New answers tagged greeks
3
The greeks are non-linear and only give you the instantaneous rate of change. The larger the change in underlying is, the less accurate the change based on the greeks will be. A doubling of the underlying will certainly not be predictable by the greeks alone. Delta and Gamma can be used to estimate the new price using a second-order taylor series ...
4
In addition to the good answer given below: think of options with almost no maturity left (i.e. 1-day before expiry): for these options, the time value of the option is almost zero, so the option value as a function of the underlying has almost fully converged to the "step-function" of the pay-off at maturity (i.e. the classical "hockey-stick&...
4
Your greeks are derivatives, not absolute price differences, when your underlying changes by 1%. Also, the put price and delta are not linear in the underlying (in fact in your case they are highly non linear) and you cannot expect for a large change (which the 1% change is) that the delta increases by the linear amount that you expected.
You need to look at ...
0
I'm not sure what you mean by "cross" effects - the only correlation is that they both are functions of the change in underlying ($\Delta S$)
Delta PnL is $\Delta * (\Delta S)$
Gamma PnL is $(1/2) \Gamma * (\Delta S)^2$
Essentially the first and second terms of a taylor expansion
Vega and Theta are sensetivities to volatility and time, respectively,...
4
There is no difference in real world settings.
Market practitioners usually always price a digital as a tight call spread to capture skewness. For example, setting strikes at $$𝐾± = 𝐾 ±1/2𝑑𝐾,$$ in the limit of $𝑑𝐾 → 0$, the payoff approaches that of a digital. The reason is that a tight call spread, by using two vanilla options, effectively accounts ...
3
Here is another solution using Plotly.
First of all let me correct a typo in your code
def Vanna_(S, K, T, r, sigma):
lista = []
d1 = (np.log(S / K) + (r + 1/2 * sigma ** 2) * T) / (sigma * np.sqrt(T))
d2 = d1-sigma*T**(1/2)
return (1 / np.sqrt(2 * np.pi) * S * np.exp(-d1 ** 2 * 1/2) * np.sqrt(T))/S * (1- d1/(sigma*np.sqrt(T)))
Then let me ...
5
Something like this?
from mpl_toolkits import mplot3d
from itertools import product
S = np.linspace(100,120)
vols = np.linspace(0.05,0.7)
combs = list(product(S, vols))
values = [Vanna_(underlying, K, T, r, sigma) for underlying, sigma in combs]
x, y = np.hsplit(np.array(combs), 2)
fig = plt.figure()
ax = plt.axes(projection="3d")
ax.scatter3D(x,...
0
First, let's go back to basics to answer why theta can be both positive and negative, and why it's referred to as time decay? At it's core, an option's value is composed of two components:
intrinsic value, and
time value.
As time passes, the proportion of the 'time value' gradually decreases until the option is worth exactly its intrinsic value at its ...
4
It’s just the effect of interest. If you are long a deep ITM European put, it is worth the PV of K minus the stock price. But one day later the PV of K has grown a bit. That’s it. It’s the opposite for calls because you have to pay the K, so bringing the date closer costs you money. This is all assuming interest rates are positive.
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