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18

Gamma is the second partial derivative of the change in the price of the option wrt to the change in the underlying. Said another way, it is the change in delta. If you write down the Black-Scholes pricing formula, you's see the gamma term: $$...\frac{1}{2}\frac{\partial^2C}{\partial S^2}(\Delta S)^2...$$ Notice that the $\Delta S$ (change in stock price) ...


15

Find the topic of model-independent properties of option prices very interesting as well. Here are some results that I am aware of and the respective references in the literature. Some are already contained in your initial list as well. Plain Vanilla Prices are Convex in the Strike Theorem 4 in Merton (1973). Delta is Bounded by the Slopes of the Payoff ...


13

You are absolutely right to point out that most proactive participants in options markets prefer to be long gamma, and it is typically reactive market makers who take the opposite side of their trades. While the typical options trader (I find it difficult to call anyone trading options an "investor") does not hedge his position, market makers will attempt ...


13

This is in fact a tricky matter. As you say one way is to calculate delta by an analytic formula, i.e. calculate the first derivative of the option pricing formula you are using with respect to the underlying's spot price. The second way is to do it numerically, i.e. change the spot price by a small value $dS$, calculate the value of the option and then ...


13

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.). ...


9

You need to compute your greeks as finite differences, but the full procedure may be pretty tricky. I will use vega $\aleph$ as the example here. Let's begin by designating your Monte Carlo estimator as a function $V(\sigma,s,M)$ where $\sigma$ is the volatility as usual, $s$ is the seed to your random number generator, and $M$ is the sample count. To ...


8

VIX is calculated from a basket of SPX options, and VIX futures expire into following expiration, e.g. September VIX futures that will expire next Wednesday will use SPX October options chain to calculate settlement value. If $B$ is the value of the basket then VIX value at expiration is $\sqrt{ B }$. Then VIX futures price is the expectation of the basket $...


8

Think of moving volatility in the other direction. As volatility approaches zero, any call strike strictly smaller than the ATM strike, $K<K_{ATM}$, will have zero probability of ending in the money, and the corresponding option value will be zero. An infinitesimally small change in stock price will not move $K$ past $K_{ATM}$, so the option value ...


8

Automatic Differentiation (aka AD) is a family of methods that are used to evaluate the derivative of a coded function. These methods are far more accurate than finite differences, since they are theoretically exact in the absence of floating point roundoff error. AD is, however, subtly different than symbolic differentiation. The key difference here is ...


8

Under the Black-Scholes model, \begin{align*} Gamma &= \frac{N'(d_1)}{S \sigma \sqrt{T-t}}\\ Vega &= SN'(d_1) \sqrt{T-t}. \end{align*} Then, it is easy to see that \begin{align*} Vega = S^2 \sigma (T-t) Gamma. \end{align*}


8

For any process with independent increments, by the very fact of statistical independence the variance of $x_{t3}-x_{t1}$ is going to be the sum of the variances of $x_{t2}-x_{t1}$ and $x_{t3}-x_{t2}$ for $t1\leq t2 \leq t3$. Many processes have independent increments, including ABM, GBM, Poisson, etc. Then if you add a homogeneity assumption (the ...


7

Short gamma is being of the view that realized volatility would be less than the implied volatility for the period in which an option is valid. So if you think realized volatility in the future would be consistently lesser than implied volatility at present, then you'd be short gamma. The premium one would receive by selling an option (call or put) is a ...


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

No, you should not expect such a relationship to hold in general. The reason is that American options have an "exercise barrier" which European options don't, and this results in different prices and greeks. In the case of put options (with interest rate $r>0$) as the spot price falls, at some point it becomes optimal to exercise early and take the cash. ...


6

Short Answer : Futures don't have Greeks Long Answer : Assuming a non strictly mathematical (i.e. false) point of view. Well, having Greeks on VIX Futures is not relevant, VIX value is itself a "Greek" (and Futures don't have Greeks). Sensitivity to Price of the Underlying : Insensitive (ν = 0) Volatility of the Underlying : Delta Δ = 1 (to Volatility of ...


