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## Hot answers tagged options

4

Derivatives pricing models are not predictive. They simply extract information about the market’s expectations embedded in the prices of market instruments to which they are calibrated. This information can then be used to price other more complex derivatives. Calibration is important for hedging purposes. Suppose you ask your model for the hedging ratio of ...

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overall gamma is second derivative of whole portfolio over underlying. adding any function (such as underlying*constant) which second derivative is 0 does not alter overall second derivative.

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You can compute expectation of drifted processes as well and derive same pricing formulas,but usually its more complicated (compare derivation of Black Scholes using martinglaes and through PDE. PDE proof ,where drift is explicit, is much longer) With martingale representations you have more analytical mathematical tools/formulas available (e.g. barrier ...

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How about just defining the maturity date as todays date (or any other start date) ajusted by a period of T x 365 days? Here is an example: T = 0.5 today = ql.Date().todaysDate() maturity = today + ql.Period(f"{int(T*365)}d")

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To hedge your long call option (which as a delta between 0 and 1), you had to short sell some stock. If the stock price crashes the option you are long of is less in the money (or further out of the money). Therefore you are overhedged by your short stock position and need to buy back some stock. You can also think in terms of gamma. Your long vanilla ...

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for an intuitive answer, if we start with a vanilla call as our base, then with an up & out call, we would like the underlying to go up in price yes. But as the price increases, we also increase the probability of kicking out and losing our payout - so we don't want it to go up too much. If the barrier is so far away that the probability of reaching it ...

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Standard call options are trivially more expensive than up/down and out call options. However, for high strikes, down and out options will very likely never be knocked out, therefore their prices should be close to standard call options. For low strikes, down and out call options are almost worthless, therefore , the down and out call options curve price ...

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The digital option pays $H$ at time $T$ if $S_T \geq K$ , so its option time at time $t$ is given by $$V_t=E_t\left[e^{-r(T-t)}H 1_{\{S_T \geq K\}}\right]=e^{-r(T-t)}H* P_t(S_T \geq K)$$ The model used is Black-model, that $$dS_t=rS_tdt+\sigma dW_t$$ or $$S_T=S_te^{\left(r-\frac12 \sigma^2\right)(T-t)+\sigma (W_T-W_t)}{}$$ Calculate $P_t(S_T \geq K)$ ...

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Recall that the price of your contract is \begin{align*} V_t = e^{-r(T-t)} \mathbb{E}^\mathbb{Q} [H1_{\{S_T>K\}}|\mathcal{F}_t] \end{align*} because your option always pays $H$ if $S_T>K$. Next, \begin{align*} V_t &=He^{-r(T-t)} \mathbb{E}^\mathbb{Q} [1_{\{S_T>K\}}|\mathcal{F}_t] \\ &= He^{-r(T-t)} \mathbb{Q} [{\{S_T>K\}}|\mathcal{F}_t] \\...

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Your question is in fact one on the linearity of the replication cost of an option. Let formulate it a general way: once you can express the replication cost $C$ of a payoff as a function of several factors $X$, the strike $S$ and the volatility $\sigma$ that you assume to be a function of the strike, you are asking if \frac{1}{N}\sum_\ell C\big(X, S_\ell,...

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I assume you work in the Black Scholes framework. Then, \begin{align*} P(S_0,K,T) = Ke^{-rT}\Phi(-d_2)-S_0\Phi(-d_1), \end{align*} where \begin{align*} d_1 &= \frac{\ln\left(\frac{S_0}{K}\right)+\left(r+\frac{1}{2}\sigma^2\right)T}{\sigma\sqrt{T}}, \\ d_2 &= \frac{\ln\left(\frac{S_0}{K}\right)+\left(r-\frac{1}{2}\sigma^2\right)T}{\sigma\sqrt{T}}= ...

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$N\left(d_2\right)$ is the risk-neutral probability that the spot is greater than the strike at maturity, therefore the RN probability that you get your payoff.

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