# Tag Info

16

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 ...

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. ...

13

You have a multidimensional problem - there isn't an answer of "this is what the greeks look like" for all cases, because it depends on the various levels of the different parameters. For example, if we limit ourselves purely to KO Call options, where the spot is 100, and there is no drift, with a time to maturity of 1 year (changing this is equivalent to ...

13

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. Actually, you don't need the ...

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*}

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

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

They are not the same, but they are related. Gamma is sensitivity to realized volatility. Vega is sensitivity to implied volatility. Vanilla options are always long gamma and long vega, so they are "long vol" and saying "I am a buyer of vol/gamma/vega" means that you are taking a position that benefits from a rise in volatility (either ...

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

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.

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

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

Vega is the partial derivative of the option price (as a function of parameters -- current stock price $S_t$, strike price $K$, implied volatility $\sigma$, etc.) with respect to $\sigma$ -- holding other parameters fixed: $$vega = \frac{\partial}{\partial \sigma} V(S_t,K,\tau,r,\sigma)$$ You are confusing the stochastic process with the parameter $S_t$ ...

4

Here is a document that will answer some of your concerns. There are many other good reads out there but this one is a nice one to get started with. In case the link is broken at the time one reads this answer, the document is called "Just what you need to know about variance swaps" by Sebastien Bossu (JPMorgan 2005).

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

No, you are incorrect. A deep in the money option is long vega. It's not just about the probability of being in the money, it's about how far in the money it is. Your reasoning is correct if we are talking about digital options which pay a fixed amount if the option expires in the money, but incorrect for regular options. One way to prove this ...

4

Well , complete elimination of even Delta is not possible, forget about Vega. When I say this , I'm talking about the trouble you'd face if you keep dynamically hedging your position from time to time. I mean it's not practical , however theoretically feasible it may seem. But anyway if you're interested, below ways could be of your help. You might want to ...

4

Constant Vega Requires Options Weighted Inversely Proportional to the Square of the Strike. E.g. if you have the following portfolio of options: \begin{equation} \int_{S_i(t)}^{\infty}\frac{2\Big(1-\log[\frac{K}{S_i(t)}]\Big)}{K^2}C_i(t,\tau,K)dK+\int_{0}^{S_i(t)}\frac{2\Big(1-\log[\frac{K}{S_i(t)}]\Big)}{K^2}P_i(t,\tau,K)dK \end{equation} You have a ...

4

IV is one of the inputs for your option pricing model, vega measures the actual impact (e.g. in Dollars, Euros...) of any change in IV. Intuitively IV is the price of the option while vega is the sensitivity to IV. Bottom line: There is a clear distinction!

4

Volatility is in effect what you are trading in options and the one unknown quantity in options valuation. The other inputs in option valuation: (1) Spot Price: observed in the markets, (2) Strike: defined by contract terms, (3) time: defined by maturity of contract, and (4) interest rates: also observed in the markets. So in effect volatility is ...

4

It seems like he is assuming that the shorter term volatilities change more than the longer term ones and the relatively sensitivity is proportional to $1 / \sqrt{T}$. Thus, this hedge is not against a parallel shift of the surface. This is not an uncommon assumption and the corresponding vegas are often referred to as "time weighted vegas".

4

Let $V$ denote the variance and $v$ the volatility, i.e. $V=v^2$. The natural argument for the option price under a stochastic volatility model is typically the variance, i.e. $C_\text{SV}=C_\text{SV}(S_0,V_0,...)$. However, using the chain rule, we can compute vega in terms of the volatility: \nu=\frac{\partial C_\text{SV}}{\partial v}=\frac{\partial C_\...

4

Vega is the option's price sensitivity to the volatility (i.e. IV). In the graph below, vega is shown to be a strictly positive function in volatility, which means that at any point in the graph (i.e. for any value of IV, irrespective of whether the option is OTM, ATM or ITM), the option price: will increase in value if IV goes up (because Vega is positive) ...

4

VIX almost always only spikes when SPX goes down as @Jan Stuller also mentions in a comment. Insofar the question is a bit counterfactual. I frequently use twin axis in the charts that follow. The position of the label corresponds to the axes the ticker belongs to. These are essentially two question in real world scenarios. 1 ) VIX and Vega: VIX up, IVOL up,...

3

Intuitive, no math explanation: Imagine two call options, option A expiring tomorrow and option B expiring in two months. Both of the options are way out of the money and have the same strike price. Due to some event the implied volatility of the stock spikes. Let's assume stock price stays the same. Does the chances of option A expiring in the money ...

3

Instead of just considering a parallel shift of the whole volatility surface, you can decompose the surface into maturities/strikes domains, so called buckets and consider Vega buckets which are sensitivities wrt to bumps of each of these domains. The vol smile is often inter/extra-polated using a model calibrated to market prices, e.g. the SABR model or ...

3

In general only non-linear instruments, like options, posses vega. Vega is always positive, no matter the directional component. So when you are long either a call or a put option you are long vega and when you are short either a call or a put option you are short vega. Thereby it becomes clear that you can go long and short different option positions to ...

3

The reason is that in many common models including geometric Brownian motion, the variance of the logarithmic returns is proportional to time. Thus, their standard deviation/volatility is proportional to the square root of time. Consider for example the class of Levy models where $X$ is is the logarithmic stock price process such that $S_t = S_0 e^{X_t}$. ...

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