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1

You can see fairly quickly that an exact answer to this question is not going to be feasible because your functional transformation is to take the square root of $\sigma_t^2$, and the square root function has a countably infinite number of derivatives. This implies that a Taylor expansion is going to leave us with a countably infinite number of terms, most ...


1

"conditional volatilities from GARCH models are not stochastic since at time $t$ the volatility is completely pre-determined (deterministic) given previous values"-https://en.wikipedia.org/wiki/Stochastic_volatility $\sigma_t$ is still a random variable in the sense that it has an unconditional distribution. However, this unconditional distribution is not ...


-1

It must be : $$ E[\sigma _t]= \sqrt{ \frac{\omega}{1−\alpha_1− \beta_1}} $$ Otherwise you can model directly σt instead of σ^2(t)


0

In any finite sample, it is always possible for the Zhou estimator to return a negative number, even though we know the unobservable parameter being estimated is non-negative. This is a well known issue in the academic literature. There are several approaches to dealing with this problem: 1) Ignore it. (I don't like this one). It is particularly nefarious ...


1

I'll address your questions in order: 1a) For TSRV constructed using high frequency returns from NYSE market open to market close on a single day, the output should be numbers on the order of magnitude of 1e-4 to 1e-5. In other words, your numbers look about right. I got these number from calculating TSRV for IBM data myself using Kevin Sheppard's MatLab ...


1

The volatility smile is the result of market forces knowing form experience that out of the money option pay out more often that what would be expected by a normal (Gaussian) distribution. For years Quants speculated why the market drove the out of the money options higher that the price of the Black-Scholes model. The best theory speculates that the ...


1

As Richard mentioned, in some particular Markets for a given expiration, options whose strike price differs substantially from the underlying asset's price command higher prices than what is suggested by standard option pricing models. These options are said to be either deep in-the-money or out-the-money. Graphing implied volatilities against strike prices ...


4

There are several reasons, maybe the most important and also quite intuitive one: Implied volatility more or less assumes that the stock price is driven by Brownian motion and thus moves in a continuous fashion. What we observe is that stocks can jump (usually downwards, sometimes upwards) which needs to be modelled using something like a jump process ...


0

If you want a quick back of the envelop number, just use the IV of the nearest ATM option. Still the best would be to get the IV which is quite simple. However, three data points are definitely not enough to do this. Here's a matlab code that allows you to do it (credits to: volopta.com): % Variance swap calculation using the replication algorithm of % ...


1

This is a somewhat ill-posed question. The "components" in your question are not components, they are just different options and all have different implied volatilities - all for the same underlying. If you are looking to get single number volatility a-la VIX without the whole VIX calculation, you should use ATM (at-the-money) implied volatility, which is ...


0

As it was already mentioned. Try to read about Kalman filter and Markow Switching models. I have even seen some academic papers where authors tried to define MA length based on MSM or KF. Try to google it ...


0

In black-scholes the option price depends not on sigma^2 but on sigma^2 T. So if volatility is going to be 20% or 21% over the next 10 years (assume for simplicity no other values are possible, just these two with equal prob, but we don't know which) then that will have a bigger impact on the option value than a 20 vs 21 uncertainty for a 1 year option. That ...


2

In general, $v = \frac{\partial C}{\partial \sigma} > 0$ and $\theta = \frac{\partial C}{\partial t} < 0$. If maturity $T$ increases than $C$ increases. Suppose volatility is non-constant. Then if $T$ increases, the option value is more volatile, since the stock price is more volatile. Since $v > 0$ the option price must increase. He claims that ...


1

FVIX is not hard to compute. Just regress changes in VIX on excess returs of your base assets (it can be the 25 FF portfolios if those are what you are trying to explain) i.e run the following: \Delta VIX_t = X_t\beta+\epsilon



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