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I am currently working on a report regarding the put-call symmetry relations under the Heston model. I did all the math and managed to prove the relations using PDE approach. However, I wish to have a more intuitive interpretation of the derived relations.

Specifically, suppose a call option (European or American) with strike price $K$ and spot price $S_0$ is priced under the Heston dynamics with initial variance $V_0$:

$$ dS_t = (r-q)S_tdt+ \sqrt{v_t}S_tdW_t^1, $$

$$ dv_t = \kappa(\theta - v_t)dt + \sigma\sqrt{v_t}dW_t^2, $$

$$ \rho dt = dW_t^1dW_t^2, $$ its value will equal to the put option with strike price $S_0$ and spot price $K$ priced under the Heston dynamics with the following parameters:

$$ r_p = q, $$ $$ q_p = r, $$ $$ \kappa_p = \kappa-\rho\sigma, $$ $$ \theta_p = \frac{\kappa\theta}{\kappa-\rho\sigma}, $$ $$ V_{0,p} = V_0, $$ $$ \sigma_p = \sigma, $$ $$ \rho_p = -\rho. $$

My main question is: what is the interpretation or intuition of $$ \kappa_p = \kappa-\rho\sigma, $$ $$ \theta_p = \frac{\kappa\theta}{\kappa-\rho\sigma}, $$ and $$ \rho_p = -\rho. $$ Does anyone have an explanation for the changes in theses three parameters? What are the physical and financial implications? Thanks!

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This is a consequence of transforming a Put on $S_T$ with strike $K$ into a Call on $(K S_0)/S_T$ with strike $S_0$ under the stock measure. The new set of parameters $r_p$, $q_p$, $\kappa_p$, ... etc . are those that correspond to the Heston dynamics for the process $((K S_0)/S_t, v_t)$ under the stock measure.

General results on that kind of symmetry can be found in various papers for instance Peter Carr and Roger Lee, Put Call Symmetry: Extensions and Application (2007) http://math.uchicago.edu/~rogerlee/PCSR22.pdf.

In the Heston case you are looking at, start from the Put price as discounted expectation under the risk neutral measure $P$: $$ p = e^{-rT} E^P\left[(K - S_T)^+ \right] $$ Next rewrite it as $$ p = e^{-qT} E^P\left[\frac{e^{(q-r)T} S_T}{S_0} \left(\frac{K S_0}{S_T} - S_0\right)^+ \right]=e^{-qT} E^Q\left[\left(\frac{K S_0}{S_T} - S_0\right)^+ \right] $$ where $Q$ is the stock measure defined by the Radon Nikodym derivative $$ \frac{dQ}{dP} =\frac{e^{(q-r)T} S_T}{S_0} $$ Now apply Ito's Lemma to $X_t=\frac{K S_0}{S_t}$: $$ \frac{dX_t}{X_t}=-(r-q)dt - \sqrt{v_t} dW^1_t+ v_t dt $$ and finally apply Girsanov theorem to obtain the dynamics of $X_t$ and $v_t$ under $Q$: $$ \frac{dX_t}{X_t}=-(r-q)dt - \sqrt{v_t} dW'^1_t+v_t dt - v_t dt=-(r-q)dt + \sqrt{v_t} (-dW'^1_t) \\ d v_t = \kappa (\theta - v_t)dt + \sigma \sqrt{v_t} dW'^2_t+ \rho \sigma v_t dt = (\kappa-\rho \sigma ) \left(\frac{\kappa \theta}{\kappa-\rho \sigma } - v_t \right)dt + \sigma \sqrt{v_t} dW'^2_t$$ with $W'^1$ and $W'^2$ standard Brownian motions under $Q$ with correlation $\rho$, so that you are now pricing a call under new Heston parameters $r_p$, $q_p$, $\kappa_p$, ... etc. defined as in your post.

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  • $\begingroup$ Thank you Antoine for the reply. Now I have a better understanding of it. However, I am still a bit fuzzy on the intuition behind the change in the parameters. Or is the most intuitive explanation simply the one you stated in the first paragraph? $\endgroup$ – davidolohowski Apr 6 '18 at 18:11
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    $\begingroup$ Yes. The intuitive explanation is the 1st paragraph. If it helps, consider the case where $S_T$ represents the EURUSD FX rate instead of a stock. Then one clearly sees that a Put on EURUSD with strike $K$ is same as a Call on USDEUR with strike $1/K$, with notional $K S_0$. Now if you posit an Heston model $r$, $q$, $\kappa$, etc. for $S_t=$ EURUSD under the USD risk neutral measure, you deduce a new Heston model for USDEUR = 1/EURUSD = $1/S_T$ under the EUR risk neutral measure, with parameters $r_p$, $q_p$, $\kappa_p$, etc. as in your post, after applying Ito and Girsanov. $\endgroup$ – Antoine Conze Apr 6 '18 at 20:11
  • $\begingroup$ Thank you so much for the explanation. One more thing. I could not manage to derive the second Brownian motion for the volatility process using Girsanov, i.e. from risk neutral to stock measure. Can you possibly help me with that? Thanks again! $\endgroup$ – davidolohowski Apr 6 '18 at 20:20
  • $\begingroup$ From the Girsanov theorem the drift of any process $Y_t$ is adjusted by its quadratic covariation $d<Y_t,e^{(q-r)t} S_t/S_0>$ with the change of measure. $\endgroup$ – Antoine Conze Apr 6 '18 at 20:29

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