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To answer your question consider the following example using actual prices for SPY ETF on 7/31/15: "hopey.netfonds.no" By looking at the last 19 trades that occurred at the very last second, you will see a notable price movement on prices. If you go to Google/Yahoo Finance the Closing Price for the ETF is 210.50 (largest trade at the close?) but the very ...


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first, there is a formula for the continuously monitored case. second, if you use log coordinates the Euler discretization is exact so this should be done. third, the convergence for discretely monitored to continuously is actually very slow so you will need a lot of steps. fourth, it's actually better to draw the hitting time to the barrier rather than ...


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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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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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You asked genuine question. Remember one thing always, volatility is considered good by derivative players but not by the market as whole. More volatile is market, more riskier it is. Since American option can be exercised at any time, whereas European options can only be exercised at the maturity only, so American options leads to excess volatility that ...


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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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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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For the case where $K^* \leq 0$, the option becomes trivial - the option payoff is 0 for a put and a linear function of the underlying values for a call. The valuation is then also trivial. For coding, you need to check the cases whether $K^* \leq 0$ or not, and then treat them differently.


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A digital call option (cash-or-nothing) can be replicated with two call options with different maturity. When we make the delta infinitely small and assume we have arbitrary strike prices. We get:


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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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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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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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Let's start with the main idea, I hope you can finish the computations yourself. Whenever you want to derive a pricing equation, try the following approach: discounted value of portfolio/option/derivative must be a martingale for non-arbitrage reasons. Since you have a Markovian dynamics in variables $t$ and $S$, you assume that the price is some function ...


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One solution is to calculate the annual dividend yield implied by that. $Div_{yield}=\delta=1.5/40$ and then replace the $r$ on $d_+$ by $r-\delta$. A cleaner way would be to compute it using a binomial tree.


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Generally no. Check page 21 of the following document: http://www.frankfurt-school.de/clicnetclm/fileDownload.do?goid=000000053103AB4


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Perhaps other memeber of qSE are going to correct me, but I think the following rule of thumb is useful. Whenever you have a doubt, try to forget that a pricing measure is a probability measure. This is just a pricing tool: originally for any option/derivative/contingent claim we'd like to know its price, so we introduce a map $\pi:X\to \Bbb R$ such that ...


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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 black-scholes the option price depends not on sigma^2 but on sigma^2 T. So if volatility is going to be 20% or 21% over the next 10 years (assume for simplicity no other values are possible, just these two with equal prob, but we don't know which) then that will have a bigger impact on the option value than a 20 vs 21 uncertainty for a 1 year option. That ...


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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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Trigger contracts certainly exist and sometimes the trigger is out of the money and so yes the holder loses out. I have seen traded swaps that cancel when a reference rate is passed and the cancelling is disadvantageous. Stop thinking of the contracts as "options" and start thinking of them as derivatives.


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First Question:This derivation is a special case of a PDE for general stochastic volatility models,described in books by Lewis (2000), Musiela and Rutkowski(2011) and others. The argument is similar to the hedging argument that uses a single derivative to derive the Black-Scholes PDE. In the Black-Scholes model, a portfolio is formed with the underlying ...


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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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