Does expected convergence in total variation distance imply weak convergence?

From the definition of total variation distance, we know that convergence in total variation implies weak convergence. However, suppose we have the following, $$d_{TV}(X_n, X) = Y_n,$$ and $$\mathbb{E}[Y_n] \rightarrow 0$$, and hence, $$\mathbb{E}[d_{TV}(X_n, X)] \rightarrow 0$$. If the expectation of the total variation distance converges to $$0$$, can we still somehow conclude that $$X_n$$ converges weakly to $$X$$?

$$d_{TV}$$ refers to the total variation metric. The total variation distance between two probability measures $$P$$ and $$Q$$ on a common probability space $$(\Omega, \mathcal{F})$$ is given by, $$d_{TV}(P, Q) = \sup_{A \in \mathcal{F}} |P(A) - Q(A)|.$$

• What is the definition of $d_{TV} (X_n,X)$? Commented Jan 19, 2022 at 8:56
• added the definition in the question. thank you.
– zd_
Commented Jan 19, 2022 at 9:07
• Why is $d_{TV} (X_n,X)$ random? Isn't it the same as $d_{TV}(P_n,P)$ where $P_n,P$ are measures induced by $X_n$ and $X$? Commented Jan 19, 2022 at 9:09

First notice that $$Y_n$$ is a non-negative number, so $$\operatorname{E}[Y_n]=Y_n$$, then we have that $$Y_n\to 0$$. Now let $$P_n$$ the probability measure induced by $$X_n$$, and $$P_X$$ the probability measure induced by $$X$$ and set $$Q:=\frac1{2}P_X+\sum_{n\geqslant 1}\frac1{2^{n+1}}P_n$$, then its easy to check that $$Q$$ is a probability measure and that $$P_n\ll Q$$ and $$P_X\ll Q$$, therefore there are Radon-Nikodym derivatives $$f_n$$ and $$f_X$$ such that $$P_n=f_n\cdot Q$$ and $$P_X=f_X\cdot Q$$.

Let $$\mathcal{B}(\mathbb{R})$$ the Borel $$\sigma$$-algebra of $$\mathbb{R}$$ and note that

\begin{align*} d_{TV}(X_n,X)&=\sup_{A\in \mathcal{B}(\mathbb{R})}\left|\int_{A}(f_n-f_X)dQ\right|\\ &=\int_{\{f_n-f_X\geqslant 0\}}(f_n-f_X)dQ\\ &=\frac1{2}\int_{\mathbb{R}}|f_n-f_X|dQ \end{align*}

where the last equality follows from the fact that

$$\int_{\{f_n-f_X\geqslant 0\}}(f_n-f_X)dQ=\int_{\{f_n-f_X< 0\}}(f_X-f_n)dQ$$

as $$\int_{\mathbb{R}}(f_n-f_X)dQ=0$$. Therefore $$f_n \xrightarrow{L_1}f_X$$, what implies that $$X_n\xrightarrow{\text{dist.}}X$$ (as $$L_1$$ convergence implies weak convergence of measures).∎

An easier way to state the same is that for all $$c\in \mathbb{R}$$

$$|P_n((-\infty ,c])-P_X((-\infty ,c])|\leqslant d_{TV}(X_n,X)\to 0$$

so $$X_n\xrightarrow{\text{dist.}}X$$.