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I managed to figure out another way of doing it via change of measure as follows... We know that the dynamics of $S_t^2$ is given by, \begin{align} S_t^2&=S_0^2 \text{exp}\left( \left( 2r-2q-\sigma^2\right)t+2\sigma^2 W_t^Q \right)\\ \Rightarrow \frac{1}{S_t^2}&=\frac{1}{S_0^2} \text{exp}\left( -\left( 2r-2q-\sigma^2\right)t-2\sigma^2 W_t^Q ... 2 LetI= \mathbb{E}_t^\mathbb{Q}\left[\text{exp}(-2\sigma W_{T-t}) \cdot\mathbb{1}_{S_T\ge K}\right] = \frac{1}{\sqrt{2\pi}} \int_{\hat{d}_2}^{\infty} e^{-2\sigma x} e^{-x^2/2} dx.$$So$$ I = \frac{1}{\sqrt{2\pi}} \int_{\hat{d}_2}^{\infty} e^{-(x-2\sigma)^2/2} dx \, e^{2\sigma^2}. $$Change variables y = x-2\sigma and you are done. 0 To expand on Randor's answer, the standard Black-Scholes formula as you've given it assumes a constant continuous dividend yield of q. To adapt this to cope with discrete deterministic (absolute) dividends d_i at known times \tau_i, you could recast the formula in terms of the "dividend-free" stock price:$$S^* := S - \sum_i d_i e^{-r\tau_i}$$and ... 0 When valuing a plain index option, there are two options in terms of index dividend: (1) The underlying price is a spot price like in the FTSE 100 case (option is valued off the index): you can use continuous dividend yield. You can imply a dividend yield from a linearized call-put parity: The present value of the dividend payment is ... 1 Ftse100 would not have a smooth dividend yield, as your formula has, it would be discrete, being much higher on certain days of year than others. In pricing options on ftse, u need to take into account implied dividends (dividends that are implied by put call parity) 1 Sorry to disagree but if interest rates is 0, the binary is still not worth 1 now. Suppose spot S(0) = 100, assume x = 110 and upon touch (whenever it happens as the option has no maturity) you receive one dollar. Suppose I buy 1 stock. If the barrier hits, i sell the stock and receive 110 USD. What if I buy N stocks at t=0? upon hit of barrier i ... 0 Instead of just considering a parallel shift of the whole volatility surface, you can decompose the surface into maturities/strikes domains, so called buckets and consider Vega buckets which are sensitivities wrt to bumps of each of these domains. The vol smile is often inter/extra-polated using a model calibrated to market prices, e.g. the SABR model or ... 0 since vega is the sensitivity to a parallel shift of the entire vol surface, why do you not simply bump the entire input surface all at the same time? You can bump all your vanillas simultanous then recalibrate your model to the bumped surface. The use your model to reprice, which gives you your vega. 2 I don't believe you will necessarily find a cite-able source as, I believe, this comes from a practical rather than theoretical motivation. As you know option prices are a function of: future prices, discount rates and implied volatility, volatility surface skew and other supple/demand factors. So when you are trading these instruments, you need to ... 1 When a pay-off is piecewise linear plus jumps, it the same as the portfolio of calls and digital calls. Its price must agree with that of the portfolio by no arbitrage. Every time there is a jump we add in a digital call and every time there is a change in gradient we add in calls equal to the gradient change. Here we have a call struck at K. Just below ... 1 Using the usual arbitrage arguments, we can write option prices as discounted expectations of future values under risk-neutral probabilities. That is$$ V(S,0) = B(0,T) E\left[ V(S,T) \right] $$Start by re-writing your particular payoff as the following sum$$ C_K+aC_{2K}+KD_K $$where C_x is a call struck at x and D_x is a digital option struck at ... 2 The term of art in our industry for this type of option pricing formula is a series solution. As Farahvartish indicates in the comments, a series solution is not considered to be an "analytical solution" due to the reliance on a converging infinite sum for actual numeric output.(*) Series solutions have been employed at least since the 1990s, when they ... 1 Essentially the question is importance sampling$$ \int f(S_T) \psi_{r}(S_T) dS_T = \int f(S_T) \psi_{\alpha}(S_T) \frac{\psi_r}{\psi_\alpha}(S_T) dS_T $$Here \psi_{\mu} denotes the log-normal density with drift \mu. So when you simulate with drift \alpha each sample used is$$ f(S_T) \frac{\psi_r}{\psi_\alpha}(S_T) $$instead of f(S_T). You ... 2 You can write$$\mathbb{E}\left[ \max(a X_T + b X_S -K,0)\right] = \mathbb{E}\left[ \max(a X_S Y_{S,T} + b X_S -K,0)\right],$$with Y_{S,T} = X_T/X_S. For a given value of X_S we can write$$\mathbb{E}\left[ \max(a X_S Y_{S,T} + b X_S -K,0)\right] = X_S \mathbb{E}\left[ \max(a Y_{S,T} + b -K/X_s,0)\right], since $Y_{S,T}$ is log-normal this can ...

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"It must dominate at zero" means that when the final spot level is zero, the value of the super-replicating portfolio must be greater than or equal to the value of the payoff, which is zero. Since the super-replicating portfolio consists of some stock (which has zero value when the spot price is zero) and some bonds (which have value one), there must be a ...

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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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yes if you use the BS Model for computing deltas and the same model for evolving the stock price then you should replicate the pay-off of any contract.

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