3 개의 랜덤 변수의 상관 관계에 대한 경계

세 가지 랜덤 변수가 있습니다 . 세 변수 사이의 세 가지 상관 관계는 동일합니다. 그건,$x,y,z$

$$\rho=\textrm{cor}(x,y)=\textrm{cor}(x,z)=\textrm{cor}(y,z)$$

줄 수있는 가장 빡빡한 것은 무엇입니까 ?$\rho$

공통 상관 관계 $\rho$ 는 값 가질 수 $+1$있지만 아닙니다 $-1$. 경우 $\rho_{X,Y}= \rho_{X,Z}=-1$ , 다음 $\rho_{Y,Z}$ 동일하지 않은 수 $-1$ 하지만 사실상 $+1$ . 세 확률 변수의 일반적인 상관 관계의 가장 작은 값은 $-\frac{1}{2}$ . 보다 일반적으로, 최소한의 공통 상관$n$랜덤 변수이다$-\frac{1}{n-1}$ 벡터로 간주 할 때, 그들은 (차원 단면의 정점에있는$n-1$)에서$n$차원 공간.

$n$ 단위 분산 랜덤 변수 의 합의 분산을 고려하십시오 $X_i$. 우리는 그 여기서 ˉ ρ 인평균값의 (

$$\begin{align*} \operatorname{var}\left(\sum_{i=1}^n X_i\right) &= \sum_{i=1}^n \operatorname{var}(X_i) + \sum_{i=1}^n\sum_{j\neq i}^n \operatorname{cov}(X_i,X_j)\\ &= n + \sum_{i=1}^n\sum_{j\neq i}^n \rho_{X_i,X_j}\\ &= n + n(n-1)\bar{\rho} \tag{1} \end{align*}$$ $\bar{\rho}$ 상관 계수. 그러나 이후VAR(Σ전X전)≥0, 우리는 쉽게에서 얻을 (1)그 ˉ ρ ≥-$\binom{n}{2}$$\operatorname{var}\left(\sum_i X_i\right) \geq 0$$(1)$ $$\bar{\rho} \geq -\frac{1}{n-1}.$$

그래서, 상관 계수의 평균 값은 적어도 . 경우모든상관 계수는이같은값ρ을, 그들의 평균은 동일ρ를하고 우리가 그래서 ρ를≥-$-\frac{1}{n-1}$$\rho$$\rho$ 그것은 일반적인 상관 치되는 확률 변수 가질 수 있습니다ρ는동일을-

$$\rho \geq -\frac{1}{n-1}.$$ $\rho$ ? 예. Xi가상관되지 않은 단위 분산 랜덤 변수이고Yi=Xi-1로설정 한다고 가정합니다.$-\frac{1}{n-1}$$X_i$ . 그런 다음E[Yi]=0이고 var(Yi)= ( n − $Y_i = X_i - \frac{1}{n}\sum_{j=1}^n X_j = X_i -\bar{X}$$E[Y_i]=0$ 및 COV(YI,Y의J)=-2(N- $$\displaystyle \operatorname{var}(Y_i) = \left(\frac{n-1}{n}\right)^2 + (n-1)\left(\frac{1}{n}\right)^2 = \frac{n-1}{n}$$ 은 ρYi,Yj=cov(Yi,Yj $$\operatorname{cov}(Y_i,Y_j) = -2\left(\frac{n-1}{n}\right)\left(\frac{1}{n}\right) + (n-2)\left(\frac{1}{n}\right)^2 = -\frac{1}{n}$$ 따라서,Y나최소 공통 상관 값을 달성하는 임의의 변수는- $$\rho_{Y_i,Y_j} = \frac{\operatorname{cov}(Y_i,Y_j)}{\sqrt{\operatorname{var}(Y_i)\operatorname{var}(Y_j)}} =\frac{-1/n}{(n-1)/n} = -\frac{1}{n-1}.$$ $Y_i$ . 우연히도, ∑iYi=0이므로 벡터로 간주되는 랜덤 변수는n차원 공간의(n-1)차원 초평면에있습니다.$-\frac{1}{n-1}$$\sum_i Y_i = 0$$(n-1)$$n$

소감 가능한 바인딩입니다 . $-1/2 \le \rho \le 1$ 그러한 모든 가치는 실제로 나타날 수 있습니다.

