Given two unit vectors in a real inner product space, one can define the *correlation* between these vectors to be their inner product , or in more geometric terms, the cosine of the angle subtended by and . By the Cauchy-Schwarz inequality, this is a quantity between and , with the extreme positive correlation occurring when are identical, the extreme negative correlation occurring when are diametrically opposite, and the zero correlation occurring when are orthogonal. This notion is closely related to the notion of correlation between two non-constant square-integrable real-valued random variables , which is the same as the correlation between two unit vectors lying in the Hilbert space of square-integrable random variables, with being the normalisation of defined by subtracting off the mean and then dividing by the standard deviation of , and similarly for and .

One can also define correlation for complex (Hermitian) inner product spaces by taking the real part of the complex inner product to recover a real inner product.

While reading the (highly recommended) recent popular maths book “How not to be wrong“, by my friend and co-author Jordan Ellenberg, I came across the (important) point that correlation is not necessarily transitive: if correlates with , and correlates with , then this does not imply that correlates with . A simple geometric example is provided by the three unit vectors

in the Euclidean plane : and have a positive correlation of , as does and , but and are not correlated with each other. Or: for a typical undergraduate course, it is generally true that good exam scores are correlated with a deep understanding of the course material, and memorising from flash cards are correlated with good exam scores, but this does not imply that memorising flash cards is correlated with deep understanding of the course material.

However, there are at least two situations in which some partial version of transitivity of correlation can be recovered. The first is in the “99%” regime in which the correlations are *very* close to : if are unit vectors such that is *very* highly correlated with , and is *very* highly correlated with , then this *does* imply that is very highly correlated with . Indeed, from the identity

(and similarly for and ) and the triangle inequality

Thus, for instance, if and , then . This is of course closely related to (though slightly weaker than) the triangle inequality for angles:

Remark 1(Thanks to Andrew Granville for conversations leading to this observation.) The inequality (1) also holds for sub-unit vectors, i.e. vectors with . This comes by extending in directions orthogonal to all three original vectors and to each other in order to make them unit vectors, enlarging the ambient Hilbert space if necessary. More concretely, one can apply (1) to the unit vectorsin .

But even in the “” regime in which correlations are very weak, there is still a version of transitivity of correlation, known as the *van der Corput lemma*, which basically asserts that if a unit vector is correlated with *many* unit vectors , then many of the pairs will then be correlated with each other. Indeed, from the Cauchy-Schwarz inequality

Thus, for instance, if for at least values of , then (after removing those indices for which ) must be at least , which implies that for at least pairs . Or as another example: if a random variable exhibits at least positive correlation with other random variables , then if , at least two distinct must have positive correlation with each other (although this argument does not tell you *which* pair are so correlated). Thus one can view this inequality as a sort of `pigeonhole principle” for correlation.

A similar argument (multiplying each by an appropriate sign ) shows the related van der Corput inequality

and this inequality is also true for complex inner product spaces. (Also, the do not need to be unit vectors for this inequality to hold.)

Geometrically, the picture is this: if positively correlates with all of the , then the are all squashed into a somewhat narrow cone centred at . The cone is still wide enough to allow a few pairs to be orthogonal (or even negatively correlated) with each other, but (when is large enough) it is not wide enough to allow *all* of the to be so widely separated. Remarkably, the bound here does not depend on the dimension of the ambient inner product space; while increasing the number of dimensions should in principle add more “room” to the cone, this effect is counteracted by the fact that in high dimensions, almost all pairs of vectors are close to orthogonal, and the exceptional pairs that are even weakly correlated to each other become exponentially rare. (See this previous blog post for some related discussion; in particular, Lemma 2 from that post is closely related to the van der Corput inequality presented here.)

A particularly common special case of the van der Corput inequality arises when is a unit vector fixed by some unitary operator , and the are shifts of a single unit vector . In this case, the inner products are all equal, and we arrive at the useful van der Corput inequality

(In fact, one can even remove the absolute values from the right-hand side, by using (2) instead of (4).) Thus, to show that has negligible correlation with , it suffices to show that the shifts of have negligible correlation with each other.

