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4

Feynman–Kac Theorem: Assume that $F$ is a solution to the boundary value problem \begin{align} &F_t+\mu(t,x)F_x+\frac{1}{2}\sigma^2(t,x)F_{xx}-rF=0\\ &F(T,x)=\Phi(x), \end{align} Assume furthermore that the process $e^{-r_s}\sigma(s,X_s)F_s$ is in $\mathcal L^2$ where \begin{align} dX_s=\mu(s,x)ds+\sigma(s,x)dW_s, \end{align} then $F$ has the ...


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Your question is not clear. What you might want to say is what distribution should the futures price follow, under the risk-neutral or physical probability measure. In this sense, it will depend on your intention. For potential future exposure, you may want to use the physical measure for the price evolution, while the distribution will depend on your model ...


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Answering my own question as it could be useful for others. Actually package fOptions is vectorized. The only constraint (and that make sense) is that you can't compute at the same time 2 different greeks, or mix up calls and puts. So assuming that you want to compute the delta of a set of puts, the code will be the following: ...


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It is an attempt to make a general statement that is not entirely correct, and certainly will become increasingly incorrect as new products are introduced. It is true that individual stock options traded in the US are American exercise. And the index options that are traded on the CBOE are European type, however there are also index options traded on the ...


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Let $\{P_t \mid t \geq 0\}$ be a compound Poisson process, where \begin{align*} P_t = \sum_{i=1}^{N_t} (V_i -1), \end{align*} and $N_t$ is a Poisson process with intensity $\lambda$ and jump times $\tau_i$, $i = 1, \ldots, \infty$. Let $Y_i=\ln V_i$ and $f(x)$ be the density function. Then \begin{align*} P_t - \lambda t E(V_1) &= P_t - \lambda t ...


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In general, $v = \frac{\partial C}{\partial \sigma} > 0$ and $\theta = \frac{\partial C}{\partial t} < 0$. If maturity $T$ increases than $C$ increases. Suppose volatility is non-constant. Then if $T$ increases, the option value is more volatile, since the stock price is more volatile. Since $v > 0$ the option price must increase. He claims that ...


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Without loss of generality, we can assume $y>0$.let $x =\ln S_t$ and defining $\tilde{F}(x, t)=F(S, t)$ we have $$\int_{0}^{\infty}(\,{F}(e^yS_{t^{-}},t)-{F}(S_{t^{-}},t)\,)k(y)\,dy=\int_{0}^{\infty}(\,\tilde F(x+y,t)-\tilde F(x,t)\,)k(y)\,dy$$ Beginning,it is natural to look at the following interval first $$\int_{0}^{\Delta ...


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The "right" thing to do is to treat the options as derivative contracts. Let's say for simplicity that you are using Monte Carlo to compute VaR. Then you would simulate the equity prices on each iteration, and then apply an option-pricing formula to get the corresponding option prices on that iteration. This lets you obtain an accurate simulated portfolio ...


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In the Black-Scholes Model or Heston Model, the American option satisfies the same PDE, but with different boundaries.For an American call option $C_A(S,\tau )$, we can therefore write \begin{align} \frac{\partial {{C}_{A}}}{\partial \tau }=+\frac{1}{2}{{\sigma }^{2}}{{S}^{2}}\frac{{{\partial }^{2}}{{C}_{A}}}{\partial {{S}^{2}}}+(r-q)S\frac{\partial ...


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There is no closed formula for American put option. However, there is an analytic solution for perpetual American put option. The only difference is that the maturity of the perpetual American option is infinite. Why that makes such a difference? That's because we can determine the optimal stopping time (and therefore optimal exercise price) if we don't ...


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it all comes down to how you define analytic. If you push the definition far enough there are some. An exact and explicit solution for the valuation of American put options DOI:10.1080/14697680600699811 Song-Ping Zhu pages 229-242 However, it's an infinite sum of recursively defined double integrals.


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Yes, there is none. Quoting Higham (2004): "The mathematical problem defined by (...) is much more difficult than the Black–Scholes PDE that arose without the early exercise facility. In general, there is no closed form expression for $P^{Am}(S, t$) and we must use numerical methods to obtain approximate values." Where (...) refers to the American Option ...


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Maybe it is better to use martingale theory to characterise whether it is an equality or not. Let $S_t$ be a (right)-continuous positive martingale with $S_0 < H$. Let $\tau = \inf \{ t > 0| S_t = H \}$. The option pays 1 unit of cash at $\tau$, and there is no maturity (perpetual option). What is the price of the option? I.e. compute $$P_0 = ...


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in the continuous case, you can regard the dividend rate as the interest on a foreign bank account if we invest it so the number of shares grows at the rate $d.$ So we can think it as a call option on a foreign exchange rates. Now calls and puts are the same thing in foreign exchange just by changing viewpoint. So the pricing is just as hard for calls as for ...


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If $\mu$ is large, then it is more likely for the call to finish in the money. Your and my intuitions suggest that this means that the option is more valuable. But this is wrong. A call option is an insurance policy. A call option is useful because it protects you in the case that the value of the stock goes down. That is why call options are valuable for ...


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In optimiazation system, you have to weight the price for the different maturities in a way that reflect your confidence in each data point (influenced by liquidity). One way to do so is to weight, each price by its Black-Scholoes Vega (see Tankov (2003)). So when minimazing the squared differences of the sum your weighted option prices, you can use the ...



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