# Tag Info

## New answers tagged heston

4

The numerical approximation of the call option price in the Heston model is notoriously unstable and can easily lead to imprecise answers for extreme parameter. Several different formulas exist for computing the price with some being more stable than others. The formula you are using is arguably one of the worst ones. The most precise algorithm I know of is ...

1

The Feller condition is not verified in your case: $2\kappa\theta>\xi ^ {2}$ If this condition is not verified, you can get negative variance as explained in this wikipedia article: https://en.wikipedia.org/wiki/Heston_model

5

Let \begin{align*} \mathrm{d}S_t&=\mu S_t\mathrm{d}t+\sqrt{v_t}S_t\mathrm{d}B_{S,t}, \\ \mathrm{d}v_t&=\kappa(\bar{v}-v_t)\mathrm{d}t+\xi\sqrt{v_t}\mathrm{d}B_{v,t}, \end{align*} where $\mathrm{d}B_{S,t}\mathrm{d}B_{v,t}=\rho\mathrm{d}t$. The market price of risk (or Girsanov kernel or Sharpe ratio) is ${\varphi}_t=\left(\frac{\mu-r}{\sqrt{v_t}},\... 2 Under Heston LSV (HLSV) dynamics, Gatheral's equality is: $$\sigma_{LV}^{HLSV}(S_t,t) = \sqrt{E^{HSLV}\left[V_tL(S_t,t)^2 | S_t \right]} = L(S_t,t)\sqrt{E^{HSLV}\left[V_t | S_t \right]},$$ as$L(S_t,t)$is$\sigma(S_t)$-measurable, where superscript$HSLV$is meant to remind us what is our dynamics we started with (in particular the joint probability ... 0 First, couple of corrections (I am not sure, just guessing):$X_T$- is it strike or price of forward underlying? Let it be strike,$X$and the underlying is$S_t$with forward:$Fwd_T=S_t/D_T$. Breeden and Litzenberger formula: No, B&L formula is this:$PDF(S_T)=D_T\cdot\frac{d^2C(X)}{dX^2}$, where$D_T$is discount factor. Finally, my recipe to ... 3 The replicate function works best when you fully define your discretization scheme within a function. Then you can simply replicate the function-call x amount of times. Also, try and keep code duplication to a minimum and improve your general syntax. This will help you and your peers that might need to review and/or change your code in the future. ... 8 Bad news: Your calculation is not quite correct As you say, the initial price of a European call option is $$C(S_0;K,T)= S_0e^{-qT}\Pi_1-Ke^{-rT}\Pi_2. \tag{\star}$$ However, the exercise probabilities$\Pi_1$and$\Pi_2$depend on the stock price$S_0\$ too! Thus, you need the product rule and the chain rule to differentiate the option price with respect ...

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