# Tag Info

0

1

let' s define a ARMA-GARCH model: $y_{t} = \mu_{t} + \epsilon_{t}$ where $\mu_{t}$ is the conditional mean process (ARMA(p,q) part, $\mu_{t} = E(y_{t}|\mathcal{F}_{t-1})$) . The errors (or mean residuals) re defined by: $\epsilon_{t} = \sigma_{t} \eta_{t}$ where $\eta_{t}$ is a white noise (0,1) Then : $Var[\epsilon_{t}]= \sigma_{t}^{2}$. next see ...

3

If your question is: "Given all the information available up to time $t$, if I compute the 1 period ahead forecast $r_{t+1}$, is the conditional volatility over $[t,t+1[$ given by $\sqrt{r_{t+1}}$?", the answer is NO. To compute the 1 period ahead conditional variance, you should use your model equations (see this post which might help you better understand ...

4

Clearly, from a theoretical point of view, a varswap is a better way of capturing volatility change, since as mentioned by Mark Joshi a varswap has, by construction, a Vega that does not vary with the stock price. For a single option on the other hand the Vega is at maximum at a stock price $S^*$ roughly comparable to the strike price X and decays in a "bell ...

5

The vega of an option is very dependent on the spot price. The vega of a variance or volatility swap is not.

5

If you take Quantuple's stuff a little further, you can really see whether you're long skew. You can pretty easily see the dependence on convexity too (though it should be obvious that you're long convexity). So first off, we need some smile parametrisation that lets us easily control convexity and skew. I just went with a made up one; $$\mathrm{convexity} ... 5 As I've mentioned in a comment, it would be wrong to think that a variance swap specifically amounts to being "long skew". What you can say however is that, in the absence of jumps (i.e. in a pure diffusion framework, see here and here for further info), the fair variance strike K_{var} at which a variance swap with notional N and payoff$$ N \times ( \...

1

When calculating the simple arithmetic mean, each observation has an equal weight: $$\hat \mu^{simple} = \frac{1}{T}\sum_{t=1}^T x_t.$$ If the observations are $i.i.d.$, $\hat \mu^{simple}$ is an efficient estimator of the population mean. When estimating the mean of a GARCH process, $\hat \mu^{simple}$ is no longer efficient. It makes sense to ...

Top 50 recent answers are included