Theorem: If the likelihood allows its gradient to be passed under the integral sign, then
\[\Ex \u(\bt^*) = \zero.\]Proof: Letting \(P\) denote the distribution function and \(p\) the density function,
\[\begin{alignat*}{2} \Ex \u(\bt^*) &= \int \nabla_{\bt} \log p(x|\bts) dP(x|\bts) \\ &= \int \frac{\nabla_{\bt} p(x|\bts)}{p(x|\bts)} dP(x|\bts) &\hspace{4em}& \href{vector-calculus.html}{\text{Chain rule}}; \nabla \log x = 1/x \\ &= \nabla_{\bt} \int \frac{p(x|\bts)}{p(x|\bts)} dP(x|\bts) && \text{If gradient can be passed} \\ &= \nabla_{\bt} \int dP \\ &= 0 && \int dP = 1 \end{alignat*}\]