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Take zero mean MVN on k dimensions: $$ p(x) = \frac 1 {\sqrt {( 2 \pi ) ^k |\Sigma|}}e^{-\frac {1} {2} x \Sigma^{-1} x}$$ we will surely agree that $p(x)\leq1$ for all $x$, including $x=0$ ($x\in\mathbb R^k$). Set $x=0$. So: $$ p(x) = \frac 1 {\sqrt {( 2 \pi ) ^k |\Sigma|}}e^{-\frac {1} {2} 0 \Sigma^{-1} 0}=$$ $$ \frac 1 {\sqrt {( 2 \pi ) ^k |\Sigma|}}\leq1\implies( 2 \pi ) ^k |\Sigma|\geq1\implies |\Sigma|\geq(2 \pi ) ^{-k}$$

But $\Sigma$ can be any positive definite matrix, its eigenvalues can be positive yet much smaller than $(2 \pi ) ^{-k}$.

What am I doing wrong?

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The density need not be $\leqslant 1$. And if the variance is small enough, it is greater than $1$ around $0$. – Daniel Fischer Mar 12 '14 at 13:01
Very interesting. – Troy McClure Mar 12 '14 at 13:04
up vote 1 down vote accepted

$p(x)\le 1$ is wrong.

The equivalent is $\forall A\subset \Bbb R^d$, $$ \int_A p(x) dx \le 1 $$but you can't say anything is $|A|$ is small.

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