In a previous question I asked if I could scale the likelihood as my MCMC process advanced, to keep the acceptance fraction within a reasonable range (~0.2-0.5). I was told that this is not a valid approach, since doing that meant that the "Markov chain is no longer time homogeneous".

But, what if I used the burn-in stage to find an appropriate scale factor for my likelihood such that the acceptance fraction is reasonable? By "scale factor", I mean simply a real value that multiplies (and thus scales) my likelihood:

lkl_scaled = scale_factor * lkl_original

In this case I wouldn't be changing this factor during the MCMC process from which I later obtain the distributions of the model parameters. I would only do so during the burn-in phase, which I later discard.

I've tried this already and the results are excellent (where the chains get stuck forever with my original likelihood, they properly explore the parameters space with the scaled likelihood). I can't really see nothing wrong with this approach, but I'd like to be sure.

Is this a valid approach? If so, are there any caveats I should be aware of?

PD: I am aware of the existence of parallel tempering MCMC, but this approach is far simpler and it allows me to use other MCMC methods that would otherwise be of no use since the acceptance fraction is generally below 1%.

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    $\begingroup$ You are no longer sampling from the posterior if you do this. You are sampling from a fractional posterior, which is more-or-less equivalent to throwing data away. If I throw all my data away, of course I can sample much more easily (say, just generating from the prior), but this is obviously not worth the price I would be paying. $\endgroup$
    – guy
    Sep 19, 2018 at 19:25
  • $\begingroup$ Could you elaborate? If I define a more-or-less arbitrary likelihood for the problem at hand, what changes if I multiply it by some factor? $\endgroup$
    – Gabriel
    Sep 19, 2018 at 19:27
  • $\begingroup$ I can confirm I am not sampling from the prior because I am using uniform priors for all my parameters and the final distributions are exactly what one would expect for the problem being analyzed. There's no price being paid here, the opposite actually. $\endgroup$
    – Gabriel
    Sep 19, 2018 at 19:29
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    $\begingroup$ As I said, you are effectively tempering the likelihood, which is equivalent to using a fractional posterior. You are effectively throwing away 100 * scale percent of your data by doing this. If you want to know more about fractional posteriors then you can find that information easily, but suffice it to say most people don’t want to use fractional posteriors. The whole point of parallel tempering is to make use of the fact that fractional posteriors are easy to sample from to get you to samples from the thing you actually want. $\endgroup$
    – guy
    Sep 19, 2018 at 19:33
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    $\begingroup$ In that case, I suppose the question is whether or not you care about actually sampling from the posterior. If you want approximate samples from the posterior, the answer with regards to "is this approximately valid?" is no. If you don't care about actually sampling from the posterior, but have some other goal in mind, then your strategy may or may not be acceptable. I will say that the fractional posterior means, if you have non-informative priors, will be roughly OK on average, but the posterior variance of these parameters will most likely be greatly over-stated. $\endgroup$
    – guy
    Sep 19, 2018 at 19:58

1 Answer 1


To give an idea for what the effect of doing this is, consider the following simple Gaussian model. $$ Y_i = \mu + \epsilon_i, \qquad \epsilon_i \sim N(0,1). $$ Suppose we put a flat prior on $\mu$. The genuine posterior in this case is $[\mu \mid Y_1, \ldots, Y_n] \sim N(\bar Y, n^{-1})$. Let's compare this to the posterior obtained from scaling the log-likelihood. $$ \pi_c(\mu \mid \mathbf Y) \propto \left[\prod_{i = 1}^N e^{-(Y_i - \mu)^2 / 2}\right]^c = \prod_{i = 1}^N e^{-c(Y_i - \mu)^2 / 2}. $$ By the usual argument, this leads to a Gaussian posterior for $\mu$ as well: $$ \mu \sim N(\bar Y, (cn)^{-1}). $$

Now, what conclusions can we draw?

  • We are certainly not drawing from the genuine posterior.

  • In this particular case, the posterior mean turns out to be the same. This suggests that, for point estimation, there are at least certain situations in which scaling the log-likelihood does not lead to a disaster.

  • The scale factor is weakening our precision, and essentially corresponds to the posterior "throwing data away." For example, setting $c = 0.5$ is sort of like using half of the data.

  • This "throwing away data" idea does not affect the posterior mean in this case because we used a flat prior, but would have also impacted the posterior mean if we had used (say) $\mu \sim N(0,\tau)$; in particular, we would have gotten more shrinkage towards zero.

  • Even though we - for this particular example - have still got the correct posterior mean, the posterior variance is off. So, if I take $c$ to be very small to improve mixing, I should expect to be artificially inflating the variance by a factor of $1/c$.

The main concern with using this approach, and tuning $c$ to get a good acceptance rate, is that you will need to take $c$ very small. This is guaranteed to mess up your posterior variance, and is likely (but, evidently, not guaranteed) to mess up the posterior mean as well.

A more subtle problem if you do have weakly-informative priors is that, as the sample size grows, you will probably need to take $c$ going to $0$ to maintain a reasonable acceptance rate. This will cause the bias induced by the use of a weakly-informative prior to never be washed away by the data, so viewing this approach by embedding it in a standard Frequentist asymptotic analysis suggests that you may end up with inconsistent estimates.

  • $\begingroup$ The only difference I see with this example and my approach is how the scaling factor is applied. In this example the likelihood is scaled to some power, which will alter its shape (to flat or 1 in the c=0 limit), while my scale factor is multiplied by the likelihood which only changes the scale and not the shape. Do the same arguments made in this answer apply regardless? $\endgroup$
    – Gabriel
    Sep 19, 2018 at 21:30
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    $\begingroup$ @Gabriel If you are multiplying the likelihood by a constant (and not the log-likelihood as I assumed) then you aren't actually doing anything. For example, all Metropolis-Hastings algorithms are invariant to multiplicative constants - this is what makes them useful, since the normalizing constant is an unknown multiplicative constant. The previous question you linked to has the scale factor coming in the log-likelihood, not the likelihood. $\endgroup$
    – guy
    Sep 19, 2018 at 21:35
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    $\begingroup$ Sorry, you are correct and I am in fact multiplying the log-likelihood. I see where the power comes from now. Thank you for the extensive explanation guy! $\endgroup$
    – Gabriel
    Sep 19, 2018 at 21:51

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