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GARCH model is used to model persistence in volatility. If you square demean exchange rate and calculate autocorrelation you will find significant autocorrelation upto many lags that indicates the clustering of volatility in data. A simple ARMA(1,0)-GARCH(1,1) model can be written as : $$y_t=\mu + \phi y_{t-1}+e_t $$ $$e_t \sim N(0, \sigma^2_t)$$ ...


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I am not sure what you are exactly asking. But usually even a simple Garch(1,1) would be the naive approach of forecasting variance using last period's variance. A very good survey of volatility modelling on the Arch/garch family is the Hansen and Lunde 2005. They show that hardly one can beat a garch(1,1), so that is a good first guess.


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You most probably don't want to estimate the covariance of prices but rather the covariance of returns. Thus for equities you can take the return of the traded price. For bonds: if the maturity is long enough (say bigger than 2 years), then you can take the returns of traded prices. The pull to par should not be too relevant here. if the maturity is short ...


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If certain broad assumptions are correct (eg, asset prices are continuous in time, markets are efficient) then asset returns must follow a Levy process. Both the Gaussian and Stable distributions are subsets of Levy processes. The question should not be whether Gaussian or Stable is better. Neither are adequate (in fact, many Stable distributions imply ...


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I think, the use of stable distributions in Finance (and, probably, in Economics) is a big mistake. It is essential that the intuitive fact that the stable distributed observations possess a large number of big deviations from empirical mean is not true (see, Lev B. Klebanov, Irina Volchenkova (2015) "Heavy Tailed Distributions in Finance: Reality or Mith? ...


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The first objective is to minimize the variance by choosing a proper control variate. First note that an expectation value is just a constant, so the covariance between an expectation value and a random variable is zero: $$\text{Cov}\left(\mathbb{E}[Y], X\right) = 0$$ Similarly for the variance of an expectation value, $\text{Var}(\mathbb{E}[Y])=0$. The ...



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