sigma_0 \varphi \left(K\right)} $$

Blue term is fair enough, but I do not see the legitimacy of introducing the random term in red within the expectation.

Second interrogation

Admitting this second expression for a moment and continuing the derivation, we get \begin{align} D & = \frac{\mathbb{E}^{\tilde{\mathbb{Q}}} \left[\frac{q \left(J_0\right)}{q \left(J_t\right)}\delta\left(J_t\right) \right] e^{\gamma^2t}}{q \left(J_0\right) \sigma_0 \varphi \left(K\right)} = \frac{\mathbb{E}^{\tilde{\mathbb{Q}}} \left[X_t\delta\left(J_t\right) \right]}{q \left(J_0\right) \sigma_0 \varphi \left(K\right)} \\ & = \frac{\mathbb{E}^{\tilde{\mathbb{Q}}} \left[\frac{\mathrm{d}\bar{\mathbb{Q}}}{\mathrm{d} \tilde{\mathbb{Q}}} e^{\frac{1}{2} \int_0^t{\varphi' \left(S_s\right) \sigma_s \frac{q' \left(J_s\right)}{q \left(J_s\right)}\mathrm{d}s}}\delta\left(J_t\right) \right]}{q \left(J_0\right) \sigma_0 \varphi \left(K\right)} \\ & = \frac{\mathbb{E}^{\bar{\mathbb{Q}}} \left[e^{\frac{1}{2} \int_0^t{\varphi' \left(S_s\right) \sigma_s \frac{q' \left(J_s\right)}{q \left(J_s\right)}\mathrm{d}s}}\delta\left(J_t\right) \right]}{q \left(J_0\right) \sigma_0 \varphi \left(K\right)} \\ & = \frac{\mathbb{E}^{\bar{\mathbb{Q}}} \left[e^{\frac{1}{2} \int_0^t{\varphi' \left(S_s\right) \sigma_s \frac{q' \left(J_s\right)}{q \left(J_s\right)}\mathrm{d}s}} \middle \vert J_t = 0\right] \times \mathbb{E}^{\bar{\mathbb{Q}}} \left[\delta\left(J_t\right)\right]}{q \left(J_0\right) \sigma_0 \varphi \left(K\right)} \end{align}

Using the fact that $J$ has the same dynamics under $\hat{\mathbb{Q}}$ and $\bar{\mathbb{Q}}$, up to the different drift of $\sigma$, they cancel out $\mathbb{E}^{\hat{\mathbb{Q}}} \left[\delta\left(J_t\right)\right]$ in the numerator and $\mathbb{E}^{\bar{\mathbb{Q}}} \left[\delta\left(J_t\right)\right]$ in the denominator. Remains to compute the expectation of the exponential term conditionned at $J_t = 0 \Leftrightarrow S_t = K$.

To my understanding, they compute it to be (at least approximately) $$ e^{\frac{\sigma_0}{2}\left[\color{blue}{\frac{1}{2} \varphi' \left(S_0\right) \frac{q' \left(J_0\right)}{q \left(J_0\right)}} \color{red}{- \rho \gamma \varphi' \left(K\right)} \right]t} $$

I interpret the first term as a freeze of the integrand at its initial value (OK if vol of vol is small, else not sure of the accuracy of it), but have no idea where the second term comes from.

Any idea?