Proof of Directional Form of the Multivariate van Trees Inequality

Suppose first that $a=0$. Then $a\cdot z=0$, so the asserted right-hand side is $0$, while the left-hand side is the expectation of a nonnegative random variable and hence nonnegative by the monotonicity of the integral. The inequality holds.

Assume now $a\ne0$, so that $a\cdot a>0$. For $j\in\{1,\dots,l\}$ define $m_{j}:Y\to\mathbb{R}$ by $m_{j}=\bigl(a_{j}/(a\cdot a)\bigr)g$. Each $m_{j}$ is measurable with respect to $\mathcal{G}$ and the Borel $\sigma$-algebra, being a constant multiple of $g$, and $m_{j}(D)=\bigl(a_{j}/(a\cdot a)\bigr)g(D)$ is square-integrable by the closure properties of square-integrability. The hypotheses of the multivariate van Trees inequality are in force, so its conclusion applies to $m_{1},\dots,m_{l}$: the error matrix $R$ with entries $R_{ij}=\mathbb{E}\bigl[(m_{i}(D)-\Theta_{i})(m_{j}(D)-\Theta_{j})\bigr]$ is a well-defined symmetric matrix and satisfies $R\succeq J^{-1}$ in the semidefinite order.

By construction $\sum_{j=1}^{l}a_{j}m_{j}(D)=\bigl(\sum_{j=1}^{l}a_{j}^{2}/(a\cdot a)\bigr)g(D)=g(D)$, since $a\cdot a=\sum_{j}a_{j}^{2}$. Hence at every point of $\Omega$

Each product $(m_{i}(D)-\Theta_{i})(m_{j}(D)-\Theta_{j})$ is integrable, being a product of square-integrable random variables, by the closure properties of the square-integrability definition, so taking expectations and using the linearity of the integral finitely many times,

the first equality being the index formula for the matrix-vector product and the dot product.

Since $R-J^{-1}$ is positive semidefinite, $a\cdot\bigl((R-J^{-1})a\bigr)\ge0$; the same index formula shows that $u\cdot(Mv)$ depends linearly on the entries of $M$, so $a\cdot(Ra)\ge a\cdot(J^{-1}a)$. Finally claim 2 of Rank-One Lower Bound for the Inverse of a Positive Definite Matrix, applied to the positive definite matrix $J$, to the nonzero vector $z$ and to the vector $a$ in place of $x$ there, gives

Combining the last three displays proves the corollary.