Background: Consider the normal means model $x=\theta+z\in\mathbb R^n$ where $z\sim N(0, I_n)$. That is, $x_i=\theta_i+z_i, z\overset {iid}{\sim} N(0, 1), i=1,...,n$. We treat $\theta\in\mathbb R^n$ as an unknown parameter and we want to estimate it based on the observation $x$. Let $\hat \theta=\hat \theta(x)$ be a general estimator.
Question: Show that $\operatorname {cov}(\hat \theta_i, x_i)=E[\partial\hat \theta_i/\partial x_i]$. Hint: Use integration by parts.
Attempt: $LHS=\operatorname{cov}(\hat \theta_i, x_i)=E(\hat\theta_i x_i)-E(\hat\theta_i)E(x_i)=E(\hat\theta_i x_i)-E(\hat\theta_i)\theta_i$
By the law of the unconscious statistician where $f$ is the density of $x_i$, $RHS=E[\partial\hat\theta_i/\partial x_i]=\displaystyle \int_{-\infty}^\infty\frac{\partial\hat\theta_i}{\partial x_i}f(x_i)dx_i=[f(x_i)\hat\theta_i]_{-\infty}^{\infty}-\int_{-\infty}^\infty\hat\theta_if'(x_i)dx_i$
where $\displaystyle \int_a^budv=[uv]_a^b-\int_a^b vdu$ and $u=f(x_i), dv=\displaystyle d\hat\theta_i, du=f'(x_i)dx_i, v=\hat\theta_i$
Is this the right direction, if so, what do I do now with the RHS?