A QMD score is centered and has finite information

Fix $v\in {\mathbb{R}}^d$ and put $f_t=\sqrt{p_{{\theta }_0+tv}}$, $f_0=\sqrt{p_{{\theta }_0}}$. By [F1]–[F2], $f_t-f_0=\frac{t}{2}(v^{T}s)f_0+r_{tv},\mathrm{\parallel }{r}_{tv}{\mathrm{\parallel }}_2=o(\mathrm{\mid }t\mathrm{\mid })\mathrm{.}$ Since each density integrates to one, $\mathrm{\parallel }{f}_t{\mathrm{\parallel }}_2=\mathrm{\parallel }{f}_0{\mathrm{\parallel }}_2=1$, and $0=2⟨f_0,f_t-f_0⟩+\mathrm{\parallel }{f}_t-f_0{\mathrm{\parallel }}_2^2\mathrm{.}$ This identity uses the unit norms of the square-root densities.

The increment has norm $O(\mathrm{\mid }t\mathrm{\mid })$, so its squared norm divided by $t$ tends to zero. By Cauchy–Schwarz, the residual inner product $⟨f_0,r_{tv}⟩\mathrm{/}t$ also tends to zero. Divide the identity in step 1.1 by $t$ and let $t\to 0$; the remaining term is $E_{{\theta }_0}(v^{T}s)=0$. Since this holds for every $v$, each coordinate of $E_{{\theta }_0}s$ is zero.

Each $s_j$ is square-integrable under a probability law, so $E\mathrm{\mid }{s}_j\mathrm{\mid }<\mathrm{\infty }$ by Cauchy–Schwarz. Also $E\mathrm{\mid }{s}_j{s}_k\mathrm{\mid }\le (E{s}_j^2{)}^{1\mathrm{/}2}(E{s}_k^2{)}^{1\mathrm{/}2}<\mathrm{\infty }$; hence every entry of $I_{{\theta }_0}$ is finite. For any $a\in {\mathbb{R}}^d$, $a^T{I}_{{\theta }_0}a=E(a^{T}s{)}^2\ge 0$, proving positive semidefiniteness, with no nonsingularity assertion.