결과에 대해 특별히 깊거나 신비로운 것이 없다는 것을 보여주기 위해이 대답은 먼저 완전한 기본 솔루션을 제시하며, 예상되는 제곱의 값이되는 편차가 음이 아니어야한다는 명백한 사실 만 요구합니다. 그 다음에는 일반적인 해결책 (약간 더 복잡한 대수적 사실을 사용함)이 이어집니다.

기본 솔루션

의 선형 조합의 분산은 음이 아니어야합니다. $x,y,z$ 수 (가) 이러한 변수의 편차를 보자 및 υ 2를 각각. 모두 0이 아닙니다 (그렇지 않으면 일부 상관 관계가 정의되지 않음). 우리가 계산할 수있는 분산의 기본 속성을 사용하여$\sigma^2, \tau^2,$$\upsilon^2$

$$0 \le \text{Var}(\alpha x/\sigma + \beta y/\tau + \gamma z/\upsilon) = \alpha^2 +\beta^2+\gamma^2 + 2\rho(\alpha\beta+\beta\gamma+\gamma\alpha)$$

모든 실수 .$(\alpha, \beta, \gamma)$

이라고 가정하면 약간의 대수적 조작은 다음과 같습니다.$\alpha+\beta+\gamma\ne 0$

$$\frac{-\rho}{1-\rho} \le \frac{1}{3} \left(\frac{\sqrt{(\alpha^2+\beta^2+\gamma^2)/3}}{(\alpha+\beta+\gamma)/3}\right)^2.$$

$(\alpha, \beta, \gamma)$$(1/3, 1/3, 1/3)$$1$$1$$\alpha=\beta=\gamma\ne 0$

$$\rho \ge -1/2.$$

의 명시 적 예$n=3$ below (involving trivariate Normal variables $(x,y,z)$) shows that all such values, $-1/2 \le \rho \le 1$, actually do arise as correlations. This example uses only the definition of multivariate Normals, but otherwise invokes no results of Calculus or Linear Algebra.

General solution

Overview

Any correlation matrix is the covariance matrix of the standardized random variables, whence--like all correlation matrices--it must be positive semi-definite. Equivalently, its eigenvalues are non-negative. This imposes a simple condition on $\rho$: it must not be any less than $-1/2$ (and of course cannot exceed $1$). Conversely, any such $\rho$ actually corresponds to the correlation matrix of some trivariate distribution, proving these bounds are the tightest possible.

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Derivation of the conditions on $\rho$

Consider the $n$ by $n$ correlation matrix with all off-diagonal values equal to $\rho.$ (The question concerns the case $n=3,$ but this generalization is no more difficult to analyze.) Let's call it $\mathbb{C}(\rho, n).$ By definition, $\lambda$ is an eigenvalue of provided there exists a nonzero vector $\mathbf{x}_\lambda$ such that

$$\mathbb{C}(\rho,n) \mathbf{x}_\lambda = \lambda \mathbf{x}_\lambda.$$

These eigenvalues are easy to find in the present case, because

1. Letting $\mathbf{1} = (1, 1, \ldots, 1)'$, compute that

   $$\mathbb{C}(\rho,n)\mathbf{1} = (1+(n-1)\rho)\mathbf{1}.$$

2. Letting $\mathbf{y}_j = (-1, 0, \ldots, 0, 1, 0, \ldots, 0)$ with a $1$ only in the $j^\text{th}$ place (for $j = 2, 3, \ldots, n$), compute that

   $$\mathbb{C}(\rho,n)\mathbf{y}_j = (1-\rho)\mathbf{y}_j.$$

Because the $n$ eigenvectors found so far span the full $n$ dimensional space (proof: an easy row reduction shows the absolute value of their determinant equals $n$, which is nonzero), they constitute a basis of all the eigenvectors. We have therefore found all the eigenvalues and determined they are either $1+(n-1)\rho$ or $1-\rho$ (the latter with multiplicity $n-1$). In addition to the well-known inequality $-1 \le \rho \le 1$ satisfied by all correlations, non-negativity of the first eigenvalue further implies

$$\rho \ge -\frac{1}{n-1}$$

while the non-negativity of the second eigenvalue imposes no new conditions.

