# Tag Info

16

This is an interesting and not so easy question. Here's my 2 cents: First, you should distinguish between mathematical models for the dynamics of an underlying asset (Black-Scholes, Merton, Heston etc.) and numerical methods designed to calculate financial instruments' prices under given modelling assumptions (lattices, Fourier inversion techniques etc.). ...

11

If your working modelling assumptions are such that the dynamics of the log price process $\ln(S_t)$ is space homogeneous, you have that the price of a European vanilla option is itself a space-homogeneous function of degree one. You can then appeal to Euler theorem to get the relationship you need. More specifically, define the price at time $t$ of the ...

11

A better, clearer, answer is to compute Lambda (leverage) of the option (link) and see if it is bigger or smaller than 1. Lambda is $\Delta \frac{S}{V}$ so we test $$\Delta \frac{S}{V} \lessgtr 1$$ which is what Joshi is saying in words: compare $\frac{1}{S}$ (delta to price for the stock, the delta of the stock is 1) to $\frac{\Delta}{V}$ (delta to price ...

10

Forward delta is 1 (defined as change in the value of the forward with respect to an instantaneous change in the price of the underlying, holding everything else constant). However for a meaningful discussion of the differences in forward and futures pricing, the forward price delta of forwards should be considered and it is exp(r(T-t)).Though the delta ...

10

the problem is that the pay-off has discontinuous first derivative. Try a contract with pay-off that is twice differentiable and it will probably work. The problem is that all the value comes from the tiny number of paths within $\Delta S$ of the strike, and these paths have huge value. This is a well-known problem. As the bump size goes to zero, the ...

9

Options have an asymmetric payoff profile: The payoffs are zero for almost all cases and positive else (as we well know). If the option is OTM, most of its payoffs are zero. A rise in volatility will hence increase the likelihood for instead positive payoffs from a change in the underlying price (i.e. delta increases). If the option is already ITM, most(...

9

Options on interest rates futures in the listed markets are always traded 1-yield (100-yield) just like the futures which are traded 1-yield. So negative rates aren't an issue and its always black volatility. In the OTC market, both normal and black volatility are quoted, but the common practice is to use black volatility is what is way more frequently used....

8

Gamma and vega have the same general shape , peaking at ATM and tapering to the tails. But gamma concentrate as the option gets closer to expiry (when vega is small). For options a long way from maturity, vega increases and gamma is small. Consequently for short dated options, if the price is close to strike, the option will have to be rehedged often (...

8

Intuitively, in a (log)-space homogenous diffusion model $$S_t \propto S_0, \forall t \geq 0$$ such that implied volatilities will only depend on the moneyness level and not on the absolute spot level, which is precisely the definition of sticky delta. Mathematically, consider a (log)-space homogeneous diffusion model (be it stochastic or not) \frac{... 7 [Mathematically] Risk-neutral pricing means that \begin{align} C_0(K,T) &= \mathbb {E}_0\left[\frac{1}{B_T} (S_T - K)^+\right] \\ &= \mathbb {E}_0\left[\left(\frac {S_T}{B_T} - \frac {K}{B_T}\right)^+\right] \end{align} Now simply notice that the dynamics of\tilde{S}_t := \frac {S_t}{B_t},\ \forall t \geq 0$$is independent of r (see the ... 7 I feel like your notations are not accurate enough to write what you would like to write. Let \Sigma(S;K,T) denote the implied volatility of a European vanilla of strike K and maturity T now that the underlying spot price is worth S. Sticky strike translates to$$ \Sigma(S+\delta S;K,T) = \Sigma(S;K,T) \iff \color{blue}{\frac{\partial \Sigma}{\...

7

You are long a vanilla option, so long gamma (positive gamma). If the stock price decreases, so does the delta of your option. Since you short-sold the stock to hedge, you now have short-sold too much since delta has decreased. As a consequence, you must buy back some stock.

7

You would be over hedged in your call position if it was delta neutral before the stock cratered. Since you are long delta on the call, you would have shorted stock to make the original position delta neutral. When the stock fell, your long call delta would have fallen, and you would buy to cover some of your short stock hedge. However, being long the ...

6

It is false. Here is an example. Let $$dS_t = rS_t dt + f(S_0) S_t dW_t,$$ $$dB_t = r dt.$$ The price is then the Black-Scholes price with volatility $f(S_0).$ The delta is the BS delta plus $$f'(S_0) \times \operatorname{BS Vega}.$$ Picking $f$ appropriately, we can make the Delta as big as we like. Note that the example is highly artificial in ...

