Relating Heston Characteristic Functions to Forward Variance Curves

# How to use the Forward Variance Curve in AFV models


# How to use the Forward Variance Curve in AFV models












We know that the normal Heston model is given by $$ \varphi(u;T) = E\bigl[e^{i u\ln S_T}\bigr] = \exp\!\Bigl( i\,u\,\ln S_0 \;+\;D(T)\,V_0 \;+\;C(T) \Bigr). $$ Where $$ D(t) = \psi_2(t) = \frac{b - d}{\sigma^2} \;\frac{1 - e^{-d\,t}}{1 - g\,e^{-d\,t}}, \\[6pt] C(t) = \int_{0}^{t}\kappa\,\theta\,\psi_2(s)\,ds = \frac{\kappa\,\theta}{\sigma^2} \Bigl[(b - d)\,t \;-\;2\ln\!\frac{1 - g\,e^{-d\,t}}{1 - g}\Bigr], $$

We also know that we can write it in Forward Variance form with the following dynamics $$ dS_t = \sqrt{\xi_t(t)}\,S_t\,dW_t^{1,Q}, \\ d\xi_t(T) = \sigma\,e^{-\kappa_Q\,(T-t)}\,\sqrt{\xi_t(t)}\;dW_t^{2,Q}.\\ d\bigl\langle W^{1,Q},\,W^{2,Q}\bigr\rangle_t = \rho\,dt. $$ With $$ \xi_t(T) = E[\nu_T|F_t] = e^{-\kappa_Q (T-t)}\,\xi_t(t) + \theta_Q\bigl(1-e^{-\kappa_Q (T-t)}\bigr) $$ (Note: am I making a mistake here? Should it actually be $\xi_t(t)$? See derivation #1 below)

Then, we can let the CGF be $$ \varphi(u;T) =E\bigl[e^{iu\ln S_T}\bigr] =\exp\!\Bigl(iu\ln S_0\;+\;\int_0^T F\bigl(\psi_1(s),\psi_2(s)\bigr)\, g_0(T-s)\,\mathrm{d}s\Bigr). $$ With $$\psi_1(s)=iu,\qquad \psi_2(s)=\int_0^sF\bigl(iu,\psi_2(r)\bigr)\,\mathrm{d}r $$ and $F(\psi_1,\psi_2) =\tfrac12(\psi_1^2-\psi_1) +(\rho\sigma\,\psi_1-\kappa)\,\psi_2 +\tfrac{\sigma^2}{2}\,\psi_2^2. $

Now, let $g_0(T-s) =e^{-\kappa(T-s)}V_0+\theta\bigl(1-e^{-\kappa(T-s)}\bigr) $. We now have $V_0 = \xi_0(0)$, correct?

Writing \begin{equation} \psi_2(T)=\int_0^Te^{-\kappa(T-s)}\,F\bigl(iu,\psi_2(s)\bigr)\,ds, \end{equation} And then noting \begin{equation} \begin{aligned} \psi_{2}'(s) &= -\kappa\,\psi_{2}(s) + F(s)\\ \quad\Longrightarrow\quad \int_{0}^{T}F(s)\,ds &= \psi_{2}(T)-\psi_{2}(0) + \kappa\int_{0}^{T}\psi_{2}(s)\,ds = \psi_{2}(T) + \kappa\!\int_{0}^{T}\psi_{2}(s)\,ds, \end{aligned} \end{equation}

one finds that we have \begin{equation} \int_{0}^{T} F(s)\,g_{0}(T - s)\,\mathrm{d}s =V_0\!\int_{0}^{T} e^{-\kappa (T - s)}\,F(s)\,\mathrm{d}s \;+\;\theta\!\int_{0}^{T} F(s)\,\mathrm{d}s \;-\;\theta\!\int_{0}^{T} e^{-\kappa (T - s)}\,F(s)\,\mathrm{d}s. \end{equation} So we get \begin{equation} \int_{0}^{T}F(s)\,g_{0}(T-s)\,ds=V_0\,\psi_{2}(T) + \kappa\,\theta\!\int_{0}^{T}\psi_{2}(s)\,ds. \end{equation}

Hence \begin{equation} \varphi(u;T) =\exp\!\Bigl(iu\ln S_0 +V_0\,\psi_2(T) +\kappa\,\theta\!\int_0^T\psi_2(s)\,ds\Bigr), \end{equation} Then setting $\psi_2(T)=D(T)$ and $\kappa\theta\!\int_0^T\psi_2=C(T)$, is exactly the standard Heston CF with $K\equiv1$.

However, I now don't understand the purpose of the forward variance curve anymore. Say this looks like

Should I simply replace $V_0$ = $\xi_0(0)$ with the value of the forward variance curve at $Maturity = 0$, always? That doesn't make a whole lot of sense? In case i was wrong, and it is not $V_0 =\xi_0(0)$ but $V_0 = \xi_0(T)$, then how do we solve the convolution below?

$$ \int_{0}^{T}F(s)\,g_{0}(T-s)\,\mathrm{d}s =\int_{0}^{T}F(s)\bigl[e^{-\kappa (T-s)}\,\xi_0(T-s? ) +\theta\bigl(1-e^{-\kappa (T-s)}\bigr)\bigr]\,\mathrm{d}s, $$ We clearly can't easily pull out $xi_0(T-s)$ anymore?

Derivation #1 is as follows:

Let $\xi_0(t) = E\big[\nu_T | F_0\big]$. We note first that going from $d\nu_t$ to $\nu_t$ involves using the Ito product rule, given by $d(XY) = YdX + XdY + dYdX$. We work backwards from the solution, by taking $X_s = e^{\kappa_Q s}$ and $Y=\nu_s$ Then, we have $d(X_sY_s) = d(e^{\kappa_Q s}\nu_s) = e^{\kappa_Q s}d\nu_s +\nu_s\kappa_Q e^{\kappa_Q s}ds$ \begin{equation} \begin{aligned} & d\nu_s + \kappa_Q\nu_sds = \kappa_Q\theta_Qds + \sigma\sqrt{\nu_t}\,dW^{2,Q}_s\\ & e^{\kappa_Q s}( d\nu_s + \kappa_Q\nu_sds) = e^{\kappa_Q s}(\kappa_Q\theta_Qds + \sigma\sqrt{\nu_t}\,dW^{2,Q}_s)\\ & d(e^{\kappa_Q s}\nu_s) = e^{\kappa_Q s}(\kappa_Q\theta_Qds + \sigma\sqrt{\nu_t}\,dW^{2,Q}_s)\\ \end{aligned} \end{equation} Then integrating this, where we note $ \int_0^T e^{\kappa_Q s}ds = \frac{1}{\kappa_Q}\bigl(e^{\kappa_Q T}-1\bigr)$ we get \begin{equation} \begin{aligned} & e^{\kappa_Q T}\,\nu_T - \nu_0 = \kappa_Q\,\theta_Q\int_0^T e^{\kappa_Q s}\,ds + \sigma\int_0^T e^{\kappa_Q s}\sqrt{\nu_s}\,dW_s^{2,Q}.\\ &\nu_T = e^{-\kappa_Q T}\,\nu_0 + \theta_Q\bigl(1-e^{-\kappa_Q T}\bigr) + \sigma e^{-\kappa_Q T}\!\int_0^T e^{\kappa_Q s}\sqrt{\nu_s}\,dW_s^{2,Q}.\\ % &\nu_T \end{aligned} \end{equation}