# Tag Info

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

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 $$\mathcal{V}=S_t\sqrt{\tau}N'(d_1)$$

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

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 ...

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