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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 the Heston Model we have \begin{align} C(t\,,{{S}_{t}},{{v}_{t}},K,T)={{S}_{t}}{{P}_{1}}-K\,{{e}^{-r\tau }}{{P}_{2}} \end{align} where,for $j=1,2$ \begin{align} & {{P}_{j}}({{x}_{t}}\,,\,{{v}_{t}}\,;\,\,{{x}_{T}},\ln K)=\frac{1}{2}+\frac{1}{\pi }\int\limits_{0}^{\infty }{\operatorname{Re}\left( \frac{{{e}^{-i\phi \ln K}}{{f}_{j}}(\phi ;t,x,v)}{i\phi ...

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Take a look at Hull's Appendix of the Volatility Smiles chapter. (Chapter 16 in my version). It gives a method to calculate the probability density function based on option prices: $$g(K) = e^{rT} \frac{\partial ^2 c}{\partial K^2}$$ This result comes from the Breeden Litzenberger 1978 paper.

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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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The answer is: yes. We can consider a model that assumes there is only one jump with distribution $p$, and otherwise the stock value does not change. Then for $p$ to be a martingale measure the only condition is on expectation of $p$. Hence, any distribution with desired expectation can be a marginal of some pricing measure.

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I would also say that the pricing of some exotic products require to compute expectations of functions of the random variable at consideration, and these functions may grow more than linearly : you need finite moments in order for the prices of these exotic derivatives to be bounded.

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$S_0$: The stock price today. $p$: The probability of a price rise. $u$:The factor by which the price rises. $d$: The factor by which the price falls. Three equations are required to be able to uniquely specify values for the three parameters of the binomial model. Two of these equations arise from the expectation that over a small period of time the ...

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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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