# What is the statistical logic behind using logarithms in analyzing threshold elevations?

In my field, a lot of experiments are done comparing performance between a baseline condition, and a more difficult condition. Typically, thresholds are measured at each of these two conditions, and a threshold elevation is calculated.

So, suppose for a particular observer, the contrast detection threshold (e.g. the luminance of a circular patch on perfectly black background at which an observer is able to detect whether it is present or absent 75% of the time) is found to be 0.001 cd/m^2. And suppose that in a more challenging condition, where before each trial, a bright flash occurs, that same observer's threshold rises to 0.01 cd/m^2, meaning that more light is required for successful detection.

The threshold elevation can be expressed in at least two ways:

1) as a ratio between the baseline and more challenging condition: in this case it would be 0.01/0.001 = 10.

2) as a difference of the logs, so log(0.01)-log(0.001) = (-2) - (-3) = 1.

From what I have been told, the advantage of using the log transform is that it stabilizes the variances. So, if you had several observers performing this experiment, there would be a certain spread of values for each of the two conditions. If you were to use the ratio approach (as in (1) above), then the variances also get multiplied. But with a log transform, the variances do not grow, since the two conditions are added, rather than multiplied.

Am I on the right track with my interpretation? I'm still not comfortable with this yet, for the following reasons:

1) so what if the variances are multiplied - why does this threaten the integrity of any statistical calculations done later (such as whether the threshold elevation is significantly different from a value of "1")?

2) Doesn't doing statistics on a log transform somehow mean that you're not actually doing statistics on the underlying concept of the threshold elevation (which, to me, seems to be more of a ratio concept)? Or have I got it backwards, and is a log space actually more representative of the logarithmic nature of perception?

I'd really appreciate any guidance here - I've tried looking for resources that explain, in a textbook fashion, the logic behind this approach, but I'm coming up short.

thanks.

## 3 Answers

There are at least two reasons for using a log transform in this situation. One, as you alluded to it, is to stabilize the variance. This applies to many different types of data, and the reasons and caveats described in your link above apply.

For psychophysical data like yours, however, there's also a second reason to log-transform the data. Many perceptual phenomenon follow an interesting relationship: the perceived intensity of a stimulus (e.g., how bright or loud a stimulus appears) is proportional to the log of its physical magnitude (e.g., measured in lux or sound pressure). This called Fechner's law. In some cases, you may want to log the physical units so that they more closely resemble the subjects' percepts.

• Thanks for the reply Matt. Yep, after reading through that 1995 paper, I have a greater appreciation for how a log transform is more appropriate for data that involves ratios, such as in the case of threshold elevations. However, I'm still trying to grasp the variance stabilization idea. I get that if you multiply or divide two distributions (as you would if you didn't log transform the data), you end up doing the same to the variances, but why exactly is this a bad thing? – spacediver Jan 23 '15 at 5:49

There is nothing wrong with transforming data before statistical analysis. A log transformation is highly useful when dealing with geometric response curves, which otherwise are not well behaved.

• Well, a reciprocal transform inverses the relationship between the numbers compared to the original data set, so that kinda changes the meaning. I did, however, just come across this, which may help me better understand my situation: nki.nl/media/837444/log.pdf – spacediver Jan 23 '15 at 2:20

Ok so I think I may have part of an answer. Suppose you have a baseline condition and two other conditions, and you want to compare the latter two conditions with each other (each expressed as a ratio relative to baseline). One of the advantages of a log transform is that it renders the variances of data sets more similar, which will mean they are more likely to meet the assumptions of homogeneity of variance of a statistical test.