Does weak convergence imply strong convergence for parameterized measure?

Define for some density $$q_{\kappa_n}$$ (with respect to the Lebesgue measure) on $$\Theta$$, parameterized by $$\kappa_n \in \mathbb{R}^d$$. Next, define the sequence of measures $$\mu_n$$ as

$$\mu_n(A) = \int_{A}q_{\kappa_n}(\theta)d\theta, \text{ for all measurable } A\subset \Theta.$$

Suppose that (i) $$\mu_n \overset{w}{\to} \delta_{\theta^{\ast}}$$, i.e. the sequence of measures $$\mu_n$$ weakly converges to a Dirac measure (point mass) at $$\theta^{\ast}$$ and that (ii) $$\kappa_n \to \kappa^{\ast}$$, where the latter convergence occurs in the standard Euclidean topology.

Does this imply strong convergence? (I.e., does it hold that $$\|\mu_n - \delta_{\theta^{\ast}}\|_{\text{TVD}} \to 0$$?)

• What is the dependence $\kappa \mapsto q_\kappa$? Is it continuous (if yes, in which sense)? Dec 8, 2019 at 11:20
• Oh right -- yes, $q_{\kappa}$ is a standard continuous density in $\kappa$. I am not sure what you mean by 'in which sense', but you can treat $\kappa$ like the parameters $\kappa = (\mu, \sigma)$ of a normal density $q_{\kappa}$. Dec 8, 2019 at 11:24

Clearly not. The point mass measures are all at TV distance 1 from all measures with density functions, and the set $$A=\{\theta^*\}$$ attains the maximum of $$|\mu_n(A)-\delta_{\theta^*}(A)|$$, which evaluates to $$1$$.