Let $x \in \mathbb{R}^d$, and suppose that $A$ is symmetric positive definite with $A = \mathbb{E}[xx^T]+ \delta I$, $\delta$ is a small positive constant. The diagonal of $A$ is $D = \text{diag}(A)$.

What is the relation between $\text{det}(A)$ and $\text{det}(D)$? I have tested it by program, and found that $\text{det}(D) \geq \text{det}(A)$. If true, how to prove it?


Let $A = U^TU$ be the Cholesky factorization of $A$.

Notice that $A_{ii}=\left\|u_i\right\|^2$, where $\left\|u_i\right\|$ is the L2 norm of the $i$-th column vector of $U$.

By Hadamard's inequality,

$$\det(A) =\det(U)^2 \leq \prod_{i=1}^n \|u_i\|^2 =\prod_{i=1}^n A_{ii}$$


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