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For which $n \times n$ correlation matrix $C$ can one construct Bernoulli random variables $(B_1, \ldots, B_n)$ with correlation $C$ ?

Following the approach described in this MO thread, one can think of the following construction. Define independent Bernoulli random variables $B_{k_1, \ldots, k_n}$ for $(k_1, \ldots, k_n) \in \mathbb{Z}^n$ and another independent $\mathbb{Z}^k$-valued random variable $I=(I_1, \ldots, I_n)$. Then $(B_{I_1}, \ldots, B_{I_n})$ is a correlated Bernoulli vector.

1: Is there any example of correlation structure that cannot be obtained this way ?

2: Any easy example of correlation matrix $C$ that cannot be the correlation matrix of a $\{0,1\}^n$ valued random vector ?

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Not sure this is important, but I guess your Bernoulli random variables are $\{0,1\}$-valued. Do you want them to have mean 1/2? Also, out of curiosity, why is this community wiki? –  Mark Meckes Mar 26 '10 at 13:29
Oh, you said correlation matrix, not covariance matrix. So it doesn't matter whether you assume $B_i$ are uniform in $\{0,1\}$, or have a more general distribution supported on two values. –  Mark Meckes Mar 26 '10 at 13:33
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3 Answers

up vote 7 down vote accepted

Here's a pretty general construction. Take unit vectors $v_1,\dots,v_n$ in $\mathbb{R}^n$ and let $u$ be a random unit vector (chosen with the uniform probability measure on the unit sphere). Define $B_i$ to be 1 if the inner product of $u$ and $v_i$ is positive and -1 otherwise. Then the correlation between $B_i$ and $B_j$ is the inner product of $v_i$ and $v_j$. (I haven't checked that carefully but I think it's true.)

Added in the light of the comment below: OK, I should have checked. It's actually not the inner product but π minus twice the angle between the two vectors all over π. I.e., it depends linearly on the angle between the two vectors, is 1 when that angle is zero and -1 when it is π. The angle is the inverse cos of the inner product, which gives us a formula.

So it gives us a fairly big supply of matrices -- I can't quite decide whether it gives us all (or rather all for which the variables take two values, each with probability 1/2).

A more general-looking construction is this. Take any probability space and let $A_1,...,A_n$ be sets of measure 1/2. Pick a random point x and let $B_i$ be 1 if x is in $A_i$ and -1 otherwise. But that becomes trivial, because if you have any set of Bernoulli variables taking the values $\pm 1$ with probability 1/2, then you can set $A_i$ to be the set where $B_i=1$.

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you mean that if a Gaussian vector $G = (\xi_1, \ldots, \xi_n)$ has correlation matrix C then the Bernoulli vector $B = (sgn(\xi_1), \ldots, sgn(\xi_n))$ also has C$ as correlation matrix ? It does not seem to be true. –  Alekk Mar 26 '10 at 16:22
This is de Finetti's construction right? Or what is called his "angle theorem". Google brought up this article www.emis.de/journals/JEHPS/Decembre2008/Lawrence.pdf –  Gjergji Zaimi Apr 7 '10 at 6:08
thanks for the reference: this paper is pretty much what I was looking for. –  Alekk Apr 7 '10 at 9:04
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Re 1: Not quite sure that I understand your construction but apparently it can never yield negative correlations?

Re 2: Consider the matrix $C_3=\pmatrix{1&-1/2&-1/2\cr -1/2&1&-1/2\cr -1/2&-1/2&1}$.

This is a legitimate correlation matrix (for instance of the Gaussian vector $(N,-N/2+N'\sqrt3/2,-N/2-N'\sqrt3/2)$ with $(N,N')$ i.i.d. and Gaussian) but neither $C_3$ nor any matrix whose $C_3$ is a submatrix are correlation matrices of Bernoulli vectors.

To see this, assume that $C_3$ is the correlation matrix of a random vector $X=(X_1,X_2,X_3)$ with values in $\{0,1\}^3$ and for every $i$ let $x_i^2=E(X_i)=E(X_i^2)$. Then $X_1/x_1+X_2/x_2+X_3/x_3$ is almost surely constant because the sum of the coefficients of $C_3$ is $0$. Hence $X_1/x_1+X_2/x_2$ takes exactly two values. For non degenerate $\{0,1\}$ valued random variables $X_1$ and $X_2$, this means that $x_1=x_2$. Likewise, $x_1=x_3$, hence $S=X_1+X_2+X_3$ is almost surely constant. Now $S=0$ or $S=3$ means that $X=(0,0,0)$ or that $X=(1,1,1)$, respectively, hence these cases are excluded. By the symmetry $X_i\to1-X_i$, one can assume that $S=1$ almost surely. This means that $X$ is concentrated on the three points $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ and, furthermore (recalling the relations $x_1=x_2=x_3$), that $X$ is uniformly distributed on these three points. Thus the correlation of $X_1$ and $X_2$ is $-1/3$ and not $-1/2$ as it should be.


I guess the same reasoning excludes every $n\times n$ matrix $C_n$ with diagonal entries $1$ and off-diagonal entries $-1/(n-1)$, for $n\ge3$.

By the way, one sees that $C_3$ cannot be obtained through Gaussian random variables and hyperplanes as in Gowers' answer because, if it was, it would be produced by the $3\times3$ matrix with diagonal entries $\sin(1\cdot\pi/2)=1$ and off-diagonal entries $\sin((-1/2)\cdot\pi/2)=-1/\sqrt2$, which is not definite positive. (The same applies to $C_n$ for every $n\ge3$.)

Correlation matrices $C=(C_{i,j})$ of Bernoulli random vectors might be exactly those such that the matrix $(\sin(C_{i,j}\pi/2))$ is definite positive, in which case the Gaussian-cut-by-hyperplanes construction would yield them all.

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I have a question about Didier Piau's answer: Why does it follow that "the correlation of $X_1$ and $X_2$ is $−1/3$ and not $−1/2$ as it should be"?

Perhaps I misunderstood the setting, but if $X$ is uniformly distributed over the points $(1,0,0)$, $(0, 1, 0)$ and $(0,0,1)$, then $E(X_i) = 1/3$, $E(X_iX_j)=0$, and hence $Var(X_i) = 2/9$ and $Cov(X_iX_j) = -1/9$. Therefore, the correlation between $X_i$ and $X_j$ is $(-1/9)/(2/9) = -1/2$, as it should be. In fact I think that, in this setting, the correlation matrix of $X_1, X_2, X_3$ is exactly the matrix $C_3$.

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