ADVI initialisation

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John Reid

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Feb 12, 2016, 11:48:41 AM2/12/16
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Thanks again for Stan! I'm finding the ADVI algorithm very useful for exploring my model. However I wonder if there is any way to initialise the means of the meanfield Gaussians in the transformed space? For some of my parameters I have some good guesses that I'd like to try as starting points. In fact also how does the initialisation work? Is it random?

Thanks,
John.

Dustin Tran

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Feb 12, 2016, 12:09:06 PM2/12/16
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Hi John, There’s an init argument which enables you to specify the means in the original (possibly constrained) latent variable space; see CmdStan manual for more details. I don’t believe we currently support initialization in the transformed/unconstrained space(?). It should be the same as if MCMC initializations supports it.

If you know the transformations beforehand, which are given in the manual, you can of course apply the inverse of the functions to your desired transformed parameters, and use them as initializations on the original parameter space.

Dustin
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John Reid

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Feb 12, 2016, 12:15:23 PM2/12/16
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OK I'll check the CmdStan manual. I'm using rstan 2.9 and I didn't see anything in the help for vb(). I was assuming the transformations would be the constrain_pars() and unconstrain_pars(). Am I right? And thinking about it initialization in the constrained space is more useful in any case.

Thanks for the quick reply,
John.
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Dustin Tran

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Feb 12, 2016, 12:43:02 PM2/12/16
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Yep, those two functions work as well.

Dustin

Ben Goodrich

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Feb 12, 2016, 1:33:59 PM2/12/16
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On Friday, February 12, 2016 at 12:15:23 PM UTC-5, John Reid wrote:
OK I'll check the CmdStan manual. I'm using rstan 2.9 and I didn't see anything in the help for vb(). I was assuming the transformations would be the constrain_pars() and unconstrain_pars(). Am I right? And thinking about it initialization in the constrained space is more useful in any case.

The rstan on CRAN has an init argument that takes a list of values in the constrained space, but you may have to install it from source because the binaries are not fully mirrored out yet.

Ben

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