Covariance matrix of score of random vector with independent entries is diagonal

Question

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

Answer 1

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 $\int_{-\infty}^\infty |p_i'(x)|\,dx<\infty$.)