Here is a basic application of the van der Corput inequality:

Proposition 2 (Weyl equidistribution estimate)Let be a polynomial with at least one non-constant coefficient irrational. Then one haswhere .

Note that this assertion implies the more general assertion

for any non-zero integer (simply by replacing by ), which by the Weyl equidistribution criterion is equivalent to the sequence being asymptotically equidistributed in .

*Proof:* We induct on the degree of the polynomial , which must be at least one. If is equal to one, the claim is easily established from the geometric series formula, so suppose that and that the claim has already been proven for . If the top coefficient of is rational, say , then by partitioning the natural numbers into residue classes modulo , we see that the claim follows from the induction hypothesis; so we may assume that the top coefficient is irrational.

In order to use the van der Corput inequality as stated above (i.e. in the formalism of inner product spaces) we will need a non-principal ultrafilter (see e.g this previous blog post for basic theory of ultrafilters); we leave it as an exercise to the reader to figure out how to present the argument below without the use of ultrafilters (or similar devices, such as Banach limits). The ultrafilter defines an inner product on bounded complex sequences by setting

Strictly speaking, this inner product is only positive semi-definite rather than positive definite, but one can quotient out by the null vectors to obtain a positive-definite inner product. To establish the claim, it will suffice to show that

for every non-principal ultrafilter .

Note that the space of bounded sequences (modulo null vectors) admits a shift , defined by

This shift becomes unitary once we quotient out by null vectors, and the constant sequence is clearly a unit vector that is invariant with respect to the shift. So by the van der Corput inequality, we have

for any . But we may rewrite . Then observe that if , is a polynomial of degree whose coefficient is irrational, so by induction hypothesis we have for . For we of course have , and so

for any . Letting , we obtain the claim.

## 27 comments

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5 June, 2014 at 12:14 pm

Fred LunnonSingle bars in 3 out of 5 displayed formulae preceding (1), perhaps?

5 June, 2014 at 9:42 pm

Emmanuel KowalskiThe non-transitivity of correlation also comes up in representation theory, with characters (or approximate versions such as trace functions) as unit vectors; I’ve taken to interpret “X is correlated with Y” as meaning “X has something in common with Y”, which to me carries the right intuition.

It can be interpreted algebraically rigorously with the “common parts” being common irreducible subrepresentations.

10 June, 2014 at 9:11 am

Emmanuel KowalskiAh! And now that I have Jordan’s book, I see that this more or less exactly the interpretation he suggests!

6 June, 2014 at 2:24 am

Basil KRecently I’ve come across the following property for an element a in some “tolerance space” (A,T) (a set A with a reflexive and symmetric relation T):

for all b, b’ in A, if (b T a) and (a T b’) then (b T b’) .

I was wondering what would make a reasonably “natural” example for such special elements a, which facilitate transitions in an otherwise not necessarily transitive setting, and I’ve been asking everyone I know about it – and even people I don’t know.

I understand that the spirit of this post is analytic rather than algebraic (as my Math Stackexchange question is), but surely the cone intuition seems to be common to both, so I thought I’d share.

7 June, 2014 at 2:16 am

George ShakanThat is quite an elegant proof of Proposition 1!

I think I found a couple typos. Of a single unit vector should be of a single unit vector . Also I in Proposition 1, probably the should be for some positive integer parameter .

[Corrected, thanks – T.]7 June, 2014 at 7:59 am

aquazorcarsonThanks for making Van der Corput so well-motivated. Very enlightening read.

7 June, 2014 at 9:16 am

dzakoI dont get the definition of the ultrafilter product $_p$. This is defined for complex sequences and $e(P)$ is clearly not a complex seq.

7 June, 2014 at 10:24 am

Terence Taomaps each natural number to the complex number , and is thus a complex sequence.

