3
$\begingroup$

I know the following info for values of a particular column for a dataset:

[mean=4.989209978967438, stddev=2255.654165352454, count=2400088]

Given just this, is it possible to approximate what the underlying distribution might be?

$\endgroup$
4
  • 2
    $\begingroup$ Any distribution that has a defined mean and standard deviation can be transformed by trivial means to have arbitrary mean and SD (linear transformation). $\endgroup$ Aug 1, 2019 at 21:08
  • $\begingroup$ I think the consensus is that no it is not possible. However, as a contrarian I must ask is "approximate the underlying distribution" well defined? Because if it is not, then the answer is yes - if you want it to be yes, and no otherwise. $\endgroup$
    – emory
    Aug 2, 2019 at 1:06
  • $\begingroup$ @emory: I am not sure what you meant by "approximate ..." being well defined. From what I understand from the general consensus, it is not possible. But if I understand you correctly, we might as well be able to "guess" a sample distribution with such characteristics. Is this possible? $\endgroup$ Aug 2, 2019 at 5:37
  • $\begingroup$ I mean we accept that Normal(0, 1) is a good approximation for the t-distribution with large degrees of freedom. Who decided that? What is large? I think for some purposes yes we can approximate everything with an appropriately parameterized normal distribution and for other purpose we can not. $\endgroup$
    – emory
    Aug 2, 2019 at 13:30

2 Answers 2

10
$\begingroup$

Not unless you already know what the distribution is. And likely not even then, unless it's a normal or lognormal distribution (which can be completely described by those two values).

You can calculation the mean and standard deviation from any set of numbers sampled from any distribution, so you cannot recreate a distribution based on them alone.

They do contain some information, of course. Your summaries indicate that the distribution is either heavily skewed or allows negative values (or both).


Edit: Since the distribution is skewed, it's definitely not a normal distribution, though it might be lognormal.

h/t to @ChrisHaug for pointing out my oversight about the lognormal

$\endgroup$
3
  • 2
    $\begingroup$ There aren't negative values, it's definitely highly skewed. $\endgroup$ Aug 1, 2019 at 15:06
  • 2
    $\begingroup$ Ok, but the normal distribution is not unique in being completely described by its mean and variance (take the lognormal for a strictly positive, right-skewed example). $\endgroup$
    – Chris Haug
    Aug 1, 2019 at 23:50
  • $\begingroup$ @ChrisHaug Thanks for catching that error. I've edited the answer to mention the lognormal; are there other distributions that are completely described by mean & variance? $\endgroup$
    – mkt
    Aug 2, 2019 at 10:04
0
$\begingroup$

You cannot conclude just from the first two moments what is the generating distribution. However, you can narrow the generating candidate distribution, depending on whether the number of observations is considered enough to converge to the Gaussian. If you assume, that your sample approximates the Gaussian, then you can say that the underlying distribution would belong in the family of alpha stable distributions with parameter $\alpha$ that allows the law of large numbers to work. Essentially, that describes the following work:

The Law of large numbers under fat tails by Nassim Taleb

Note however that you have some indicators of whether sample distribution approaches the Gaussian. Otherwise, you are blind

$\endgroup$
1
  • $\begingroup$ Alrighty, I'll check it out. $\endgroup$ Aug 2, 2019 at 5:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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