I'm quite enamoured with likelihood ratios as a means of quantifying relative evidence in scientific endeavours. However, in practice I find that the raw likelihood ratio can get unprintably large, so I've taken to log-transforming them, which has the nice side-benefit of representing evidence for/against the denominator in a symmetric fashion (i.e. the absolute value of the log likelihood ratio represents the strength of evidence and the sign indicates which model, the numerator or denominator, is the supported model). Now, what choice of logarithm base? Most likelihood metrics use log-base-e, but this strikes me as a not very intuition-friendly base. For a while I used log-base-10, which apparently was dubbed the "ban" scale by Alan Turing and has the nice property that one can easily discern relative orders of magnitude of evidence. It recently occurred to me that it might be useful also to employ log-base-2, in which case I thought it might be appropriate to use the term "bit" to refer to the resulting values. For example, a raw likelihood ratio of 16 would transform to 4 bits of evidence for the denominator relative to the numerator. However, I wonder if this use of the term "bit" violates its conventional information theoretic sense. Any thoughts?

  • $\begingroup$ If you're already ok with ban, then you should also be fine with bit. (i.e., no need for the full justification I gave below). Going from ban to bits is just a change of units, log base 2 instead of base 10. (Similarly, go with "nats" if you like base e). $\endgroup$
    – jpillow
    Jul 11, 2011 at 9:24

1 Answer 1


I think it's perfectly well justified. (In fact, I've use this convention in papers I've published; or you can call them "nats" if you prefer to stick with logarithms of base $e$).

The justification runs as follows: the log-likelihood of the fitted model can be viewed as a Monte Carlo estimate of the KL divergence between the "true" (unknown) data distribution and the distribution implied by the fitted model. Let $P(x)$ denote the "true" distribution of the data, and let $P_\theta(x)$ denote the distribution (i.e., the likelihood $P(x|\theta))$ provided by a model.

Maximum likelihood fitting involves maximizing

$L(\theta) = \frac{1}{N}\sum_i \log P_\theta(x_i) \approx \int P(x) \log P_\theta(x) dx$

The left hand side (the log-likelihood, scaled by the # datapoints $N$) is a Monte Carlo estimate for the right hand side, i.e., since the datapoints $x_i$ were drawn from $P(x)$. So we can rewrite

$L(\theta) \approx \int P(x) \log P_\theta(x) dx = \int P(x) \log \frac{P_\theta(x)}{P(x)} dx + \int P(x) \log P(x)dx$

$ = -D_{KL}(P,P_\theta) - H(x)$

So the log-likelihood normalized by the number of points is an estimate of the (negative) KL-divergence between $P$ and $P_\theta$ minus the (true) entropy of $x$. The KL divergence has units of "bits" (if we use log 2), and can be understood as the number of "extra bits" you would need to encode data from $P(x)$ using a codebook based on $P_\theta(x)$. (If $P = P_\theta$, you don't need any extra bits, so KL divergence is zero).

Now: when you take the log-likelihood ratio of two different models, it should be obvious that you end up with:

$\log \frac{P_{\theta_1(x)}}{P_{\theta_2}(x)} \approx D_{KL}(P,P_{\theta_2}) - D_{KL}(P,P_{\theta_1})$

The entropy $H(x)$ terms cancel. So the log-likelihood ratio (normalized by $N$) is an estimate of the difference between the KL divergence of the true distribution and the distribution provided by model 1, and the true distribution provided by model 2. It's therefore an estimate of the number of "extra bits" you need to code your data with model 2 compared to coding it with model 1. So I think the "bits" units are perfectly well justified.

One important caveat: when using this statistic for model-comparison, you should really use LLR computed on cross-validated data. The log-likelihood of training data is generally artificially high (favoring the model with more parameters) due to overfitting. That is, the model assigns this data higher probability than it would if it were fit to an infinite set of training data and then evaluated at the points $x_i \dots x_N$ in your dataset. So the procedure many people follow is to:

  1. train models 1 and 2 using training data;

  2. evaluate the log-likelihood ratio of a test dataset and report the resulting number in units of bits as a measure of the improved "code" provided by model 1 compared to model

The LLR evaluated on training data would generally give an unfair advantage to the model with more parameters / degrees of freedom.

  • $\begingroup$ But the approximation to KL divergence is approximate, no? It's only good up to a constant scaling factor. So if the log likelihood ratio is 2, that only tells you that there is a difference of $2x$ bits, where $x$ is unknown. So it tells you essentially nothing. Plus, the approximation assumes infinite sample size, and for finite sample sizes, the KL divergence should not depend on sample size but log likelihood will. I don't see how what you're saying could possibly work. $\endgroup$ Jul 11, 2011 at 2:46
  • $\begingroup$ No, you're mistaken. Look at the statement: It's an estimate of the KL divergence. It's a Monte Carlo estimate: it converges asymptotically in accordance with the CLT. (Look closely, there's a factor of 1/N in front). So its accuracy will depend on the number of samples, but this will be true for any quantity computed from samples. In the limit, the log LLR will be (as stated) the difference between the number of extra bits needed to encode an average sample with model #2 vs. with model #1. $\endgroup$
    – jpillow
    Jul 11, 2011 at 3:03
  • 1
    $\begingroup$ @jpillow great answer, thanks ... Do you think it would also make sense to divide the LLR by the number of samples (ie the length of the data), and then to present the results as "bits / sample"? In my case the sample is an experimentally meaningful unit, not an arbitrary bin size. $\endgroup$
    – cxrodgers
    Jan 2, 2020 at 15:48
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    $\begingroup$ yes absolutely! We often divide by #samples, or in some cases by # seconds (e.g. when the samples come at a fixed rate) to obtain bits / s. $\endgroup$
    – jpillow
    Jan 2, 2020 at 19:17

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