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Proof of sufficiency of the conditions

The implications work in both directions: provided $-1/(n-1)\le \rho \le 1,$ the matrix $\mathbb{C}(\rho, n)$ is nonnegative-definite and therefore is a valid correlation matrix. It is, for instance, the correlation matrix for a multinormal distribution. Specifically, write

$$\Sigma(\rho, n) = (1 + (n-1)\rho)\mathbb{I}_n - \frac{\rho}{(1-\rho)(1+(n-1)\rho)}\mathbf{1}\mathbf{1}'$$

for the inverse of $\mathbb{C}(\rho, n)$ when $-1/(n-1) \lt \rho \lt 1.$ For example, when $n=3$

$$\color{gray}{\Sigma(\rho, 3) = \frac{1}{(1-\rho)(1+2\rho)} \left( \begin{array}{ccc} \rho +1 & -\rho & -\rho \\ -\rho & \rho +1 & -\rho \\ -\rho & -\rho & \rho +1 \\ \end{array} \right)}.$$

Let the vector of random variables $(X_1, X_2, \ldots, X_n)$ have distribution function

$$f_{\rho, n}(\mathbf{x}) = \frac{\exp\left(-\frac{1}{2}\mathbf{x}\Sigma(\rho, n)\mathbf{x}'\right)}{(2\pi)^{n/2}\left((1-\rho)^{n-1}(1+(n-1)\rho)\right)^{1/2}}$$

where $\mathbf{x} = (x_1, x_2, \ldots, x_n)$. For example, when $n=3$ this equals

$$\color{gray}{\frac{1}{\sqrt{(2\pi)^{3}(1-\rho)^2(1+2\rho)}} \exp\left(-\frac{(1+\rho)(x^2+y^2+z^2) - 2\rho(xy+yz+zx)}{2(1-\rho)(1+2\rho)}\right)}.$$

The correlation matrix for these $n$ random variables is $\mathbb{C}(\rho, n).$

Contours of the density functions $f_{\rho,3}.$ From left to right, $\rho=-4/10, 0, 4/10, 8/10$. Note how the density shifts from being concentrated near the plane $x+y+z=0$ to being concentrated near the line $x=y=z$.

The special cases $\rho = -1/(n-1)$ and $\rho = 1$ can also be realized by degenerate distributions; I won't go into the details except to point out that in the former case the distribution can be considered supported on the hyperplane $\mathbf{x}.\mathbf{1}=0$, where it is a sum of identically distributed mean-$0$ Normal distribution, while in the latter case (perfect positive correlation) it is supported on the line generated by $\mathbf{1}'$, where it has a mean-$0$ Normal distribution.

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More about non-degeneracy

A review of this analysis makes it clear that the correlation matrix $\mathbb{C}(-1/(n-1), n)$ has a rank of $n-1$ and $\mathbb{C}(1, n)$ has a rank of $1$ (because only one eigenvector has a nonzero eigenvalue). For $n\ge 2$, this makes the correlation matrix degenerate in either case. Otherwise, the existence of its inverse $\Sigma(\rho, n)$ proves it is nondegenerate.

Your correlation matrix is

$$\begin{pmatrix} 1&\rho&\rho\\ \rho&1&\rho\\ \rho&\rho&1 \end{pmatrix}$$

The matrix is positive semidefinite if the leading principal minors are all non-negative. The principal minors are the determinants of the "north-west" blocks of the matrix, i.e. 1, the determinant of

$$\begin{pmatrix} 1&\rho\\ \rho&1\end{pmatrix}$$

and the determinant of the correlation matrix itself.

1 is obviously positive, the second principal minor is $1-\rho^2$, which is nonnegative for any admissible correlation $\rho\in[-1,1]$. The determinant of the entire correlation matrix is

$$2\rho^3-3\rho^2+1.$$

The plot shows the determinant of the function over the range of admissible correlations $[-1,1]$.

You see the function is nonnegative over the range given by @stochazesthai (which you could also check by finding the roots of the determinantal equation).

There exist random variables $X$, $Y$ and $Z$ with pairwise correlations $\rho_{XY} = \rho_{YZ} = \rho_{XZ} = \rho$ if and only if the correlation matrix is positive semidefinite. This happens only for $\rho \in [-\frac{1}{2},1]$.