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I was wondering what relations and differences are between pivotal statistic versus distribution free statistic?

  1. From Wikipedia

    a pivotal quantity or pivot is a function of observations and unobservable parameters whose probability distribution does not depend on the unknown parameters 1 (also referred to as nuisance parameters).

  2. If I understand correctly, a statistic T is said to be distribution free, if the distribution of T(X) doesn't depend on the distribution of X. Examples are Kolmogorov-Smirnov test statistic.

My understanding about their differences are: a pivotal statistic still depends on the form of the distribution, but not on the value of its parameter? A distribution free statistic doesn't depend on either form or parameter of the distribution. Am I correct?

My question comes from a reply at MSE. I also appreciate that if you could answer my question there.

kjetil b halvorsen
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Tim
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1 Answers1

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In the first one, the distribution of the quantity involves an unknown parameter or parameters and DOES depend on the distribution of the data. What it doesn't depend on is that parameter or parameters.

So for $X\sim N(\theta,1)$, if we take $Q(\underline{x};\theta) = \bar{x}-\theta$, the distribution of $Q$ is $N(0,1/n)$. This is useful, because you can immediately write down an interval for $Q$ and hence back out an interval for $\theta$.

Note that if $X$ had a different distribution, $Q$ would no longer be $N(0,1/n)$. It's NOT distribution free, its distribution is free of $\theta$ (but note that $Q$ itself is still a function of $\theta$).

In the second one, the distribution of the statistic, $T(\underline{x})$ (which statistic doesn't involve any unknown parameters) - for some specific given value of some population quantities/parameters (so that you're under a null, not an alternative*) - doesn't depend on the distribution of the data.

So, for example, under the null, the distribution of the statistic in a sign test doesn't depend on the distribution the data were drawn from (it's always binomial, as long as the data are continuous, independent, etc). But said distribution sure as heck changes if you change the median difference from zero (i.e. move away from the null).

* if its distribution didn't depend on whatever population quantity or effect the test was trying to pick up under the alternative, it would be useless as a test - the power would always be the significance level

There may well be some circumstances where $Q = T(x-\theta)$ is pivotal at say a large class of location families, and where $T(x)$ is distribution free when $\theta=0$. I think it should work in a variety of circumstances; perhaps the aforementioned sign test would be an obvious place to start.

Glen_b
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    Thanks! What does the underline in $\underline{x}$ mean? – Tim Mar 09 '13 at 15:52
  • Is a distribution-free statistic always a pivot? – Tim Mar 10 '13 at 00:05
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    The underline emphasizes that I'm referring to the sample as a vector, not a single value from it. – Glen_b Mar 10 '13 at 01:09
  • No, not least because $T(x) \neq T(x-\theta)$. Even with that finessed away, I don't think it's generally the case that the same form of statistic might be used in both. It was surprising to me to consider that it might happen at all - I'd never contemplated it before your question - because of the way $Q$ is generally tied to the distribution of the data and $T$ is - by definition - not. – Glen_b Mar 10 '13 at 01:14
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    I guess perhaps my question wasn't clear. Given a statistic (as a measurable mapping, a statistic doesn't explicitly depend on the distribution of the sample), if it is distribution-free, is it ancillary? Note a statistic is called ancillary, if it is a pivot quantity. – Tim Mar 10 '13 at 05:02