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If stock returns ($r_t$) are not auto correlated why is that the squared term of the returns (volatility) exhibit serial correlation? Does heteroskedacity, by its nature, imply that time varying volatility is autocorrelated and therefore volatility today is influenced by its lagged values?

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2 Answers 2

To your first question: The first and the second moment are independent, so even if returns are not autocorrelated the size of returns can be.

To your second question: Autocorrelated volatility implies volatility not being constant but varying, which is heteroscedasticity or volatility clustering.

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Does Autocorrelation imply Heteroskedacity? –  jessica Apr 26 '13 at 20:08
    
Autocorrelated volatility implies heteroskedacity. –  vonjd Apr 28 '13 at 8:14
    
no, this is not true. this is very possible that then there is heteroskedacity but it is not invariant –  0d0a Apr 28 '13 at 14:47
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@cf16: Can you back up this claim? Heteroscedasticity is nothing but volatility clustering which is caused by autocorrelated volatility. See e.g. this paper proba.jussieu.fr/pageperso/ramacont/papers/clustering.pdf (p. 1-2). –  vonjd Apr 28 '13 at 15:24
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yes, then this is true –  0d0a Apr 28 '13 at 20:51

The first and the second moment are independent, so even if returns are not autocorrelated the size of returns can be.

of course. Example: generate path of variable with binomial distribution that takes value of 1 or -1, that is

$\sum_{i=1}^n{} x_i,x_i=\{1,-1\}$

now you can generate another path $\sum_{i=1}^n{} x'_i,x'_i=\{1,-1\}$ choosing values of $x'$ with freedom, and this way paths will be correlated or not, and squares of it will be correlated all the time, as two constant series of 1,1,1,...,1

regarding you second question

Does Autocorrelation imply Heteroskedacity?

No. These are two independent concepts. When you take matrix of varince-covariance of time series homoscedasticity is related to the elements on a diagonal, still with assumption that pairwise coefficients are 0. If they are not 0 then there is autocorrelation.

see here for details on this

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But autocorrelated volatility implies heteroskedacity. –  vonjd Apr 28 '13 at 8:13
    
no, this is not true –  0d0a Apr 28 '13 at 14:48
    
Can you back up this claim? Heteroscedasticity is nothing but volatility clustering which is caused by autocorrelated volatility. See e.g. this paper proba.jussieu.fr/pageperso/ramacont/papers/clustering.pdf (p. 1-2). –  vonjd Apr 28 '13 at 15:25
    
no, again, not. note: you can have positive and negative autocorrelation –  0d0a Apr 28 '13 at 15:40
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yes, then this is true –  0d0a Apr 28 '13 at 20:52

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