On a non-parametric confidence interval for the regression slope

Róbert Tóth; Ján Somorčík

doi:10.1007/s40300-017-0109-z

METRON

December 2017, Volume 75, Issue 3, pp 359–369 | Cite as

On a non-parametric confidence interval for the regression slope

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Róbert Tóth
Ján SomorčíkEmail author

Article

First Online: 17 April 2017

Received: 17 August 2016

Accepted: 08 April 2017

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Abstract

We investigate an application of the Tukey’s methodology in Theil’s regression to obtain a confidence interval for the true slope in the straight line regression model with not necessarily normal errors. This specific approach is implemented since 2005 in an R package; however, without any theoretical background. We illustrate by Monte Carlo, that this methodology, unlike the classical Theil’s approach, seriously deflates the true confidence level of the resulting interval. We provide also rigorous proofs in case of four (in general) and five data points (under some additional conditions); together with a real life usage example in the latter case. Summing up, we demonstrate that one should never combine statistical methods without checking the assumptions of their usage and we also give a warning to the already wide community of R users of Theil’s regression from various fields of science.

Keywords

Theil’s regression Tukey’s confidence interval Walsh averages Software R

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Notes

Acknowledgements

The authors would like to thank the Editor-in-Chief and an anonymous referee for their helpful suggestions that improved the readability of the paper. The work was supported by the VEGA grant No. 2/0047/15 of the Science Grant Agency of the Slovak Republic.

Appendix

Proof of Theorem 1

Part a): We prove the equivalent statement “\(w_9<s_2\) iff \(s_1+s_9 < 2s_2\)”. Start with \(w_9<s_2\) and consider the 9 smallest Walsh averages \(w_1<w_2<\cdots <w_9\). Each of them is of the form \((s_i+s_j)/2\) for some \(i \le j\) and since \(s_2=(s_2+s_2)/2\), the assumption \(w_9<s_2\) means that \((s_i+s_j)/2 < (s_2+s_2)/2\). This sharp inequality, together with \(s_1< s_2< \cdots < s_{10}\), immediately implies, that \(i=1\) and the 9 smallest Walsh averages \(w_1<w_2<\cdots <w_9\) have to be of the form \((s_1+s_1)/2< (s_1+s_2)/2< \cdots < (s_1+s_9)/2\). Therefore, the inequality \(w_9<s_2\) can be rewritten as \((s_1+s_9)/2 < (s_2+s_2)/2\) and the first part of the proof is complete.

Now, start with \(s_1+s_9 < 2s_2\), i.e. \((s_1+s_9)/2 < s_2\). Since the 8 Walsh averages \((s_1+s_1)/2< (s_1+s_2)/2< \cdots < (s_1+s_8)/2\) are even smaller then \((s_1+s_9)/2\), we see that there are at least 9 Walsh averages smaller than \(s_2\). Therefore, also the 9-th smallest Walsh average, i.e. \(w_9\), is smaller than \(s_2\).

Part b) follows from part a) by symmetry arguments. \(\square \)

Proof of Theorem 2

Split the whole probability space into these four disjoint random events: \(A=\{s_2\le w_9 \wedge w_{47}\le s_9\}\), \(B=\{ s_2\le w_9\wedge s_9< w_{47}\}\), \(C=\{ w_9<s_2\wedge w_{47}\le s_9\}\), \(D=\{ w_9<s_2 \wedge s_9<w_{47}\}\). Denote by U the random event \(\{\beta _{1} \in (w_9,w_{47})\}\). Note that the minimum and the maximum of all slopes \(s_i\) and their Walsh averages \(w_i\) are \(s_1\) and \(s_{10}\), respectively. This together with Theorem 1 implies that \(P(U) = P(U \cap A) + P(U \cap B)+P(U \cap C)+P(U \cap D) \le P( \{\beta _{1}\in (s_2,s_9)\} \cap A) + P( \{\beta _{1}\in (s_2,s_{10})\} \cap B) + P( \{\beta _{1}\in (s_1,s_9)\} \cap C) + P( \{\beta _{1}\in (s_1,s_{10})\} \cap D) = p_1+p_2+p_3+p_4\). \(\square \)

Proof of Theorem 3

By contradiction, let \(S_{ac}\) be the greatest of \(S_{ab}\), \(S_{ac}\), \(S_{bc}\) – the case that \(S_{ac}\) is the smallest can be treated analogously. Since \(S_{ij} = \beta _1 + (\varepsilon _i-\varepsilon _j)/(x_i-x_j)\) and \(x_a<x_b<x_c\), one observes that \(S_{ac}>S_{ab}\) and \(S_{ac}>S_{bc}\) are equivalent to \((\varepsilon _a-\varepsilon _c)(x_a-x_b)>(\varepsilon _a-\varepsilon _b)(x_a-x_c)\) and \((\varepsilon _a-\varepsilon _c)(x_b-x_c)>(\varepsilon _b-\varepsilon _c)(x_a-x_c)\), respectively. By summing the latter pair of inequalities we obtain \((\varepsilon _a-\varepsilon _c)(x_a-x_c)>(\varepsilon _a-\varepsilon _c)(x_a-x_c)\), which is impossible. \(\square \)

Proof of Theorem 4

Theorem 3 implies that \(s_1=S_{i,i+1}\) and \(s_{10}=S_{j,j+1}\) for some distinct i and j from \(\{1,2,3,4\}\). Since \(|i-j| \le 3\), one obtains by straightforward algebra, that \(s_{10}-s_1=|i-j| \cdot |S_{i,j} - S_{i+1,j+1}| \le 3(s_9-s_2)\) under Condition 1 and the assumption that both \(S_{i,j}\) and \(S_{i+1,j+1}\) belong to \(\{s_2,s_3,\ldots ,s_{9}\}\). However, summing the inequalities \(s_1+s_9 < 2s_2\) and \(2s_9 < s_2+s_{10}\) appearing in the definition of \(p_4\) yields the opposite inequality, hence \(p_4=0\).

It remains the case when not both \(S_{i,j}\) and \(S_{i+1,j+1}\) belong to \(\{s_2,s_3,\ldots ,s_{9}\}\). This happens iff \(|i-j|=1\). Without loss of generality, we will suppose that \(j=i+1\) and \(i \le 3\).

(a): The case \(i \le 2\). Because \(s_2 \le S_{i+2,i+3}\), \(S_{i+1,i+3} \le s_9\) and \(s_{10}=S_{i+1,i+2}\), the inequality \(2s_9 < s_2+s_{10}\) appearing in the definition of \(p_4\) would imply that \(2 S_{i+1,i+3} < S_{i+2,i+3} + S_{i+1,i+2}\), which is equivalent to \(0<0\) under Condition 1, hence \(p_4=0\).
(b): The case \(i=3\). Compare \(S_{2,4}\), \(S_{2,3}\), and \(S_{3,4}\) with \(s_2\), \(s_9\), and \(s_1\), respectively, to obtain a contradiction to the inequality \(s_1+s_9 < 2s_2\) appearing in the definition of \(p_4\).\(\square \)

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Róbert Tóth
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Ján Somorčík
- 2
Email authorView author's OrcID profile

1.Tangent WorksBratislavaSlovakia
2.Comenius University BratislavaBratislavaSlovakia

On a non-parametric confidence interval for the regression slope

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