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6

Consider the GARCH(1,1) process \begin{align} r_{t+1} &= \sigma_{t+1} z_{t+1} \\ \sigma^2_{t+1} &= \omega+\alpha r^2_t +\beta \sigma^2_{t} \end{align} for the returns $r_t$, with ${z_t} \sim N (0,1)$ IID. In what follows, let us distinguish the conditional return variance $$ V [ r_{t+1} \vert \mathcal{F}_t ] = \sigma^2_{t+1} $$ from the ...


5

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


4

Not sure why you're multiplying by 24? EDIT: got confused by your STDEV * 24, you meant (and calculated) STDEV * SQRT(24) If $X_i$ is random variable representing daily log returns, and assuming that log returns are i.i.d. then volatility of monthly return is $\sigma_{monthly}=\sqrt{21}\times\sigma_{daily}$ (assuming 252 days a year, i.e. $252/12=21$ per ...


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 ...


4

Assume that your stationary time series (here a daily close-to-close log-returns' series) is modelled as follows $\forall t \in \mathcal{T}=\{1,...,N\}$ \begin{align} r_t &= E_{t-1}[r_t] + \epsilon_t \\ &= E_{t-1}[r_t] + \sigma_t z_t \end{align} with $z_t \sim N(0,1)$ and $\{z_t\}_{t \in \mathcal{T}}$ are IID. The above equations suggest that, ...


3

The n=100 specifies the number of periods (rolling) for the vol estimate - see the original link https://web.archive.org/web/20100326215050/http://www.sitmo.com/eq/409 where the n is referred to as Number of historical prices used for the volatility estimate An example in R: library(TTR) library(quantmod) getSymbols("AAPL") nrow(AAPL) # we have 2384 price ...


2

Two volatility processes yield a higher flexibility of the model. This is of greater importance if one tries to price derivatives with different maturities in one single model. A additional volatility component helps to capture the term structure of volatility, which can depend greatly on time to maturity. See for example the VIX term structure from CBOE: ...


2

I think,the additional volatility factor,$v_2(t)$, provides more flexibility in modeling the volatility surface.We know $\rho$ controls the slope of the implied volatility.In the single-factor Heston model, $\rho$ is constant over maturities,In deed $$Corr[{dS}/{S\,,\,dv]}\;=\rho \,$$ which means that model has trouble providing an adequate fit to market ...


2

Loosely speaking, it can be seen as inserting an additional degree of freedom in the underlying's dynamics. This can be useful from a static perspective: with an additional lever to play on, one can hope to better capture the short term implied volatility smile, which "naive" stochastic volatility models (single volatility factor, no jumps) are known to be ...


2

The short answer: Your observation is caused by some sort of central limit theorem. The long answer: The reason for the volatility smile/skew is the non-normality of the assumed return distribution. If the implied volatility of out-of-the-money put options is higher than of at-the-money put options, it implies that the market assumes that the risk-neutral ...


2

This can be due to various effects. I will list you 2 of them off the top of my head: Jumps/Crashes : assume you were to price a put option which expires in a few days from now. Your diffusion model tells you this option should be worth $\$3.25e^{-7}$ since it is very out of the money as of today. What price will you quote? Well, in order to be ...


2

Volatility (often defined in terms of standard deviation of returns, or in terms of implied volatility from option markets) is indeed one measure of risk, but like any single measure of risk, it is incomplete. Part of the reason for this is that in financial markets, the returns are not normally distributed but rather have "fat tails." This means that ...


2

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 ...


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 ...


1

It isn't strictly speaking possible to convert a log vol to a normal vol, although it may be possible to get a rough idea. I am assuming you only have the vol of log returns but not the actual time series here. If you had the original time series, then you would just calculate the standard deviation of the prices to get the normal vol. I assume this is ...


1

If I have read the question correctly then I will assume that $a$, $b$, $c$, $d$, $T_i$, and $k_i$ are constants. If this is the case then the only term which we need to show is bounded is $$ \big(a + b(T_i - t)\big)\exp\big(-c(T_i-t)\big). $$ If we assume that we are only considering the temporal domain $0 \leq t \leq T_i$ such that $T_i - t \geq 0 $ then ...


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 ...



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