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## New answers tagged volatility

0

The answer to your question is $$\text{Var}(w'R_1+w'R_2)=w'\Sigma_1w + w'\Sigma_2w + 2 w' \Sigma_{12} w$$ where: \begin{align} \Sigma_1(i,j) &= \text{Cov}(R_1 (i), R_1 (j)) \\ \Sigma_2(i,j) &= \text{Cov}(R_2 (i), R_2 (j)) \\ \Sigma_{12}(i,j) &= \text{Cov}(R_1 (i), R_2 (j)) \end{align} are positive definite $N \times N$ covariance matrices. ...

1

If you have options data with long enough history you could always construct a comparable index by computing the implied volatilities and using a similar weighting methodology to VIX or looking at the implied volatility of the 1 month call/put with strike closest to the price at the observation date (i.e. one closest to 100% moneyness). If you want an ...

4

These are 2 completely different ways of estimating volatility. GARCH models are calibrated on historical time series i.e. information provided under the real-world measure $\mathbb{P}$. Although you can obviously use them for forecasting, the core information which is used to build the model is backward-looking in nature (historical behaviour of the stock)....

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 ...

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I have found the mistake. The ugarchfit function sets automatically non negativity constraints for all coefficients- This makes sense since the alpha in our case shouldn't be negative. However, when releasing the constraint to negative values you get the right results. The only explanation I can think of is that in the course of optimisation, temporarily ...

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 ...

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 ...

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 ...

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, ...

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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 ...

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A limitation of both papers is they focus on point estimates, i.e they compare $\sigma_{t}$ with $h_{t}$ in the loss functions of the SPA Tests. A possible suggestion to overcome it, is to use a loss function based on density forecast, in order to capture the whole forecast density distribution and not only a single point. This may have important ...

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I can think two techniques that may possible be of help. The first technique is the moving average adjusted returns originally proposed by Andersen et. al(2001). See Hansen et al. (2008) for details. In order to account for serial correlation an MA(1) is fitted to to the intraday returns data, the residual of which is then squared and aggregated over the ...

0

[Answer] Well yes, this comes from the interpretation of an option price as the initial endowment of a perfect replicating portfolio. Here the replication consists in holding a self-financing portfolio of shares + foreign/domestic bonds where one dynamically re-balances the delta continuously (assuming all usual assumptions hold: i.e. no transaction consts ...

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When you are hedging through FX then there are two factors influencing the impact that currency will have on the portfolio: the volatility of currency relative to that of the underlying asset and the interaction between currency and the underlying asset. The larger the volatility ratio ( volatility foreign currency/volatility asset ), the greater the impact ...

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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 ...

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

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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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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 ...

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At $t=0$, you have a vol surface $(T,K)\to\sigma(t=0,S_t=S_0,T,K)$ the hard question is which dynamics i.e $\sigma$ seen as $\sigma: (t,S_t)\to ((T,K)\to \sigma(t,S_t,T,K))$ and even, if you imagine that behind this, there is a deterministic function, you still have to suppose a dependence with respect to $S_t$ Here are two examples 1) $\sigma(t=0,S_t=... 0 There's nothing wrong with your formulation, in my opinion. If you model the rate z_30 with a fixed mean, then indeed the forward ZCB price is long vega. This means that the forward interest rate is short vega (i.e. the 30yr into 10yr forward rate goes down when vol goes up). This is self-consistent. In most textbooks, however, the forward interest ... 2 The short answer: Your observation is caused by some sort of central limit theorem. The long answer: The reason for the volatility smile/skew is the non-normality of the assumed return distribution. If the implied volatility of out-of-the-money put options is higher than of at-the-money put options, it implies that the market assumes that the risk-neutral ... 2 This can be due to various effects. I will list you 2 of them off the top of my head: Jumps/Crashes : assume you were to price a put option which expires in a few days from now. Your diffusion model tells you this option should be worth$\$3.25e^{-7}$ since it is very out of the money as of today. What price will you quote? Well, in order to be ...

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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 ...

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