17 June, 2014 at 10:51 pm

AnonymousThe paragraph after equation 1 seems to have several inaccuracies. First it was not assumed that all the pairs have nonnegative correlations. Then i wasnt able to follow the pigeonhole estimates for both examples. Maybe I am missing something.

18 June, 2014 at 7:50 am

Terence TaoBoth arguments are valid even in the presence of some negative correlations. For instance, once one has , one must have for at least pairs , for if this were not the case, one would have for all but fewer than pairs , and if one uses the trivial upper bound for these exceptional pairs, one obtains a contradiction. Note that at no point in this argument is any non-negativity hypothesis on the required. Similarly for the second argument.

At a more intuitive level, any negative correlation between one pair of will only cause the other inner products on the RHS of (1) to become even

morepositive, if (1) is to hold (but the correlation of each pair maxes out at , so there has to be a lot of pairs that share in this positive correlation): making a pair of vectors at an obtuse angle will squeeze a lot of other angles to become more acute. So negative correlation is not actually an enemy here, and is in fact helpful in some ways. (For similar reasons, the pigeonhole principle is still mathematically valid – and even “stronger”, in some sense – if some pigeonholes are allowed to have a negative number of pigeons, despite the breakdown of the physical pigeon metaphor in this setting.)1 February, 2017 at 3:47 pm

KodluI believe the $\varepsilon^3$s should be $\varepsilon^4$s in the above comment, as in the main text below equation (2).

[Corrected, thanks – T.]18 June, 2014 at 8:35 pm

AnonymousThanks for your enlightening answer. That part is completely clear now. Now I am confused by the paragraph after equation (2). First if the inner product is standard Euclidean, then the set of vectors positively correlated with v is simply a half space, which is not so “narrow”. Also I don’t understand the explanation for why the equation is independent of dimension. Why does it help to know the weakly correlated pairs are exponentially rare as dimension goes up?

18 June, 2014 at 9:13 pm

Terence TaoHere, one should think of “positively correlated” as meaning “correlation at least epsilon for some fixed epsilon independent of dimension”. Then all the vectors will be squashed into a cone of aperture about , and the angular measure of this cone decays exponentially with the dimension. This reduces the number of opportunities for pairs of vectors in this cone to be orthogonal or to make obtuse angles with each other.

22 June, 2014 at 8:52 pm

AnonymousThe triangle inequality seems to give slightly weaker bounds than a determinant based approach (for eg.

http://math.stackexchange.com/questions/147374/correlations-between-3-random-variables) some people have used. For eg. if and are both 0.87, the triangle inequality gives > 0.48, but the determinant based approach gives > 0.5138. Is this alternate approach correct?

22 June, 2014 at 9:01 pm

Anonymousthe second to last line wasnt posted fully by the comment engine. restating that line, given 3 variables 1,2,3 and pairwise correlations x, y, z, triangle inequality gives the minimum bound as 0.48, while determinant > 0 gives a bound around 0.51.

23 June, 2014 at 4:21 pm

Terence TaoThe triangle inequality for angles , which gives the same bound as the determinant approach (as one can verify numerically after using the identity ), is slightly stronger than the bound coming from the triangle inequality for norms.

25 June, 2014 at 10:46 am

AnonymousThanks for clarifying this.

2 July, 2014 at 1:58 am

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15 December, 2014 at 1:30 am

hammyhamsterPresumably there are implications for investment porfolios here. It’s well-known that adding a high-volatility (aka variance of return) stock to a portfolio can decrease the portfolio’s overall variance if the high-vol stock is negatively correlated to the others. But lack of transitivity presumably increases the challenge of doing this for large portfolios ?

11 June, 2015 at 4:52 pm

AnonymousSorry if I’m being silly, but I don’t understand the explanation below equation (2). For example. let be a positive integer and , that is, all the remaining terms be such that there is not correlation between them and . In this case, what is stated in the paragraph below isn’t true. I think I’m missing something here, please help me out!

[Oops, there was a typo – the exponent 3 should have been a 4. Corrected now – T.]13 July, 2015 at 9:00 am

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