Article

# Multivariate Chebyshev Inequalities

The Annals of Mathematical Statistics 12/1960; DOI: 10.1214/aoms/1177705673
Source: OAI

ABSTRACT If $X$ is a random variable with $EX^2 = \sigma^2$, then by Chebyshev's inequality, \begin{equation*}\tag{1.1}P\{|X| \geqq \epsilon\} \leqq \sigma^2/\epsilon^2.\end{equation*} If in addition $EX = 0$, one obtains a corresponding one-sided inequality \begin{equation*}\tag{1.2}\quad P\{X \geqq \epsilon\} \leqq \sigma^2/ (\epsilon^2 + \sigma^2)\end{equation*} (see, e.g., [8] p. 198). In each case a distribution for $X$ is known that results in equality, so that the bounds are sharp. By a change of variable we can take $\epsilon = 1$. There are many possible multivariate extensions of (1.1) and (1.2). Those providing bounds for $P\{\max_{1 \leqq j \leqq k} |X_j| \geqq 1\}$ and $P\{|\max_{1 \leqq j \leqq k} X_j \geqq 1\}$ have been investigated in [3, 5, 9] and [4], respectively. We consider here various inequalities involving (i) the minimum component or (ii) the product of the components of a random vector. Derivations and proofs of sharpness for these two classes of extensions show remarkable similarities. Some of each type occur as special cases of a general theorem in Section 3. Bounds are given under various assumptions concerning variances, covariances and independence. Notation. We denote the vector $(1, \cdots, 1)$ by $e$ and $(0, \cdots, 0)$ by 0; the dimensionality will be clear from the context. If $x = (x_1, \cdots, x_k)$ and $y = (y_1, \cdots, y_k)$, we write $x \geqq y(x > y)$ to mean $x_j \geqq y_j(x_j > y_j), j = 1, 2, \cdots, k$. If $\Sigma = (\sigma_{ij}): k \times k$ is a moment matrix, for convenience we write $\sigma_{jj} = \sigma^2_j, j = 1, \cdots, k$. Unless otherwise stated, we assume that $\Sigma$ is positive definite.

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