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I'd like to calculate Bayes Factor for two exponential datasets, testing $H_0: \lambda_1 = \lambda_2$ against $H_1: \lambda_1\not=\lambda_2$, and here comes the prior probability $\pi$ for all $\lambda$'s.
Question. How I choose a prior $\pi$ for $\lambda$'s if I want it to be non-informative?

My exponential data models depressive symptoms total score so I guess that prior for $\lambda$'s should have some kind of connection to this, right? I was thinking about flat prior $\pi$ ~ $U[0,1]$.
I am aware of the existence of Jeffreys prior, but as I have learnt from here they're not to be applicable here, 'cause this prior is improper (or 'too improper' for computing Bayes Factors).

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  • $\begingroup$ When I think of a non-informative prior, I often think of a uniform prior, so I think that should work $\endgroup$ – Andrew Marderstein Sep 25 '17 at 18:33
  • $\begingroup$ To my understanding, the Bayes factor doesn't require priors (it just comes from the ratio of two likelihoods). Are you after a ratio of posteriors (which would be a BF times a ratio of priors)? $\endgroup$ – Glen_b -Reinstate Monica Sep 26 '17 at 0:37
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You can use a Gamma Prior that is almost flat (very little informative) with some specific hyper-parameters if you want a conjugate prior.

Another possbility is to use, for example, an $U[0,k]$ with very large $k$. In both cases how large or small the parameters must be to become very little informative depends on the scale of your data.

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