Consider Borel probability measures on $\mathbb R^d$. The the Prokhorov metric, $d_P$, metrizes weak convergence of probability measures, i.e., $d_P(Q_n,Q)\to 0$ if and only if $Q_n\to Q$ weakly. Let $d(\cdot,\cdot)$ be some distance function, like total variations and Kullback-Leibler divergence, implying weak convergence. That is, $$d(Q_n,Q)\to 0\Rightarrow d_P(Q_n,Q)\to 0.$$
Are the following true?
- given $Q$ and $0<\alpha<\infty$, there is a $\beta<\infty$ such that $$\{Q': d(Q',Q)\leq \alpha\}\subset\{Q': d_P(Q',Q)\leq \beta\}.$$
- if the above is true, then in the case of $\alpha$ going to $0$, we can also pick $\beta\to 0$.
I forgot when I had such an impression and got no clues when asked by a friend. I tried to show it but failed. My initial idea was to show any $Q'$ in the first set is also bounded in terms of $d_L$. So are these true? or any quick counter example? Thanks.
For the second one, can we go like this: wlog, assume $\alpha=1/n$. For every $\alpha_n$, write $$\beta_n=\sup_{Q':d(Q',Q)\leq\alpha_n}d_P(Q',Q).$$ and also $Q'$ achieving this sup as $Q_n$. If such $Q_n$ in $\{Q': d(Q',Q)\leq \alpha_n\}$, then we have $d(Q_n, Q)\leq 1/n$, which implies $d_P(Q_n,Q)\to 0$. So at some point after $n_0$, we do have $\beta_n$ is bounded.
Key here is to ensure that $Q_n$ lies in the set $\{Q': d(Q',Q)\leq \alpha_n\}$. My thinking is that if $d$ is indeed a metric, then $\{Q': d(Q',Q)\leq \alpha_n\}$ would be closed in the weak topology metrized by $d_L$.
Does this work?