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In the SVI-JW parametrization, we have

$$ w(k; a, b, \rho, m, \sigma) = a + b \left [ \rho(k-m) + \sqrt{(k-m)^{2} + \sigma^{2}} \right ] $$

Which gives us

$$ \begin{align*} \sigma_{BS}(k) &= \frac{1}{\sqrt{t}}\sqrt{a + b \left [ \rho(k-m) + \sqrt{(k-m)^{2} + \sigma^{2}} \right ]} \\ \\ \\ \frac{\partial \sigma_{BS}}{\partial k} &= \frac{b\left [\rho + \frac{(k - m)}{\sqrt{(k-m)^{2} + v^2}}\right ]}{2\sqrt{t}\sqrt{a + b \left [ \rho(k-m) + \sqrt{(k-m)^{2} + \sigma^{2}} \right ]}} \end{align*} $$

We can evaluate ATM variance $v_{t}$ by setting $k=0$ in $w(k; a, b, \rho, m, \sigma)$, we can evaluate ATM skew $\psi_{t}$ by evaluating $\frac{\partial \sigma_{BS}}{\partial k}|_{k=0}$ and we can evaluate minimum implied variance $\tilde{v}_{t}$ by setting $\frac{\partial \sigma_{BS}}{\partial k} = 0$ and plugging $k$ into $w(k)$.

How can we find the put and call slopes $p_{t}$ and $c_{t}$? I assumed it would be the limit of the variance $\frac{w(k)}{t}$ when $k \rightarrow \pm \infty$ but this gives me $\frac{b}{\sqrt{t}}(\rho \pm 1)$ which does not match Gatheral's results in his original paper.

Link to original paper Arbitrage-free SVI volatility surfaces by Jim Gatheral, Antoine Jacquier here.

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$(p_t,c_t)$ are respectively related to the put/call slopes of the total implied variance, not variance $$ w(k,t)=\sigma^2(k,t) t $$

Under SVI $$ w(k) = a + b \left(\rho(k-m) + \sqrt{(k-m)^2 + \sigma^2} \right) $$ such that $$ \frac{\partial w}{\partial k}(k) = b \left( \rho + \frac{k-m}{\sqrt{(k-m)^2+\sigma^2}} \right) $$ and $$ \lim_{k \to \pm \infty} \frac{\partial w}{\partial k}(k) = b \left( \rho \pm 1 \right) $$ (see also here end of p.5)

Now, remembering should you define: $$ p_t := \frac{1}{\sqrt{w_t}} b (1-\rho) $$ $$ c_t := \frac{1}{\sqrt{w_t}} b (1+\rho) $$ with $w_t$ the ATMF total implied variance ($w_t = v_t t$ in the JW space) then you have that indeed:

  • $p_t$ is proportional to the opposite of the put slope of total implied variance and is expected to be positive (because $w_t$ and $b$ are positive)
  • $c_t$ is proportional to the call slope of total implied variance and is expected to be positive (because $w_t$ and $b$ are positive)
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  • $\begingroup$ However the original paper has definitions for ($c_{t}$, $p_{t}$) of $\frac{b}{\sqrt{w_{t}}} (1 \pm \rho)$ where $w_{t} = w(0, t)$. Where is the $\frac{1}{\sqrt{w_{t}}}$ term coming from? And why is the sign different for $p_{t}$? $\endgroup$ – Ruse Sep 28 at 12:09
  • $\begingroup$ Hi, I've edited my answer to make it clearer. The term $1/\sqrt{w_t}$ is basically a scaling term. And indeed, $p_t \propto -\text{put slope}$ and $c_t \propto +\text{call slope}$. $\endgroup$ – Quantuple Sep 28 at 12:29

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