6
$\begingroup$

This question is an extension of a related question about adding small probabilities. Suppose you have log-probabilities $\ell_1 \geqslant \ell_2$, where the corresponding probabilities $\exp(\ell_1)$ and $\exp(\ell_2)$ are too small to be distinguished from zero in the initial computational facility being used (e.g., base R). We want to find the log-diffference of these probabilities, which we denote by:

$$\ell_- \equiv \ln \big( \exp(\ell_1) - \exp(\ell_2) \big)$$

Questions: How can you effectively compute this log-difference? Can this be done in the base R? If not, what is the simplest way to do it with package extensions?

$\endgroup$
  • $\begingroup$ "where the corresponding probabilities exp(ℓ1) and exp(ℓ2) are too small to be distinguished from zero" - I think you mean one instead of zero? $\endgroup$ – Don Hatch Dec 18 '18 at 7:20
  • $\begingroup$ @Don Hatch: In the context of small probabilities it will usually be the case that $\ell_1$ and $\ell_2$ are high-magnitude negative numbers, so $\exp(\ell_1) \approx \exp(\ell_1) \approx 0$. The former are the log-probabilities and the latter are the actual probabilities, which are near zero. $\endgroup$ – Reinstate Monica Dec 18 '18 at 7:25
  • $\begingroup$ Why convert log probabilities rather than just take the difference of the logarithms of probabilities, which is the same as examining the log of the ratio of probabilities? This latter would seem more logical because if we did a log transformation for good reason, we should not abandon that good reason by back transforming before doing a calculation, but rather do testing in the preferred transformed system. $\endgroup$ – Carl Dec 18 '18 at 13:44
  • $\begingroup$ @Carl: There are many instances when one wishes to add or subtract small probability values. In these cases, performing an entirely different operation (e.g., ratio) is not helpful. The purpose of the logarithmic transformation is to be able to store very small probabilities on a scale where they are distinguishable from zero. $\endgroup$ – Reinstate Monica Dec 18 '18 at 22:39
  • $\begingroup$ The point is that if the ratio is significantly different from 1, the probabilities are then not the same even if the gross probabilities do not differ from zero. $\endgroup$ – Carl Dec 19 '18 at 3:47
10
$\begingroup$

To see how to deal with differences of this kind, we first note a useful mathematical result concerning differences of exponentials:

$$\begin{equation} \begin{aligned} \exp(\ell_1) - \exp(\ell_2) &= \exp(\ell_1) (1 - \exp(-(\ell_1 - \ell_2))). \\[6pt] \end{aligned} \end{equation}$$

This result converts the difference to a product, which allows us to present the log-difference as:

$$\begin{equation} \begin{aligned} \ell_- &= \ln \big( \exp(\ell_1) - \exp(\ell_2) \big) \\[6pt] &= \ln \big( \exp(\ell_1) (1 - \exp(-(\ell_1 - \ell_2))) \big) \\[6pt] &= \ell_1 + \ln (1 - \exp(-(\ell_1 - \ell_2))). \\[6pt] \end{aligned} \end{equation}$$

In the case where $\ell_1 = \ell_2$ we obtain the expression $\ell_+ = \ell_1 + \ln 0 = -\infty$. Using the Maclaurin series expansion for $\ln(1-x)$ we obtain the formula:

$$\begin{equation} \begin{aligned} \ell_- &= \ell_1 - \sum_{k=1}^\infty \frac{\exp(-k(\ell_1 - \ell_2))}{k} \quad \quad \quad \text{for } \ell_1 \neq \ell_2. \\[6pt] \end{aligned} \end{equation}$$

Since $\exp(-(\ell_1 - \ell_2)) < 1$ the terms in this expansion diminish rapidly (faster than exponential decay). If $\ell_1 - \ell_2$ is large then the terms diminish particularly rapid. In any case, this expression allows us to compute the log-sum to any desired level of accuracy by truncating the infinite sum to a desired number of terms.


Implementation in base R: It is possible to compute this log-difference accurately in base R using the log1p function. This is a primitive function in the base package that computes the value of $\ln(1+x)$ for an argument $x$ (with accurate computation even for $x \ll 1$). This primitive function can be used to give a simple function for the log-difference:

logdiff <- function(l1, l2) { l1 + log1p(-exp(-(l1-l2))); }

Implementation with VGAM package: Machler (2012) analyses accuracy issues in evaluating the function $\ln(1-\exp(-|x|))$, and suggests that use of the base R functions may involve a loss of accuracy. It is possible to compute this log-difference more accurately in using the log1mexp function in the VGAM package. This gives you the an alternative function for the log-difference:

logdiff <- function(l1, l2) { l1 + VGAM::log1mexp(l1-l2); }
$\endgroup$
3
$\begingroup$

The following workaround is often very useful for these sorts of problems:

  1. subtract the smaller of l1 and l2 from each of l1 and l2 (effectively, we have multiplied the probabilities by some constant z = exp(min(l1,l2)))
  2. Now compute the sum using standard functions (you can use log1p and expm1 if you want).
  3. Afterwards, add the quantity you subtracted in step 1.

A simple example:

l1 <- -2000 ## exp(-2000) is computational zero
l2 <- -2002
z <- min(l1,l2)
l1 <- l1 - z
l2 <- l2 - z
## now we are guaranteed that one of them is zero
y <- log(exp(l1) - exp(l2))
y + z

returns the correct value -2000.145, which is equal to -2000 + log(1-exp(-2)).

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.