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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 $\|\mu_n - \delta_{\theta^{\ast}}\|_{\text{TVD}} \to 0$?)

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    $\begingroup$ What is the dependence $\kappa \mapsto q_\kappa$? Is it continuous (if yes, in which sense)? $\endgroup$
    – PhoemueX
    Dec 8, 2019 at 11:20
  • $\begingroup$ 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 $q_{\kappa}$. $\endgroup$
    – Jeremias K
    Dec 8, 2019 at 11:24

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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 $1$.

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    $\begingroup$ Thanks a lot! You made me feel stupid in a really good way ;D (i.e., your answer was very clear and in fact retrospectively so obvious that I am confident I won't forget it) $\endgroup$
    – Jeremias K
    Dec 8, 2019 at 15:30

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