= Solution
Apply the preceding argument to the two product observation laws and the real-valued <estimator> $T=\widehat\psi_n$. Alternatively, directly use
$$
(T-\theta)^2+(T-\tau)^2
=2\left(T-\frac{\theta+\tau}{2}\right)^2+\frac{(\theta-\tau)^2}{2}
\geq\frac{(\theta-\tau)^2}{2}.
$$
The average of the two <mean squared errors> is at least $(\theta-\tau)^2/4$ times the overlap of the two joint <probability density functions>. Their overlap equals
$$
\int\min(p_f^{(n)},p_g^{(n)})\,d\mu^{(n)}
=\frac12\int(p_f^{(n)}+p_g^{(n)}-|p_f^{(n)}-p_g^{(n)}|)\,d\mu^{(n)}
=1-\frac12\|P_f^{(n)}-P_g^{(n)}\|_1.
$$
Taking the infimum over <estimators> therefore yields
$$
\boxed{R_n^*\geq\frac{(\theta-\tau)^2}{4}\left(1-\frac12\|P_f^{(n)}-P_g^{(n)}\|_1\right).}
$$
Here $R_n^*$ is the <minimax risk> for this functional. The factor $\frac12\|P_f^{(n)}-P_g^{(n)}\|_1$ is the <total variation distance>, so the same <metric squared-loss two-point bound> applies to a functional even when distinct density parameters have the same functional value.
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