# Why autocovariances could fully characterise a time series?

I read from textbook that 'Autocovariance can fully charaterise the time series' joint distribution', I do not fully understand the connection between covariance and joint distribution here. Please explain to me, someone?

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What is the textbook you are reading? –  Seth Jun 19 '12 at 17:09
@seth, I am reading John Cochrane's 'Time Series for Macroeconomics and Finance' –  Flying pig Jun 19 '12 at 17:11
I would bet he assumes the process is Gaussian, right? –  whuber Jun 19 '12 at 17:25
@whuber, yes, he uses ARMA model to illustrate, and assume error term always as white noise. –  Flying pig Jun 19 '12 at 17:36
White noise by itself does not guarantee the result you need; you need Gaussian white noise. –  Dilip Sarwate Jul 4 at 18:19

A stationary Gaussian process is completely characterized by the combination of its mean, variance and autocorrelation function. The statement as you read it is not true. You need the following additional conditions:

1. The process is stationary
2. the process is Gaussian
3. the mean $μ$ is specified

Then the entire stochastic process is completely characterized by its autocovariance function (or equivalently its variance $σ^2$ + autocorrelation function).

This simply relies on the fact that any multivariate Gaussian distribution is uniquely determined by its mean vector and its covariance function. So given all the conditions that I stated above the joint distribution of any $k$ observations in the time series has a multivariate normal distribution with mean vector having each component equal to $μ$ (by stationarity) each component has variance $σ^2$ (again by stationarity) and the covariance components are given by the corresponding lagged covariances in the autocovariance function (again stationarity comes in because the autocovariance only depends on the time difference (or lag) between the two observations whose covariance is being taken.

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(+1) I think this is said implicitly in condition (1) but you also require that $\mu$ is constant, right? –  Macro Jun 19 '12 at 18:05
@Macro Yes stationarity, even weak sense (covariance) stationarity requires a constant mean and a constant variance. –  Michael Chernick Jun 19 '12 at 18:16
@MichaelChernick, then we could reproduce the joint distribution of stochastic process (or simulate stochastic process itself) by having its mean and autocovariance? –  Flying pig Jun 19 '12 at 19:05
@Flyingpig Yes for any subset of the variables as long as it is a stationary Gaussian process. it doesn't have to be an AR, MA or ARMA process. It only has to be a stationary Gaussian process. It should not be a surprise. This is a well known property for multivariate normal distributions. –  Michael Chernick Jun 19 '12 at 20:19
@Macro I guess the condition constant mean is redundant in the required conditions that I gave. I just mentioned it because to completely characterize the stochastic process you need to know what the value is for the mean and variance and not just that they are both constant. –  Michael Chernick Jun 19 '12 at 20:26