5
$\begingroup$

Is it possible to generate numbers from Gamma distribution (with parameters shape=10, scale=15, say) which also follow a AR(1) process, simultaneously? If it's possible, than how to do that?

$\endgroup$
2
$\begingroup$

I don't think this sort of thing will generally be possible without changing some aspects of the usual definition of an AR.

[most of original answer removed, since I have better information]


Richard Hardy points out in the comments below that probably meant a Gamma marginal distribution (... and rereading your question, that makes sense); Richard thought to use an innovation-approach to create the model. While it's not possible to pick just any distribution for innovations, there is at least one such innvation-model that does work (see the Lawrance reference below).

Here's some references that will prove useful:

Grunwald G.K., Hyndman R.J., Tedesco, L. and Tweedie, R. L. (2000)
"Non-Gaussian Conditional Linear AR(1) Models,"
Australian & New Zealand Journal of Statistics, 42:4 (Dec) p479-495 DOI: 10.1111/1467-842X.00143 (Working paper here: http://robjhyndman.com/papers/clar.pdf)

This and the research report below discuss a variety of approaches to non-Gaussian AR(1) models. [The above paper gives a general formulation of non-Gaussian AR(1) models that includes nearly all published non-Gaussian AR(1) models, while the report below is partly a survey paper as well as beginning the synthesis of the paper above.]

Grunwald, G.K., Hyndman, R.J. and Tedesco, L.M. (1995),
"A unified view of linear AR(1) models,"
Research Report, Department of Statistics, University of Melbourne
(http://robjhyndman.com/papers/ar1.pdf)

Lawrance, A.J. (1982),
"The innovation distribution of a gamma distributed autoregressive process,"
Scand. J. Statist., 9, 234–236


[You could perhaps more readily construct an AR in the log of a gamma. Fitted to log-data, that would correspond to a shifting scale in the conditional distribution for the gamma.]

$\endgroup$
  • 1
    $\begingroup$ The $\varepsilon_t$ does not belong in the formula since its expectation is zero. Also, could you elaborate a little on how an AR(1) process seems to imply a need for a location-family? $\endgroup$ – Richard Hardy Nov 4 '15 at 9:11
  • $\begingroup$ @Richard Thanks for pointing out the epsilon. The effect of the previous observation shifts the mean linearly with $y_{t-1}$ but leaves the conditional variance unaltered. Under the condition that one also wants to have a particular distribution family (as was asked for in the question, which to begin with I interpreted as a requirement on the conditional distribution) the need to be able to shift the mean without changing the variance withing a distribution family suggests a location family for that conditional distribution (which would only change the location). $\endgroup$ – Glen_b -Reinstate Monica Nov 4 '15 at 11:05
  • $\begingroup$ Similar, but slightly different issues arise if we try to deal with the unconditional distribution being gamma. $\endgroup$ – Glen_b -Reinstate Monica Nov 4 '15 at 11:20
  • $\begingroup$ I thought the unconditional distribution had to be gamma, and I was interested how one could reject this intuitively, i.e. intuitively motivate why outcomes of AR(1) could not be unconditionally Gamma-distributed.. $\endgroup$ – Richard Hardy Nov 4 '15 at 11:51
  • $\begingroup$ @Richard Hmm. Are you anticipating that the shape of the conditional distribution changes from observation to observation in order to end up with a Gamma? (without that, there's a pretty obvious problem) $\endgroup$ – Glen_b -Reinstate Monica Nov 4 '15 at 12:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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