# Tag Info

14

For an option with price $C$, the P$\&$L, with respect to changes of the underlying asset price $S$ and volatility $\sigma$, is given by \begin{align*} P\&L = \delta \Delta S + \frac{1}{2}\gamma (\Delta S)^2 + \nu \Delta \sigma, \end{align*} where $\delta$, $\gamma$, and $\nu$ are respectively the delta, gamma, and vega hedge ratios. Then it is clear ...

14

Partly because it's hard to get a hold of, the Arslan et. al. paper is starting to assume mythical proportions. As said by Dimitri Vulis, the general idea of the paper is set out in (one or two of) Peter Carr's papers. For the benefit of the OP and others I will try to summarize the most salient points of the paper below and also point out the assumptions ...

12

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

You can't lose more than you invested by writing covered puts, because you keep enough cash to cover any potential losses from the puts. That's not to say that your losses can't be substantial, of course. The below chart shows the drawdown profile of the PutWrite index - you would have lost nearly 40% of your investment at one point. So how did the ...

10

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*}

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 ...

10

What you have to do is to show that the dollar gamma satisfies the Black-Scholes PDE. Using Feynman-Kac it then follows that the dollar gamma is an expectation of a "payoff", just like the Black-Scholes claim price is an expectation of a payoff. And if something is the expectation of a payoff then it's a martingale. I'll leave the above for you to carry out....

8

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

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

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 (...

7

The conjecture is true when the interest rate is zero. Note that, from this question, under the Black-Scholes model, \begin{align*} \Gamma(t,S_t) &= \frac{N'(d_1(t))}{S_t \sigma \sqrt{T-t}}\\ Vega(t,S_t) &= S_tN'(d_1(t)) \sqrt{T-t}, \end{align*} where \begin{align*} d_1(t) = \frac{\ln \frac{S_t}{K} + \big(r+\frac{1}{2}\sigma^2\big)(T-t)}{\sigma \...

7

$$\frac{1}{2} \frac{\partial^2 f}{\partial S^2} dS^2 \approx \frac{1}{2} \sigma^2 S^2\frac{\partial^2 f}{\partial S^2} dt$$ (for small $dt$, ignoring $(dt)^2$ terms ) $\sigma$ is embedded in $dS = \mu S dt + \sigma S dW$ and $$dS^2 = \mu^2 S^2 dt^2 + 2\mu \sigma S^2 dt dW + \sigma^2 S^2 dt \approx \sigma^2 S^2 dt$$ You picked up $1/2\Gamma \sigma^2$ from ...

6

Assume you buy a plain vanilla call option at the price $V$ and the spot $S$. You immediately delta hedge buy selling $\partial V / \partial S$ units of the underlying asset. The underlying asset now instantaneously jumps form $S$ to $S' = S + \Delta S$. The new value of the call option is $V'$. Your total p&l is \begin{equation} \text{P&L} = V' - ...

6

Not sure this is a valid question! Gamma p/l is by definition the p/l due to realized volatility being different from implied. Vega p/l is by definition the p/l due to moves in implied volatility. The second part of the question you have answered yourself. Short dated options have more gamma exposure, long dated options have more vega exposure.

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

Gamma is the sensitivity of the delta with respect to infinitesimal changes in the price of the underlying asset (in whatever unit your underlying is nominated, typically dollar, pounds, euros, ...). So, it is not a percentage change. Instead, the percentage change (option elasticity) equals $\Delta\frac{S}{V}$. This quantity, for instance, gives you the ...

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

I think what you are missing is simply the Vega-Gamma relation in the Black-Scholes model. Namely: $$Vega = \frac{\partial v}{\partial \sigma} = \sigma(T-t)S^2 \frac{\partial^2 v}{\partial S^2} = \sigma \tau S^2 \Gamma$$ Plugging this into your coverage error, you get its expression in terms of the Vega which is the most natural measurement of your ...

5

Assuming all else remains equal (implied vol has not changed and very little time decay has occurred), Gamma scalping can best be explained by Gamma (or realized volatility) enhancing the value of a delta hedged portfolio. For example: If you are long an at-the-money call option, you are long 0.5 Delta and long Gamma. If you hedge this position, you will ...

5

We work in a Black-Scholes world. Consider the following delta-hedged portfolio: $$\Pi_t=V_t-\frac{\partial V}{\partial S}S_t$$ We assume the portfolio is self-financing$^{\text{(a)}}$, therefore: \begin{align} \text{d}\Pi_t &= \text{d}V_t-\frac{\partial V}{\partial S}\text{d}S_t \\[3pt] \tag{1} & = \left(\frac{\partial V}{\partial t}\text{d}t+\... 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 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 Usually vega and gamma go in the same direction, but you can have opposite exposure in a calendar spread. For an ATM option, vega decreases closer to maturity while gamma increases. If you implement the following: -long a 1 month ATM option -short a 2 months ATM option you should be long gamma and short vega. 4 As long as you live in a world where implied and realized vol are the same, there is no net profit (or loss) from gamma scalping. However, if they are different, then you make a gain or loss which is not path dependent. This is all still in a hypothetical world of course with continuous trading. In reality when rehedging less frequently, pnl becomes random ... 4 Gamma scalping (being long gamma and re-hedging your delta) is inherently profitable because you make 0.5 x Gamma x Move^2 across the move from your option. (You get shorter delta on downmoves, so you buy underlying to hedge, you get longer on upmoves, so you sell on upmoves, etc.) Because it's inherently profitable across any move, you must pay for the ... 4 There are serious issues with how this graph is drawn, which impede understanding. The y axis is unlabeled and should be labeled "profits" or \pi on a hedged position. The other axis should be labeled "stock price change" or \Delta S, not stock price, and the middle point of the axis (where the parabola has a minimum) should be labeled zero. Points to ... 4 Using Taylor polynomials of 2nd order:V(r+h)\approx V(r) + \frac{\partial{V}}{\partial{r}}h +\frac{1}{2}\frac{\partial^2{V}}{\partial{r}^2}h^2V(r-h)\approx V(r) - \frac{\partial{V}}{\partial{r}}h +\frac{1}{2}\frac{\partial^2{V}}{\partial{r}^2}h^2$$The sum of the previous 2 equation will give us gamma as:$$Gamma = \frac{\partial^2{V}}{\partial{r}^2} ...

4

Many traders build a spreadsheet of how their delta changes across a range of market moves (-10%, -8%, ,....+8%, +10%) for example. That is a lot more useful than a single gamma number. It also means that the answers are finite and useful.

4

I can argue your case as follows, consider a portfolio such that The value of $\Pi$ of a portfolio satisfies the differential equation given by: $$\frac{\delta \Pi}{\delta t}+rS\frac{\delta \Pi}{\delta S}+\frac{1}{2}\sigma^{2}S^{2}\frac{\delta^{2}\Pi}{\delta S^{2}}=r\Pi$$ From the differential equation, $$\Theta=\frac{\delta \Pi}{\delta t}$$ \Delta=\frac{...

4

Think of this in terms of Taylor series. Let's say the option price today is $C\left(S,t\right)$ where S is the underlying price and t time. Let's say the underlying price changes by $\Delta S$ in a time interval $\Delta t$, so your P/L will be: $\mathrm{P/L}=C\left(S+\Delta S,t+\Delta t\right)-C\left(S,t\right)$ Use Taylor series to first order in t and ...

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