0
$\begingroup$

I am trying to understand some chemical concentration data I have measured. I am taking the log of the ratio of two concentrations. The ratio itself is from oscillating timeseries data and is the (local maxima in conc):(local minima in conc). I then take the log of this ratio. When I do this and plot the distribution is appears to my eyes as being roughly log-normal with the bulk of the probability at around 0.75 (using log base 2) and a long tail skewed-right (despite the plot there is no mass less than zero as the local max>local min).
enter image description here

I would like to find outliers in this distribution as well as find mean+SD cutoffs to threshold it (comparing to another similar dataset from a treatment condition and I want to characterize what the common variation is in this distribution as a control). Is it best to transform this to a log-normal distribution to do something like this? Should I check other distributions as well? Any relatively straightforward suggestions would be appreciated.

Here is the actual sorted data that was used to plot the kernel density seen above. I have many other realizations of distributions of log ratios similar to this one:

{0.34, 0.35, 0.38, 0.42, 0.45, 0.47, 0.47, 0.53, 0.54, 0.56, 0.59, 0.6, 0.61, 0.61, 0.62, 0.65, 0.71, 0.72, 0.8, 0.84, 0.9, 0.92, 0.95, 0.96, 0.96, 1.68, 1.81, 2.03, 2.03, 3.19, 3.19, 3.37, 3.79, 4.65, 4.75}

$\endgroup$
4
  • 1
    $\begingroup$ It is a common mistake (I made it early in my career) to view a mixture of two or more symmetric distributions as lognormal. For physical reasons we would expect your logarithms not to follow a lognormal distribution; and indeed, four beautiful normal-looking components are clearly evident in your plot. However, I suspect this may be an artifact of your graphics: you seem to have plotted a kernel density smooth of your data, rather than the actual distribution as you stated. It would be better to show us the actual data. $\endgroup$ – whuber Feb 20 '17 at 23:18
  • $\begingroup$ @whuber Thanks for the suggestion. I have added the log ratio data used to generate the kernel density. I have been doing exploratory data analysis on this with Mathematica and when I use FindDistribution[] it generally returns log normal distribution, mixture distribution, gamma or weibull. $\endgroup$ – user13999 Feb 21 '17 at 0:08
  • $\begingroup$ @whuber Btw, mixture distributions found are generally mixture of normal distributions. $\endgroup$ – user13999 Feb 21 '17 at 0:17
  • 1
    $\begingroup$ In your case it is unlikely that components of a mixture would be normal. That is because if you do have a mixture, it probably reflects an underlying discrete variable that has not been included in the model. Within each mixture the distribution is the log of the ratio of two extremes. It will never be the case that the log of that ratio is truly normally distributed (although it's possible that a normal approximation would suffice). I would expect each of the mixture components to have a slight, but evident, positive skewness. $\endgroup$ – whuber Feb 21 '17 at 18:08
2
$\begingroup$

Thanks for posting the data. Here is a slightly tougher check on whether the data are lognormal, a normal quantile plot of the logged ratios.

enter image description here

Considerable caution is indicated:

  1. The sample size is 35. Easy to say, but that is a small sample for this kind of exercise.

  2. The grouping is suggestive, or may be just a quirk as can be expected in any sample of this size. Certainly you should check whether there is anything distinctive underlying the 10 highest values.

  3. The fit is middling, but I didn't search through other distributions to try to find a better fit.

I don't see why mixtures are expected to be mixtures of normals. It's my impression that that is the most common kind of mixture fitted to data, a different point.

I used natural logarithms, but using log base 2 would clearly just change axis labels, and nothing fundamental.

$\endgroup$
5
  • $\begingroup$ Thanks for your reply. Much of the data looks something like this. Any idea how to generate a threshold in the low end of the data? I am comparing this control peak:trough to peak:trough dist for a treatment condition and want to make a threshold for some minimal log ratio needed to be considered as similar to control. I need something like typical (mean - 3*SD) but that seems unhelpful here because of skew. Thought maybe log-normal would allow me to do the same but should I just go with an empirical quantile threshold value working with distribution as is? $\endgroup$ – user13999 Feb 21 '17 at 1:52
  • $\begingroup$ You seem to be asking a new question in the comment. The approaches seem quite contradictory. A lognormal fit implies that all the data fit into a single distribution: splitting the data at a threshold implies something else. You might need to explain more in a new or revised question on why that makes sense. $\endgroup$ – Nick Cox Feb 21 '17 at 8:23
  • $\begingroup$ I'm just trying to come up with simple cutoff in the left-end of the distribution to say that any peak:trough log ratios less than this cutoff observed in a treatment group are extremely unlikely to have come from this population. Similar to a one-tail test if I recall correctly. $\endgroup$ – user13999 Feb 21 '17 at 16:07
  • 1
    $\begingroup$ I see no similarity to any standard test here. A test compares groups that are known in advance to be distinct. You want to find a division into groups. I don't see that or how distribution fitting can help here. You'd need a way to compare parts of the data that are and are not well fitted by the distribution and that requires prior identification of the very cut-off you are seeking. An iterative approach is not ruled out but I suspect that a greater injection of chemical principles might be more helpful. I am not the person to provide it! $\endgroup$ – Nick Cox Feb 21 '17 at 16:16
  • $\begingroup$ It's moot, on the evidence that you have provided, whether a lognormal is a good reference distribution at all. $\endgroup$ – Nick Cox Feb 21 '17 at 16:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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