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The option payoff is equivalent to $Z_{\tau \wedge T}-1$ where $\tau=\inf\{t | Z_t = 1\}$ provided that $Z_t$ is assumed to be continuous. Since $Z_t=S_t/P_t$ is a martingale under $Q_P$, we have $E_P[Z_{\tau \wedge T}]=Z_0$ and the option value is $P_0 (Z_0 - 1)=S_0-P_0$ regardless of the model.

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I solved it the following way, just want make sure I'm not missing something obvious. Set up a portfolio $PF$ consisting of long $S$ and short $P$ at time $t = 0$. Choose arbitrary time $0 < t < T$. If $S_t > P_t$ then $PF_t = S_t - P_t$ which coincides with the value of the option. If $S_t$ hits $P_t$ from above, then dissolve the portfolio by ...

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The option payoff at maturity $T$ is defined by \begin{align*} (S_T-P_T)1_{\left(\inf_{0 \le t <T}\frac{S_t}{P_t}\right) > 1}. \end{align*} Let $Q$ be the risk-neutral probability measure and $E$ be the corresponding expectation operator. Let $Q_p$ be a probability measure defined by \begin{align*} \frac{dQ_p}{dQ}\big|_t = \frac{P_t}{e^{rt} P_0}. ...

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Thr second sum should be: $$\sum_{i=6}^{10}u_i = 0.10039773$$ This gives a mean of $0.0067648$ and a standard deviation of $\sigma=.028836$. To avoid these errors you should use something to automate your calculations. Something like a spreadsheet.

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This is what a exercise frontier would look like in American option. A more common name is "exercise region". This is the region where it's optimal your option. L is the optimal exercise price. It's a convex function of maturity. Far away from maturity, the optimal price is significantly lower than K because we'd expect a deep in-the-money intrinsic value ...

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The early exercise boundary (or frontier) for American puts is the level $S^*(t)$ where it is optimal to exercise the put if $S(t)<S^*(t)$. There is no known analytical formula for it, but it can be approximated in various ways.

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When the adjusted strike goes to zero or negative, it can be proven that the call option will always be exercised, therefore the price of a call is given by the discounted of the underlying and strike (as also mentioned by Gordon). This is like a forward therefore there is no need to compute d1 and d2.

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The payoff can be decomposed as \begin{align*} \phi(S) &= 100 \, I_{50 \le S_T < 100}\\ &= 100 \, \big(I_{S_T \ge 50} - I_{S_T \geq 100}\big). \end{align*} Note that, under the risk-neutral measure $P$, \begin{align*} E(I_{S_T \ge K} \mid \mathcal{F}_t) &= P(S_T \ge K \mid \mathcal{F}_t)\\ &= N(d_2), \end{align*} where \begin{align*} d_2 = ...

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In general, the implied volatility is based on vanilla European or American options. In your case, since the positions depend on only a single underlier, if you can have an analytical formula, or approximation, for each individual position, then, in principle, you can compute an implied volatility based on the market price of your portfolio. However, note ...

