8
$\begingroup$

I read in John Cochrane's Time Series for Macroeconomics and Finance that:

Autocovariance can fully charaterize the time series [joint distribution].

I do not fully understand the connection between covariance and joint distribution here. Can someone please explain that?

Improve this question
$\endgroup$
3
  • 4
    $\begingroup$ I would bet he assumes the process is Gaussian, right? $\endgroup$
    – whuber ♦
    Commented Jun 19, 2012 at 17:25
  • $\begingroup$ @whuber, yes, he uses ARMA model to illustrate, and assume error term always as white noise. $\endgroup$
    – Flying pig
    Commented Jun 19, 2012 at 17:36
  • 1
    $\begingroup$ White noise by itself does not guarantee the result you need; you need Gaussian white noise. $\endgroup$
    – Dilip Sarwate
    Commented Jul 4, 2013 at 18:19

1 Answer 1

Reset to default
12
$\begingroup$

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.

Improve this answer
$\endgroup$
5
  • $\begingroup$ (+1) I think this is said implicitly in condition (1) but you also require that $\mu$ is constant, right? $\endgroup$
    – Macro
    Commented Jun 19, 2012 at 18:05
  • $\begingroup$ @Macro Yes stationarity, even weak sense (covariance) stationarity requires a constant mean and a constant variance. $\endgroup$
    – Michael R. Chernick
    Commented Jun 19, 2012 at 18:16
  • $\begingroup$ @MichaelChernick, then we could reproduce the joint distribution of stochastic process (or simulate stochastic process itself) by having its mean and autocovariance? $\endgroup$
    – Flying pig
    Commented Jun 19, 2012 at 19:05
  • $\begingroup$ @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. $\endgroup$
    – Michael R. Chernick
    Commented Jun 19, 2012 at 20:19
  • $\begingroup$ @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. $\endgroup$
    – Michael R. Chernick
    Commented Jun 19, 2012 at 20:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.