6

As far as PDEs (deterministic) are concerned we have the notion of a "strong solution" (directly solving the differential operator in the strong formulation of the problem) and the "weak solution" that deals with a weak formulation of the problem. For the strong formulation, finite differences are the way to go since they are the natural discretization of ...


6

First, my notation. $K$ is the strike price, $S$ is the stock price, $r$ is the continuously compounded risk-free rate, $T$ is time at expiration, $t$ is time at issue, $\sigma$ is volatility, $\delta$ is continuously compounded dividend rate. The Black-Scholes formula for a European call is $C = Se^{-\delta (T-t)} N(d_1) - Ke^{-r(T-t)} N(d_2)$ $d_1 = \...


5

For non-interest rate derivatives with not-so-long maturities worrying about rho is uncommon. Think about it: interest-rates do not change that often relative to options expiring next week, next month or at most next year. LEAPS are obviously another turf. You could think about gamma, but the intimate relation of gamma and vega (at least in BS model) makes ...


5

Joseph de la Vega wrote Confusion of Confusions in 1688, probably the World's first descriptive text on stock market processes and volatility. I'm not sure that this is why Vega is thus named, but I like to think it's in his honour.


5

FDMs represent PDEs over a simple grid shape; the different implementations are just different recurrence relations to approximate the solutions to the PDE between boundary values (e.g., for options pricing, $T=[t_\mathrm{now},t_\mathrm{maturity}]$ and $S=[\mathrm{deep\_itm},\mathrm{deep\_otm}])$. FEM is a general name for a lot of different ...


5

most models in financial maths are linear so prices and Greeks just add. This is in particular true of Black--Scholes so Yes. However, once one starts taking into account value adjustments non-linearities appear and it is a lot more complicated.


5

Simply put, no. Vega depends on a variety of factors (including the level/price of the underlying asset). However, vomma/volga/vega convexity (whatever you want to call dVega/dIV) is always positive. So as IV increases, the vega of an option increases - I think this might have been what you were getting at. It's important to understand that IV is an input ...


5

if you have a portfolio of calls and puts with the same maturity then your portfolio is gamma neutral if and only if it is vega neutral. The reasons is that the BS gamma divided by the BS vega is a function of $S$ and $T$ that does not vary with $K.$ So if you construct a linear combination that has zero gamma then the vega is zero too, and vice versa.


5

If you want to know what Greeks the market assigns to an option, i.e. the market implied Greeks, then you would use the implied volatility. And that is what traders like to look at.


5

The risk exposures/sensitivities of long and short positions always have different signs. This has to hold since derivatives are zero sum games. Vega is always positive for a long position in a European plain vanilla option (or any convex payoff in general). This is true even when the option is already in-the-money. As volatility increases, the probability ...


5

This question has been asked many times and some clarifications appear needed. As pointed out in an answer to this question, the portfolio \begin{align*} \Delta_t^1 S_t + \Delta^2_t C, \end{align*} where $\Delta_t^1 = -\frac{\partial C}{\partial S}$ and $\Delta_t^2 =1$, is, generally, neither self-financing nor locally risk-free. To derive the Black-...


5

Something went wrong in the third equality of the equation where you compute $\partial C_0 / \partial K$. Starting from the second equality, you can use that \begin{equation} S_0 \mathcal{N}' \left( d_1 \right) = K e^{-r T} \mathcal{N}' \left( d_2 \right), \end{equation} see e.g. Equation (1.29) in Wystup (2006). Alternatively, you could use the ...


4

The most general answer is to shift your input to approximate the first derivative. Given that you need Monte Carlo to price this, it may get expensive. But that's the way it goes as when you have no analytical solutions as there aint't no free lunch ...


4

If your "European vanilla options" are restricted to piece-wise linear pay-offs, then the following may help: Remark: I assume you are looking for a rule of thumb to get the profiles without the use of a computer. All piece-wise linear pay-offs can be decomposed into a sum of digital options and call options with different notional (possibly negative) and ...


4

I think you are answering your own question. Hull states: "When $\Theta$ is large and positive, $\Gamma$ tends to be large and negative and vice versa." In practice, you can expect $r(V-S \Delta)$ to be quite small.


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