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suppose you sell a K = 105 call. When the stock reaches exacty 105 you buy 1 stock at 105. Now suppose the stock moves to 104.99, using your logic you sell 1 share at 104. You lost $0.01. Again, after a while stock reaches 105 you buy 1 stock. After some time it goes up, but eventually it goes down again below 105. Thus you sell 1 share below 105. Again ... 1 Certainly, you must agree that $$C_{T}-P_{T}=\left(S_{T}-K\right)^{+}-\left(K-S_{T}\right)^{+}=S_{T}-K.$$ Therefore, since $$C_{t}=e^{-r\left(T-t\right)}E_{Q}\left[C_{T}\right]\text{ and }P_{t}=e^{-r\left(T-t\right)}E_{Q}\left[P_{T}\right]$$ it follows by the linearity of$E$that$$C_{t}-P_{t}=e^{-r\left(T-t\right)}E_{Q}\left[C_{T}-P_{T}\mid ... 0 What is wrong with your broker watching your risk for you? I assume the "modest trading house" has portfolio margin in which case you already have limit up/down calculations done for you implicitly. So the output from your broker is your margin equity and you can make that your metric. Implementing a simple rule like - not to exceed 70% of total margin ... 0 Delta hedging implies, loosely speaking, buying a proportion (delta) such that small movements in underlying have no net impact. What you have done with 100% and 0% is, in effect, bought the shares to COVER your position, if the deal goes south. Let's work this out with an example. Say you have a stock trading at \$1 and you WRITE a call with strike \$10. ... 0 The value of an cash-or-nothing option is just the discounted expected payoff of the option. So the value of such a call should be exp[-r(T-t)]NP(S>K), where P(S>K) = N(d2), and N is the cash agreed to be paid. The asset-or-nothing is a bit more complicated since it is exp[-r(T-t)]E[S|S>K]. The last term is the expected value of stock price given that S > ... 0 Noir : I'm studying the same coursera course as you and wonder why you used a multiplier "q" = 0.7483 in calculating the futures lattice eg first row of second column from left 175.42 = 0.7483 * 178.77 + (1-0.7483) * 165.45 and not the calculated value of q equal to 0.4925. 4 It is not the fact that volatility is time varying that creates the skew per se, but the fact that volatility is negatively correlated with the spot. That is to say, as the stock/index price declines volatility will tend on average to increase, and vice versa. Time varying volatility itself would create a more symmetric 'smile'. Edit: Suppose that you ... 1 Yes, you are right. It appears to be a trivial typographical error in the book. I checked the formulas on Wikipedia https://en.wikipedia.org/wiki/Greeks_%28finance%29 and they agree with yours. The signs are obvious also since N(.) is between 0 and 1, i.e. non-negative. Now, about the reasoning starting with "from a logical point of view". Are you familiar ... 0 you are confusing too much with the future state of the market. It is easy to confuse. NO one can price totally uncertain future. Future or forward prices are arbitrage free projection of the current prices. No magic. It means you can buy a forward product and hedge it using products that defined the forward curve. Fair price is model price, meaning it is ... 0 In my opnion you should you the Full revaluation historical VaR. Please readmy thread . If you need more help on the same i can gudie you .Historical Value At Risk on option portfolio 0 You need to use more dimensions. If the number of dimensions (i.e. steps) is large, you may also have to use a Brownian bridge as described in the book by Joshi or Jäckel. 0 First let me say that in the Black-Scholes model as you have it, there is of course no need for intermediate steps when pricing vanilla calls, since the SDE has the closed-form solution you included. Intermediate steps would be required for complicated payoffs or other SDEs. To answer your question though, you do need to use additional dimensions. Think ... 1 If you have many strikes of european-exercise options for two dates$T_1$and$T_2$, then the option skew$\sigma_{1,2}(x)$implies model-free risk-neutral probability distributions$p_1, p_2\$ for each of these dates, $$p_i(x) = {\left. \frac{\partial^2 }{\partial x^2}\right|} BS_{\text{Call}}(S_0, x, \sigma_i(x), r, T_i, q)$$ ...

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You should also use make_shared() instead of calling new. See http://stackoverflow.com/questions/20895648/difference-in-make-shared-and-normal-shared-ptr-in-c

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I guess your question is more about model risk and model valuation. well you already mention all the ways that you can adjust your price. from my humble opinion, you should use implied volatility to get the correct value or do some calibrations on your parameters to get exact same price as market. Think of swaption for example. you can use either use black ...

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When I saw these curves they seemed very strange to me. I believe it is a data-quality issue.I went to Bloomberg and I retrieved the implied vols for 70 near ATM strikes of the weekly SPX options expiring November 27 2015 (I believe that is the yellow curve in your diagrams i.e. November 4th week). This was today 2015-oct-27 at about 15:00 New York time. As ...

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