# Tag Info

17

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.). ...

16

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 ...

15

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. ...

11

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 ...

11

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*}

11

Consider any option, vanilla or exotic. In between fixing dates it satisfies the Black & Scholes PDE (for simplicity zero interest rate and dividends) $$\frac{1}{2} \sigma^2 S^2 \frac{\partial^2 U}{\partial S^2}(S,t)+\frac{\partial U}{\partial t}(S,t)=0$$ Let ${\cal V}(S,t) = \frac{\partial U}{\partial \sigma}(S,t)$ be the option vega. Differentiating ...

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

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

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 ...

8

Interest rate traders/quants do not really talk about rho, as in the sensitivity of the Black Scholes price to $r$. The reason, I guess, being that we use Black (not Black-Scholes) formula for options on forwards and there is no $r$ in that formula except in the discounting factor multiplying the undiscounted option value. Rho in interest rate models is then ...

8

Practically, few things in real life have convenient closed-form calculations. Instead, you price some exotic, then you bump the various inputs, one or several at a time, up and down, by various small amounts, and re-price. There are seldom any short-cuts. (Autodiff can sometimes be a shortcut.) This Wikipedia article actually has a good list of commonly ...

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 ofr(see the ... 7 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-... 7 The Black-Scholes differential equation is a second-order PDE in two dimensions and reads as \begin{align*} \frac{\partial f}{\partial t} + rx\frac{\partial f}{\partial x} + \frac{1}{2}\sigma^2 x^2 \frac{\partial^2 f}{\partial x^2}-rf&=0, \\ \Theta+rx\Delta+ \frac{1}{2}\sigma^2 x^2 \Gamma-rf&= 0, \end{align*} assuming thatf\in \mathcal{C}^{1,2}([0,...

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 ...

7

Theta on a European Put option on a non-dividend paying stock is: $$\Theta=-\frac{S_t \sigma}{2\sqrt{\tau}}N'(d_1)+rKe^{-r\tau}N(-d_2)$$ For deep in-the-money Puts, $d_1$ and $d_2$ go to negative infinity: consequently, the term $N'(d_1)$ goes to zero, whilst the term $N(-d_2)$ goes to 1. Therefore, deep ITM puts can have a positive Theta, with a limit ...

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

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

Note that, \begin{align*} \frac{\partial{C}}{\partial{\sigma}} &=\frac{S_0}{\sqrt{2\pi}}{e^\frac{-d_+^2}{2}}(\frac{-1}{\sigma})(d_-)-\frac{Ke^{-rt}}{\sqrt{2\pi}}e^{\frac{-d_-^2}{2}}(\frac{-1}{\sigma})(d_+)\\ &=\frac{1}{\sqrt{2\pi}}e^{\frac{-d_+^2}{2}}\left[-\frac{S_0 d_-}{\sigma} + \frac{Ke^{-rt}d_+}{\sigma} e^{\frac{d_+^2}{2} - \frac{d_-^2}{2}} \...

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 ...

6

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. ...

6

Yes, the VIX took a sharp downfall on 2020/03/02, from 40.11 to 33.42 (-6.69). But that is not what the 2020/04/15 Put options are based on, they are based on the 2020/04/15 VIX Futures (VIJ20), these went from 23.025 on 2020/02/28 to 23.325 on 2020/03/02 an increase of 0.3. The Vix options are based on the futures, not the spot Vix value.

6

I'll give a heuristic "proof" for general European claims which will cause mathematicians to feel sick, but which physicists / practitioners would probably be quite happy work with: Write the Black-Scholes PDE as $$\frac{\partial F}{\partial\tau}(\tau) = \mathcal{A} F(\tau)$$ with $\tau = T- t$, and the operator $\mathcal A$ is defined as $$\... 6 In the Black-Scholes-Merton model, with model option price V as a function of underlying price S_t, strike price X, continuously compounded risk-free rate r, continuously compounded dividend yield y, time-to-maturity (in year fractions) \tau and implied volatility \sigma, our \Delta is defined as$$ \Delta\equiv \frac{\partial V}{\partial S_t}...

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

Victor123, let's start from $\Delta$. This is the expected change in the price of an option if the underlying asset moves by a currency unit, say 1 USD. For the case of a call option, the Delta varies between 0 and 1. Everything else been equal, the Delta of OTM calls will approach to 0 as the price moves out of the target barrier. Conversely for the case ...

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.

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