What does it imply for standard deviation being more than twice the mean? Our data is timing data from event durations and so strictly positive. (Sometimes very small negatives show up due to clock resolution issues). We are accustomed to the following table (locally developed):

stdev / mean <= .5 : treat as normal distribution
stdev / mean >= .5 <= .75 : usually normal but might be exponential
stdev / mean >= .75 <= 2 : exponential / poisson
stdev / mean >= 2 : outside inhibitors dominate

In this case we got a ratio of 10 7 and outside inhibitors (meaningless external variables) are eliminated.

What we're trying to do is get some kind of estimate on whether the fat-tail is going to kill the estimate of the mean. The default model is noise applied to a constant time from an effectively constant load distribution, which we reject and replace with an exponential model of load distribution when the stdev gets too large. Observationally, we know that breakers almost always appear in the exponential distributions due to variables we cannot account for.

And then this thing popped up. We eliminated all external variables and still it remains. Our theoretical model for this case says it should be bi-modal normal (that is, the weighted sum of two normals) but this doesn't look like it. If it weren't for the fact we're reasonably confident we've seen the largest datapoint at just over 8 standard deviations away from the mean I'd think we haven't reached the second hump of the bi-modal distribution yet. Incidentally we have the median which is 13 times smaller than the mean.

For the fast answer, the plot does not exist because the mean and standard deviation are dominated by single outliers separated by more than the mean's value. If I set my histogram based on the median, I lose the important part of the graph off the right. If I set my histogram based on the mean, I blow the left-most bar off the top of the graph and the right-hand is still indistinguishable from noise because no histogram bar on the right is > 1.

  • 2
    $\begingroup$ In my experience it usually means that your data is overdispersed and/or heavy-tailed...perhaps even power-lawed. In this situation, the mean is biased by "outliers" or the presence of extreme value observations. Better measures of location are the median or one of the nonparametric estimators such as Lehman-Hodges... $\endgroup$ – Mike Hunter Apr 7 '16 at 16:11
  • 2
    $\begingroup$ Plot it and see is the best short answer I can give. I agree broadly with @DJohnson's "usually", but would start the list with strongly skewed before overdispersed (the term always sounds judgemental to me, but it's entrenched) and heavy-tailed. But watch out: it's not your situation at all, but many distributions with mean zero will qualify for SD $\gg$ mean (and some of those are heavy-tailed too, but e.g. a normal with mean 0 is not). $\endgroup$ – Nick Cox Apr 7 '16 at 16:19
  • 2
    $\begingroup$ It's not usual that the normal and Poisson are serious competitors, as the normal is continuous and the Poisson is discrete. $\endgroup$ – Nick Cox Apr 7 '16 at 16:22
  • 1
    $\begingroup$ Sure, the shapes can converge, but a Poisson has a variance equal to its mean, which is impossible, indeed meaningless, for a normal. Everything hinges on what you do after identification. Which calculations make sense depends on the detail. $\endgroup$ – Nick Cox Apr 7 '16 at 16:39
  • 6
    $\begingroup$ This question is completely mystifying. What are "breakers"? What is an "outside inhibitor"? What is this "thing" that "popped up"? How did a "bimodal normal" (which doesn't exist) suddenly replace an "exponential" model? What is the "second hump" (or even the first) and what does it mean to "reach" it? What does this have to do with "running totals"? What is the "plot" that does not exist and why would it be relevant? $\endgroup$ – whuber Apr 9 '16 at 19:21

Absolutely nothing.

Even in the case when you are dealing with normal distributions, these are examples of a location-scale family of distributions which means I can choose the center (mean) and spread (SD) to be anything I want it to be.

A normal probability model is a poor choice for modeling time-to-event outcomes. If the probability model is exponential, the variance is related to the square of the mean, so with an SD greater than the mean, we can infer that there is some evidence the mean is greater than "1" on whichever units you have used to measure the outcome. But that is purely ad hoc: you would do better to use maximum likelihood to estimate characteristics of the survival times directly, rather than make broad inferences.

In the case of one-sample hypothesis testing where your hypothesis is that the mean is 0, we can say a bit more. The standard deviation of the data is related to the standard error of the sample mean by the Central Limit Theorem: $SE = SD / \sqrt{n}$.

If you mean to say that the sample mean is less than 2 times the standard error, normal probability laws tell us that there is little evidence to support that the mean is nonzero.

  • $\begingroup$ We eliminate the normal distribution because the right-tail is still significant where it encounters the range limit at zero. $\endgroup$ – Joshua Apr 7 '16 at 16:28
  • 1
    $\begingroup$ @Joshua your statement doesn't quite make sense, nonetheless many time-to-event probability models have the same property of having an SD or mean be virtually anything (usu. positively valued). I still say it doesn't mean anything with neither context nor a hypothesis in mind. $\endgroup$ – AdamO Apr 7 '16 at 16:31
  • 9
    $\begingroup$ I think the first sentence is too strong. For example, the ratio SD/mean is the coefficient of variation and it does have both some theoretical and some practical use (e.g. it is well defined for the gamma distribution; with care it can be a useful descriptive measure). $\endgroup$ – Nick Cox Apr 7 '16 at 16:55
  • 2
    $\begingroup$ The emphasis (in the second half of this post) on a mean-0 hypothesis seems misdirected. The question itself explicitly states "Our data is timing data from event durations and so strictly positive." A fortiori, the mean cannot be zero, so such a test would be pointless. $\endgroup$ – whuber Apr 10 '16 at 19:20
  • 1
    $\begingroup$ The thread starts out with a very general question and then becomes more specific. It's hard to make sense of precisely what is expected. But I continue to think that your opening is needlessly dismissive. If the title question and your opening sentence are linked together with your comment just now, I think it's hard for people interested in this thread to see consistency. You're admitting what you appear to deny. $\endgroup$ – Nick Cox Apr 11 '16 at 7:21

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.