6

I think there is confusion around the forward price and the value of a forward contract. A forward contract obligates an exchange of an asset at some future time $T$. By convention, this forward contract has initial value zero (at time $0$). The forward contract, being an exchange of an asset for a set dollar amount in the future, has at some $t \in [0, T]... 6 The time to expiry is required, but it's included in the inputs: the two discounts$e^{-rT}$and$e^{-qT}$and the standard deviation$\sigma\sqrt{T}$. You might argue it could be documented more clearly, and I might agree with you. 6 This is a little more complicated than the answer provided above since this is FX and the convention for determining the strike matters. https://www.researchgate.net/publication/275905055_A_Guide_to_FX_Options_Quoting_Conventions Most pairs take premium in the foreign (i.e. left hand side) currency. This means that you are paying for an option in the ... 6 Basically, the author is saying that the delta of an option,$dC/dS = \frac{\partial C}{\partial S} + \frac{\partial C}{\partial v}\frac{\partial v}{\partial S}$, where the$\frac{\partial C}{\partial S}$is the delta assuming constant volatility, the$\frac{\partial C}{\partial v}$is the vega of the option, and the$\frac{\partial v}{\partial S}$... 6 Only constrained to be <1 in the simplified Black-Scholes setting with zero cost of carry on the underlying. In the more realistic and common setting where the cost of carry of the underlying is higher than the discounting rate, then it is entirely possible for a call to have a delta > 1. This is the case because your future costs are proportional to ... 6 Just a small addendum to @noob2's answer. The discrete shape of$\lambdais: $$\lambda \approx \frac{V_1 - V_0}{S_1 - S_0} \times \frac{S_0}{V_0}$$ which can be rewritten as $$\lambda \approx \frac{\frac{V_1 - V_0}{V_0}}{\frac{S_1 - S_0}{S_0}}$$ which is as @noob2 said just the ratio of the relative returns for option and stock. 5 The strangle vol defined in your formula \begin{align*} Strangle(∆) = 0.5[Call Vol(∆) + Put Vol(∆)] - ATM Vol \end{align*} is the smile butterfly volatility. Then you have the volatility quote. Your confusion is caused by the misuse of notations. Note that, other treatments are also available. See for example, FX Volatility Smile Construction by UWe Wystup ... 5 You are looking for the Greek commonly referred to as Charm. This is a quick visualization with a good chart I found on Google: https://www.optiontradingtips.com/greeks/charm.html 5 In FX world, the ATM strike is the delta-neutral strike, that is, the absolute delta values of a call and the corresponding put are the same. Moreover, the delta can be premium adjusted or not depending on the particular currency pair. See the linked paper as mentioned by @AntoineConze. For AUD/USD, the delta is not premium adjusted, and then the delta-... 5 As @ir7 did, I only briefly want to add to @noob2's spot-on answer. He's of course right and\Lambda=\Delta\frac{S}{V}$decides how risky the option is compared to the stock. Firstly, note that$\Lambda=\frac{\frac{\partial V}{V}}{\frac{\partial S}{S}}=\frac{\partial V}{\partial S}\frac{S}{V}$. An economist would call$\Lambda$an elasticity. It tells you ... 5 The delta is only 1 if the option is certain to be exercised. This is not the case if it is ‘slightly in the money’. If it is deep in the money, such that immediate exercise is optimal , then the delta is 1. 4 I've started thinking about this, too. My gedanken conclusion turned out to be too simple once I found what I was after: http://www.investment-and-finance.net/derivatives/o/option-beta.html, which I've confirmed in Black & Scholes (1973) p10 (eq 15). In short: $$\beta_{\text{option}} = \frac{S\cdot\Delta}{O}{\beta_S}$$ where$S$is the underlying ... 4 In any arbitrage free model, you can define the BS implied volatility$\sigma_{BS}(S;T,K)$of the model by writing call prices as $$C_{Mdl}(S;T,K) = C_{BS}(S;T,K;\sigma_{BS}(S;T,K))$$ So the model's Delta is $$\Delta_{Mdl}(S;T,K) = \partial_S C_{Mdl}(S;T,K) = \Delta_{BS} + Vega_{BS} \times \partial_S\sigma_{BS}$$ The second term is a corrective ... 4 Since the volatility is not changing, we can assume that the only change is the underlying asset price$S\$. Then \begin{align*} C(S+\Delta) &\approx C(S) + Delta \times\Delta +\frac{1}{2} Gamma \times \Delta^2 \\ &=11.50 + 0.58 \times 0.5 + \frac{1}{2}\times 2 \times (0.5)^2\\ &=12.04. \end{align*}

4

Just to strengthen the intuition in the perfect answer above: With r going very high (and hence F), all prices on cash instruments are expected to gain fast with time (to compensate for the carry) and the call-strike is expected to be deep[er] in the money; hence with a delta close[r] to one.

4

Here is a page from The Options Guide with an understandable picture. They explain, As the time remaining to expiration grows shorter, the time value of the option evaporates and correspondingly, the delta of in-the-money options increases while the delta of out-of-the-money options decreases. This happens because the shorter expiration that is deep in-...

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