I've been trying to read through the following paper:


<https://arxiv.org/pdf/0704.1751.pdf>


And I've been stuck on the proposition in the middle of page 8 which says that if a r.v. $X$ has independent entries, then:


$$\textbf{J}(X) = \text{diag}(J(X\_i))\_i$$


Where $\textbf{J}(X) := \text{Cov}(S(X))$ and $J(X):=\text{tr}(\textbf{J}(X))$.


What I have so far: Assume $X$ has independent entries, and p.d.f $f$. Then


$$S(X) = \frac{\nabla f(x)}{f(x)} = \frac{1}{\prod\_if\_i(x\_i)}\begin{bmatrix}\frac{\partial}{\partial x\_1}\prod\_if\_i(x\_i)\\ \vdots 
\\ \frac{\partial}{\partial x\_n}\prod\_if\_i(x\_i)\end{bmatrix} = \begin{bmatrix}f'(x\_1)/f(x\_1)\\ \vdots 
\\ f'(x\_n)/f(x\_n)\end{bmatrix} = \begin{bmatrix}S(X\_1)\\ \vdots 
\\ S(X\_n)\end{bmatrix}$$


From this we get $\textbf{J}(X)\_{ii} = E[S(X\_i)^2] = J(X\_i).$ Where I'm having trouble is showing that the off-diagonal entries are 0, namely that if $i\neq j$ then


$$E[S(X\_i)S(X\_j)] = 0$$


Thanks



Covariance matrix of score of random vector with independent entries is diagonal
The equality $ES(X\_i)S(X\_j)=0$ for $i\ne j$ follows because $X\_i$ and $X\_j$ are independent and $ES(X\_i)=0$, so that
$$ES(X\_i)S(X\_j)=ES(X\_i)\,ES(X\_j)=0.$$


(The equality $ES(X\_i)=0$ holds because for the density $p\_i$ of $X\_i$ we have
$$ES(X\_i)=E\frac{p\_i'(X)}{p\_i(X)}=\int\_{-\infty}^\infty\frac{p\_i'(x)}{p\_i(x)}\,p\_i(x)\,dx
=\int\_{-\infty}^\infty p\_i'(x)\,dx=0$$
under standard assumptions on $p\_i$, such as $\int\_{-\infty}^\infty |p\_i'(x)|\,dx<\infty$.)


