# How do I account for numerical overflow with Adaptive MCMC?

EDIT: I tested Forgottenscience's solution below and it works; however, note that I found the working acceptance criterion to be that if $$\log\alpha \geq \log u$$, the point is accepted, where $$u\sim\mathcal{U}(0,1)$$.

I'm using Adaptive MCMC (i.e. Haario et. al 2001, link to the paper here) to optimize the covariance matrix/Metropolis-Hastings "step sizes" for a multi-dimensional proposal distribution, in order to use this proposal distribution with a Metropolis Hastings sampler. Due to the nature of the likelihood function that I need to use, I've found that at the best-fit set of parameters, which I use as the starting point for my Markov chain for adaptive MCMC, the likelihood evaluates to numerical infinity/overflow (as well as sets of parameters near this best fit). I know that for a non-adaptive MH sampler, I could use log-likelihood methods e.g. this post, but I've found that when I follow this solution for adaptive MCMC, the MH step sizes/proposal distribution variances that Adaptive MCMC iterates to are far too small, and I'm getting a MH acceptance ratio that is way too big (about 0.98) when running the MH samplers with those step sizes. Has anyone encountered a similar problem before? Thanks!

EDIT: My likelihood is of the form

$$\displaystyle\mathcal{L}\propto \prod_{n=1}^{N}{ \left(\frac{m_{t,n}^2\Sigma_{x,n}^2+\Sigma_{y,n}^2}{m_{t,n}^2\Sigma_{x,n}^4+\Sigma_{y,n}^4}\right)^{\displaystyle w_n/2}\times\exp\left\{{-\frac{1}{2}w_n\frac{\left[y_n-y_{t,n}-m_{t,n}(x_n-x_{t,n})\right]^2}{m_{t,n}^2\Sigma_{x,n}^2+\Sigma_{y,n}^2}}\right\}}$$,

where there are $$N$$ datapoints $$(x_n,y_n)$$, weighted by $$w_n$$; $$(\Sigma_{x,n}, \Sigma_{y,n})$$ are parameters related to the combined $$x-$$ and $$y-$$uncertainties of the $$n^\text{th}$$ datapoint and the model I'm fitting to with this likelihood, and $$m_{t,n},x_{t,n},y_{t,n}$$ are parameters that are different for each datapoint. I've pinpointed the reason why this likelihood blows up; I'm working with about 700 datapoints, many of which have small values of $$(\Sigma_{x,n}, \Sigma_{y,n})$$, which is making the prefactor term $$\left(\frac{m_{t,n}^2\Sigma_{x,n}^2+\Sigma_{y,n}^2}{m_{t,n}^2\Sigma_{x,n}^4+\Sigma_{y,n}^4}\right)^{w_n/2}$$ blow up, especially if the weight $$w_n$$ is high. Essentially, I can't modify these data or parameters, so it's unavoidable that this likelihood blows up. The only solution I can think of would be to somehow instead use $$\ln\mathcal{L}$$ (a summation instead of a product) for the Adaptive Metropolis algorithm, but I'm not sure how this would work with the Metropolis Hastings acceptance ratio.

• I have hardly seen cases, where the likelihood becomes numerically infinite. Could you maybe tell us more precisely how your likelihood looks like? Then it may be easier to help you. Mar 8, 2020 at 9:36
• I edited my post to answer your question; thanks! Mar 10, 2020 at 15:44
• Does this answer your question? Metropolis-Hastings using log of the density Mar 25, 2020 at 7:21
• Yes, it does; thank you! Mar 31, 2020 at 15:49

There is no issue in converting the entire problem to log-scale. In this case, with the prior $$\pi_0$$, the new point $$z'$$ and old point $$z$$
$$\log\alpha = \min \left \{0, \log \mathcal L(z') - \log \mathcal L(z) + \log \pi_0(z') - \log \pi_0(z) + \log q(z|z') - \log q(z'|z) \right \}$$
• Thank you! So in this case, I would still be accepting $z'$ with probability $\alpha$, as usual, for Haario et. al's Adaptive MCMC + Metropolis-Hastings? However, I'm confused about your notation $q$; what are you describing with $q$? Mar 10, 2020 at 16:12
• The probability has been changed to log-space, so you should use the appropriate log-transformed uniform, $-\log u$. $q$ is just a proposal density. Mar 10, 2020 at 16:16
• I guess I'm confused about one subtlety then: for the regular, "linear-space" MH, we accept some trial $z'$ and make it the next $z$ if $\alpha \geq u$, where $\alpha=\frac{\mathcal{L}(z')\pi_0(z')}{\mathcal{L}(z)\pi_0(z)}$ and $u\sim\mathcal{U}(0,1)$; in either case, we add $z'$ to the overall sampling, regardless if it becomes the next step in the Markov Chain. Then, for this log case, If I understand correctly, we accept $z'$ if $\log\alpha\geq-\log u$, and reject otherwise? Then, as this is Adaptive MH, we update the proposal $q$ only when a new point is accepted? Mar 10, 2020 at 17:21