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I want to understand the mathematical steps done. Can someone please simplify the derivation of d(pi) from Pi? Thanks in advance.

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I write down the solution for the Heston model. You can directly generalise the result.

Let $f=f(t,s,v)\in C^{1,2,2}(\mathbb{R}_+^3)$ be a real-valued function (portfolio value) and consider the two-dimensional stochastic process $(S_t,v_t)$ with \begin{align*} \mathrm{d}S_t&=(r-q) S_t \mathrm{d}t+\sqrt{v_t} S_t \mathrm{d}W_{1,t}, \\ \mathrm{d}v_t&=\kappa(\theta-v_t) \mathrm{d}t+\xi \sqrt{v_t} \mathrm{d}W_{2,t}, \end{align*} with $\mathbb{E}[\mathrm{d}W_{1,t}\mathrm{d}W_{2,t}]=\rho\mathrm{d}t$. Then, denoting partial derivatives by subscripts, we obtain from Ito's Lemma (which byouness mentioned in his comment) \begin{align*} \mathrm{d}f &= \left(f_t + \frac{1}{2}v_tS_t^2f_{ss} + \frac{1}{2}\xi^2v_tf_{vv}+\rho\xi S_tv_tf_{sv}\right) \mathrm{d}t + f_s \mathrm{d}S_t + f_v\mathrm{d}v_t. \end{align*} Following their definition as SDEs, the changes $\mathrm{d}S_t$ and $\mathrm{d}v_t$ can also be expressed in terms of $\mathrm{d}W_{1,t}$ and $\mathrm{d}W_{2,t}$ yielding \begin{align*} \mathrm{d}f &= \left(f_t + (r-q)S_tf_s + \kappa(\theta-v_t)f_v+ \frac{1}{2}v_tS_t^2f_{ss} + \frac{1}{2}\xi^2v_tf_{vv}+\rho\xi S_tv_tf_{sv}\right) \mathrm{d}t \\ & \;\;\;\;\; + f_s \sqrt{v_t}S_t \mathrm{d}W_{1,t} + f_v\xi\sqrt{v_t}\mathrm{d}W_{2,t}. \end{align*} The rest is identical to the derivation of the Black Scholes equation, we shall assume that both sources of risks can be eliminated by dynamic hedging forcing $\mathrm{d}f$ to be proportional to $\mathrm{d}t$. (choose the values you hold in the two assets such that the stochastic terms are zero). Thus, changes in $f$ are locally risk-free and hence, $\mathrm{d}f=rf\mathrm{d}t$.

We end up with the following linear, second-order, three-dimensional PDE \begin{align*} \frac{\partial f}{\partial t}+\frac{1}{2}v_tS_t^2\frac{\partial f^2}{\partial S_t^2}+\rho\xi v_tS_t\frac{\partial f^2}{\partial S_t\partial v_t}+\frac{1}{2}\xi^2v_t\frac{\partial f^2}{\partial v_t^2}+(r-q)S_t\frac{\partial f}{\partial S_t}+\kappa(\theta-v_t)\frac{\partial f}{\partial v_t}-rf=0. \end{align*}

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    $\begingroup$ Thank you for such a detailed explanation. $\endgroup$ – Ussu Oct 14 '19 at 19:22
  • $\begingroup$ @Ussu You're welcome. Just ask if further questions should arise. If you did find it helpful, please consider accepting it as an answer such that others can easily find the answer too. Thanks a lot! $\endgroup$ – Kevin Oct 14 '19 at 19:27
  • $\begingroup$ @KeSchn a general follow up question. In the screenshot in question, if you make the $dv$ term to zero using $\Delta_1 = \partial V / \partial V_1$ it looks like $\Delta $ should always be zero. Would this be correct? Or is $\frac{\partial V}{\partial V_1} \cdot \frac{\partial V_1}{\partial S}$ not simplifiable $\endgroup$ – Slade Oct 14 '19 at 21:07

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