# Tag Info

## Hot answers tagged vega

6

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*}

4

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.

4

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 ...

4

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

3

You know $$N(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}e^{\frac{-u^2}{2}}du$$ then $$N'(x)=\frac{1}{\sqrt{2\pi}}e^{\frac{-u^2}{2}}$$ therefore $$\mathcal{V}=S_t\sqrt{\tau}N'(d_1)$$

3

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!

3

First, notice that the two greeks you mentioned in your question are simply the partial derivatives of the value of the option $V$ with respect to two different variables $S$ (the price of the underlying) and $\sigma$ (the volatility of the underlying): $$\Delta = \frac{\partial V}{\partial S} \quad \text{and} \quad \nu=\frac{\partial V}{\partial \sigma}$$ ...

2

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 ...

2

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 ...

2

The VIX is designed to "represent the implied volatility of a hypothetical at-the-money [SPX] option with exactly 30 days to expiration." (via the CBOE) The calculations are available from the CBOE in this white paper. Note that your question is wrong -- it is the implied volatility, not the vega. Moreover, you wouldn't predict a change in vega (which is a ...

2

In the Black-Scholes model the price of a binary option is $$B = e^{-r(T-t)}N(d_2)$$ with $$d_2 = \frac{\log(\frac{S}{K})-\frac12 \sigma^2 (T-t)}{\sigma\sqrt{T-t}}$$ Differentiation with respect to $\sigma$ gives our our volatility risk, or vega $$\frac{\partial B}{\partial\sigma} = e^{-r(T-t)} N^\prime(d_2)\frac{d_2+\sigma\sqrt{T-t}}{\sigma}$$ ...

1

you can find from the CBOE paper above mentioned that the value is pretty much that of a strip of vanillas, weighted by 1/K^2. Typically if spx spot goes down, then realized vol increases. Together with the increase of realized vol, implied vol gets "re-evaluated" and typically marked higher with a steeper skew etc. the remark of the implied vol surface ...

1

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 ...

1

I don't think your hypothesis is correct. If you have a very short dated ATM option, then your option will have close to infinite gamma but close to 0 vega. So this short dated ATM option is vega neutral but definitely not gamma neutral.

1

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 ...

1

Generally speaking, volatilities at all points of the vol surface are (positively) correlated in both empirical and theoretical models. So if you feel you have a prediction strategy for the VIX, you have an associated directional prediction for other volatilities, and you can take advantage of that. Directional volatility bets are most often expressed (as ...

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