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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


2

1 - The Neumann boundary condition is actually named after Carl Neumann, not John von Neumann. There is another boundary condition not often mentioned but used very often in practice in Quant finance FD solvers, which is linear (zero second spatial derivative on the boundary). This means that on the boundary the PDE $a\frac{\partial U}{\partial x} + b \frac{...


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}},\...


6

I'll give a heuristic "proof" for general European claims which will cause mathematicians to feel sick, but which physicists / practitioners would probably be quite happy work with: Write the Black-Scholes PDE as $$ \frac{\partial F}{\partial\tau}(\tau) = \mathcal{A} F(\tau) $$ with $\tau = T- t$, and the operator $\mathcal A$ is defined as $$ \...


9

Consider any option, vanilla or exotic. In between fixing dates it satisfies the Black & Scholes PDE (for simplicity zero interest rate and dividends) $$ \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 U}{\partial S^2}(S,t)+\frac{\partial U}{\partial t}(S,t)=0 $$ Let ${\cal V}(S,t) = \frac{\partial U}{\partial \sigma}(S,t)$ be the option vega. Differentiating ...


2

On page 119 in Björk (3rd edition) we have the replicating portfolio (equations 8.20 and 8.21): Hold $\frac{\partial C}{\partial s}$ of the stock and $\frac{X_{t}-S_{t}\frac{\partial C}{\partial s}}{B_{t}}$ in the bank-account. The dynamics of this portfolio is given by $$ dX_{t}=\frac{\partial C}{\partial s}dS_{t}+\frac{X_{t}-S_{t}\frac{\partial C}{\partial ...


2

Oh i think i got it. We are only considering the time step $dt$, which is why, first of all we need $\frac{1}{2} \frac{\partial^{2} G}{\partial S^{2}} d S^{2}$, and also why we remove: $\frac{1}{2} \frac{\partial^{2} G}{\partial t^{2}} d t^{2}+\frac{\partial^{2} G}{\partial S \partial t} d S d t$


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 ...


4

You could but there are difficulties associated with this approach. The main one is that $\tau$ is stochastic, ie it is different for different paths of $S$, so the standard Black-Scholes formula does not apply. For example some $\tau$s you need to check are of the form $\tau =\inf\{t : S(t) <B\}$ in which case you need to value a barrier option with ...


6

As Frido Rolloos remarks this is not the case. Let me suggest three ways to confirm this: Look at the Black-Scholes formula (from Wikipedia): $$C(S, t) = N(d_1)S_t - N(d_2)PV(K)$$ where $d_1 = \frac{1}{\sigma \sqrt{T-t}}\Big[\ln(S_t/K) + (r + \sigma^2/2)(T-t)\Big]$, $d_2 = d_1 - \sigma \sqrt{T-t}$ and $PV(K) = Ke^{-r(T-t)}$ and $N(\cdot)$ the cumulative ...


2

\begin{align} d \left(e^{-rt} \left(V^i - V^a \right)\right) &= \left(d e^{-rt} \right) \left(V^i-V^a \right) + e^{-rt} d(V^i - V^a)\\ &= (-e^{-rt} r dt) (V^i - V^a) + e^{-rt} (dV^i - dV^a) \\ &= e^{-rt} [ -r (V^i - V^a)dt + (dV^i - dV^a) ] \end{align} So multiplying everything by $e^{rt}$ gives the result.


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