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# Tag Info

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Looking at transaction prices, they would occur at the market bid if the active part is a seller, and at the ask if the active part is a buyer. With a random flow of sellers and buyers, the price will bounce between the bid and ask prices, creating a negative autocorrelation in returns. This penomenon is known as the bid-ask bounce, and has been discussed ...

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This is a partial explanation in that trading strategies with longer horizons have higher information ratios, t-statistics, slope coefficients, and R^2 in general. In other words, if information ratios for both strategies are identical then the longer-term trading strategy is already worse. John Cochrane illustrates how longer horizons have higher t-stats ...

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This question was ultimately answered on Cross Validated Here are a couple of articles that deal with this subject: Britten-Jones and Neuberger, Improved inference and estimation in regression with overlapping observations Harri & Brorsen, The Overlapping Data Problem

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The correct answer has some intuition though it doesn't generalize to continuous time very easily: Think about the paper below like this: $Var(X+Y) = Var(X) + Var(Y) + 2Cov(X,Y)$ The generalization is slightly hard because the dynamics of $\mu$ and $\sigma^2$ could be dependent for arbitrary returns. You can use a GMM estimator to derive the asymptotic ...

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The answer is that it depends. In addition to the Lo paper above, there are a number of excellent references that go into depth about annualizing or time scaling non-i.i.d. returns, one of which is Roger Kauffman, "Long-Term Risk Management", 2005 which can be found at http://www.rogerkaufmann.ch/all-Budapest.pdf. There are some well known cases where the ...

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To simplify, consider the errors rather than the returns. The variance is effectively the average of the squared errors, while absolute deviation is the average of the absolute errors. So plotting the squared errors or absolute errors over time could give an indication of whether the variance or absolute deviation is constant over time. Since variance is ...

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In terms of interpretation, an $MA$ model simply means that the time series is a function of the error from previous periods. You might find it informative to consider plotting simple $AR(1)$ models alongside various $ARMA(1,1)$ to develop a more intuitive understanding. For instance, the $AR(1)$ model (chosen as it is common for financial time series) $$x_{... 5 Apparently yes, (I haven't verified the math but have no reason to doubt it). For this simple case you can find a closed form in the following paper: Jeff A. BILMES: What HMM can do The closed form is given on part 4.4 of the paper but the whole thing is worth reading as it clearly shows the main properties of these models. You can also note that ... 5 Couple points I like to make: There exists no reliable model that can even predict future price returns (risk premiums, excess returns, whatever you want to call it) beyond a year, run as fast as you can if you hear from someone who claims he can predict risk premiums 10 years out, whether reliably or not. It makes zero sense and clearly comes from either a ... 4 Simple...because you are interested in deviations from a metric, and not whether it deviates above or below. The very definition of volatility is a "measure of deviation". Squaring returns or using the absolute values just eases the calculation to arrive at a deviation measure. Otherwise volatility would have to be calculated in other ways as positive and ... 4 Generally we use models that go so far out in a comparative sense, not as an absolute decision. You are definitely do some good reading but I believe you are thinking about these models in the wrong way. I think (and correct me if I'm wrong) you are looking at creating or finding the perfect "crystal ball" model that will predict returns/risk premiums etc. ... 4 W.l.o.g we use the initial condition S(0)=1 and define \gamma:=\mu-\frac{\sigma^2}{2}. Hence we have the dynamics$$S_t=e^{\gamma t +\sigma W_t}$$By definition Cov(X,Y)=E((X-E(X))(Y-E(Y))=E(XY)-E(X)E(Y), where the last equality follows from the linearity of the expectation. Note \gamma t+\sigma W_t is normal distributed with mean \gamma t and ... 4 I think @zer0hedge has constructed a clever example by which to demonstrate what is implied by the stylized fact by which volatility begets volatility. It is correct to conclude volatility bursts are a type of absolute autocorrelation. All volatility bursts display characteristics of autocorrelation of absolute returns, but will all types of autocorrelation ... 4 You need to compute the autocorrelation of the log returns r_t, not of the prices, p_t. The relationship of the log return series to the price series is$$ r_t = \log \frac{p_t}{p_{t-1}} $$The price series is obviously very autocorrelated, since today's price is yesterday's price plus small delta. 3 You should check for autocorrelation. However, its presence does not necessarily mean your model will produce inaccurate figures. The ARCH family of models were developed to help analyze the volatility of a time-series. This data is assumed to display a degree of heteroskedasticity. Using the GARCH model, small amounts of auto-correlation (not of practical ... 3 What is the mathematical basis to say that u^{2}_{t}/\sigma_{t}^{2} will exhibit little auto-correlation in the series? Let's r_{t} be a series of returns and let's assume (Assumption I) it follows a covariance stationary process defined as : r_{t}=\sigma_{t} z_{t} where z_{t} is i.i.d with E_{t}(z_{t})=0 and Var_{t}(z_{t})=1 ; Then  ... 3 Autocorrelation is the correlation of a series with itself. Suppose X = {X_1, X_2, X_3, ...} is your time series. Then the autocorrelation between X_t amd X_s is:$$ \frac{E[(X_t-\mu_t)(X_s-\mu_s)]}{\sigma_t \sigma_s}  This can be simplified quite a lot if the series you have is stationary (a common assumption), in which case the autocorrelation ...

