New answers tagged

5

The other answers have given a good qualitative description of what the VIX measures. In this answer, I will try to give a comprehensive quantitative overview of how the VIX formula works. What is the VIX? The CBOE Volatility Index (VIX) is an index which measures the stock market’s expectation of the future 30-day volatility of the S&P-500 index. This ...


0

Not necessarily an answer but far too long for comments. I am surprised no one asked about the underlying or what exactly you intend to backtest? Since you write option chains, I would assume it's listed products? You also write moneyness, which rules out FX in my opinion. Strictly speaking, FX would be Garman Kohlhagen and most commodity would use Black. ...


0

There is a statistically and economically significant difference between option implied volatility ("IV") and (underlying) historical asset price volatility, calculated as the standard deviation of log return time series at some time horizon. The difference is on average positive (IV > volatility). (Underlying) volatility is a measure of the ...


0

I would say it depends on the frequency of the strategy you trade. For low frequency (weeks) strategy (e.g. hold the position to maturity, delta-hedged of course) from my experience I believe it is quite accurate. For higher frequency strategy probably not, but I have no experience on that myself.


2

Writing $M_T=\max_{0\le t\le T}S_t\,,\,\, m_T=\min_{0\le t\le T}S_t$ the option payoff is \begin{align} (K-S_T)^+\underbrace{1_{\{m_T\le L\}}}_{\text{KI}}\underbrace{1_{\{M_T< U\}}}_{\text{KO}}=(K-S_T)^+(1-1_{\{m_T>L\}})1_{\{M_T<U\}}\,. \end{align} In other words, the KI-KO-option is a portfolio of a long positon in a single barrier KO option and a ...


1

Consider any function $f(S(t),K,t,T,\{x_i(t)\})$ with payoff $(S(T) - K)_+$ when $t=T$, where $\{x_i(t)\}$ are other variables/parameters so that at $t=0$ you are able to choose (i.e. calibrated) these so that your function matches the market price of the option: $f(S(0),K,0,T,\{x_i(0)\}) = C^{market}(t=0)$. As the payoff of the option does not depend on $\{...


0

If you assume that the vols $\sigma_r,\sigma_i$ are deterministic functions of time their formula (1) still holds $$\tag{1} dV(t)=\frac{1}{2}(\sigma^2_r(t)-\sigma_i^2(t))\,\Gamma^i(t)\,dt. $$ Integrating gives the accumulated hedge PnL $$ V(t)=\frac{1}{2}\int_0^t(\sigma^2_r(s)-\sigma_i^2(s))\,\Gamma^i(s)\,ds. $$ One could extend the derivation to the case of ...


0

It is relevant, it determines the stock price in a world with rational risk-neutral investors, the stock price is PV(Se^E(r)T)


Top 50 recent answers are included