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19

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 ...


14

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.). ...


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 ...


10

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 ...


9

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

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

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 ...


8

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. ...


8

Using our good friend Taylor, we know that \begin{align*} C(S+\Delta_S)\approx C(S)+\Delta_C\Delta_S+\frac{1}{2}\Gamma_C(\Delta_S)^2, \end{align*} where $\Delta_C$ and $\Gamma_C$ are the call's sensitivities and $\Delta_S$ a small change in the price of the underlying asset. In your example, $\Delta_S=1$ and thus, \begin{align*} C(52+1) &\approx 5.057387 ...


7

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 = \...


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

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

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.


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

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 ...


6

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 (...


6

Put-call parity says that a call and put (worth $C$ and $P$ respectively) with the same strike $K$ have the following relationship with the spot rate $S$, risk-free rate $r$, and time to maturity $T$ -- $$C - P = S - e^{-rT} K$$ Taking the first derivative with respect to $S$, $$ \frac{\partial C}{\partial S} - \frac{\partial P}{\partial S} = 1 $$ which ...


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

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

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

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 ...


5

With a long time to maturity, your options have a low theta because their time value decays quite slowly. If there are many months to go, the passage of one day does not change the exercise probabilities too much, whereas short life options with only a few days left have a much higher time value decay. Hence, the larger the time to maturity, the lower theta. ...


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 ...


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