= Solution
Consider any <differentiable-in-quadratic-mean path> with <score function> $g$. Put $\delta_t=\sqrt{f_t}-\sqrt f=(t/2)g\sqrt f+r_t$, where $\|r_t\|_2=o(|t|)$. The <quadratic-mean to L1 density derivative> follows from
$$
f_t-f=2\sqrt f\,\delta_t+\delta_t^2
=tfg+2\sqrt f\,r_t+\delta_t^2.
$$
By the <Cauchy-Schwarz inequality>,
$$
\|f_t-f-tfg\|_{L^1}
\leq2\|r_t\|_2+\|\delta_t\|_2^2=o(|t|).
$$
Since $a$ is a <bounded function>, multiplying this <L1 norm> bound by $\|a\|_\infty$ proves
$$
\frac{\psi(P_{f_t})-\psi(P_f)}t\longrightarrow P_f(ag)
=P_f\bigl((a-P_fa)g\bigr).
$$
The last equality uses $P_fg=0$. The <derivative> is a <bounded linear functional> of $g\in L^2(P_f)$, so the required <pathwise differentiability of a statistical functional> holds, in particular relative to the <statistical tangent set> from part (b). \b[Its <derivative> is]
$$
\boxed{D\psi_f(g)=P_f[(a-P_fa)g].}
$$
For the explicit <bounded density tilts> in part (b), this <derivative> is also obtained by direct integration, with no remainder term.
Back to article page