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I'm not a time series expert but one idea occurs to me: look at the distribution of the increments if Z(t). If the w are stochastic , that distribution should have fat tails relative to the distribution that is generating the Levy process.

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If you are using Spatial Econometrics toolbox in Matlab you could use the lrratio function which implements a sequence of such tests beginning at a maximum lag (specified by the user) down to a minimum lag (also specified by the user). (more info in http://fmwww.bc.edu/ec-p/software/matlab/mbook.pdf)

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I would advice you not to do any overlapping analysis. The results will be hard to interpret and misleading. I have seen many "practioners" looking at histograms of overlapping returns. They saw interesting patterns and found funny explanations - which were simply wrong. If you are new to econometrics then correction methods (do there exist helpful ...

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1.Is it correct, that the coefficients are now different to the coefficients of the arima output? It seems right that the ARMA coefficients are different. Indeed, in the second model, the GARCH component will capture fluctuations that the ARMA component will not have to capture, resulting in different ARMA parameter estimates. 2.This is the acf of the ...

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I can offer my opinion in response to your first two questions: 1.) Unfortunately, this is one of the problems with numbers; the answer is that if the observation is outside of the confidence interval by even a millionth of a percent, it is significant. If it is below by even the smallest amount, it is not significant. Changing your significance level or ...

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1.) Autocorrelation is the correlation of a time series against the lagged version of itself. 2). First autocorrelation is the correlation of the time series against the lag(1) version of itself. Let's look at the example below Period_Numbers = [1,2,3,4,5,6,7,8,9,10] Time_Series = [10, 20, 30, 40, 50, 60, 70, 80, 90, 100] First Autocorrelation is ...

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The high serial correlation you are getting in the first case is a spurious correlation. The correct way to do it is with returns. The price series has a unit root. You need to take diff(log(prices))) in order to have a stationary time series, on which you can then estimate autocorrelations, auto regressive coefficients, etc. properly. This was shown by ...

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TL;DR: the test statistic's distribution is $N(0,1)$ A bit more information about the Automatic Variance Ratio Test: $H_0$: ${\Delta}r_t$ is serially uncorrelated (where ${\Delta}r_t=r_t-r_{t-1}$) $H_1$: ${\Delta}r_t$ is serially correlated The test statistic is $VR=\sqrt{T/l}[\hat{VR}(l)-1]/\sqrt{2} \quad {\xrightarrow{d}} \quad N(0,1)$ The $d$ over ...

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You are correct in that the series is not stationary. The ADF test isn't designed to test for stationarity outside the center of location. You are not going to be able to use the square root rule to extrapolate because you have significant autocorrelation of the variances. I do have a suggestion on your problem by noting that returns are not data. Prices ...

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Such volatility pattern is a well-known stylized fact of financial time series (see Cont, Rama. Empirical properties of asset returns: stylized facts and statistical issues. (2001): 223-236 for more details) which is called volatility clustering. Qualitatively, it means that high returns are likely to be followed by high returns, the same applying for low ...

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Your code basically implements the assumption that you cited: The volatility of return processes is not constant with respect to time. Whether it's a single burst or some kind of a fancy function $\sigma_t$ is not important here. The fact is that your volatility is time varying. You may call it piece-wise constant, but it still is characterized as time ...

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Hi: Subtract $k$ from $z_t$ and add $k$ to $z_{t-k}$. Then you have $cov(z_{t-k,} z_{t})$ which by definition is $\gamma_{-k}$. But, by stationarity, this has to be equal to $cov(z_{t}, z_{t-k})= \gamma_{k}$ because the covariance is only a function of the lag difference.

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It depends how large the overlapping interval is. Conceptually an infinite rolling window is equivalent to the level, and no one would suggest to 'regress on levels and apply Newey West'. I think NW is 'robust' in the presence of relatively mild autocorrelation, not a panacea that will give the correct standard errors. If you use, say dailly returns ...

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