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Suppose we have a sequence of (continuous) random variables $(X_n)_{n\geq 1}$ with corresponding densities $p_n(x)$ and another sequence $(Y_n)_{n\geq 1}$ with corresponding densities $q_n(x)$ such that $X_n \Rightarrow X \sim \mathcal{N}(\mu_X,\Sigma_X)$ in distribution and $Y_n \Rightarrow Y \sim \mathcal{N}(\mu_Y,\Sigma_Y)$ in distribution. If we let $p(x)$ and $q(x)$ be the densities of $\mathcal{N}(\mu_X,\Sigma_X)$ and $\mathcal{N}(\mu_Y,\Sigma_Y)$ respectively, is it true that the Bhattacharyya coefficient between $p_n$ and $q_n$ given by $$BC(p_n,q_n) = \int \sqrt{p_n(x)q_n(x)}dx$$ converges to the BC of $p$ and $q$, $BC(p,q)=\int \sqrt{p(x)q(x)}dx$ ?

I wonder if any conclusions can be made as convergence in distribution does not even necessarily imply convergence of the respective densities.

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This took me a while but I finally got it. I went in assuming that the limiting BC would be the BC of the limit. It turns out that isn't true. Here's a counterexample to prove it:

Let $$\mathcal A_n=\bigcup_{i=-∞}^∞ A_{n,i}$$ where $A_{n,i}=[\dfrac {2i} n,\dfrac {2i+1} n)$. Then, you can define the pdf sequences as $$p_n (x)=\dfrac 1 {Z(n)} \dfrac 1 {\sqrt{2\pi}} \exp{(\dfrac {-x^2} 2)} I(x ∈ \mathcal A_n)$$ $$q_n (x)=\dfrac 1 {Z(n)} \dfrac 1 {\sqrt{2\pi}} \exp{(\dfrac {-x^2} 2)} I(x ∈ \mathcal A_n^c)$$ These are the densities of standard normal random variables conditioned on $X_n \in \mathcal A_n$ and $Y_n \in \mathcal A_n^c$, resepctively. The densities are just normal kernels rescaled by identical (due to symmetry) normalizing constants which converge to $1/2$ as $n \rightarrow \infty$. Note that for each $n$, there is no density overlap between $p_n (x)$ and $q_n (x)$. If $p_n (x) > 0$, then $q_n (x) = 0$ and vice-versa. This clearly implies that $BC(p_n,q_n) = 0, \forall n$.

It remains to show that $X_n \xrightarrow{d} X \sim \mathcal{N}(0,1)$ and $Y_n \xrightarrow{d} Y \sim \mathcal{N}(0,1)$. Without loss of generality, we can restrict attention to the $(X_n)$ sequence. To prove convergence in distribution, we simply need to show that $\lim_{n \rightarrow \infty} F_{X_n} (t) = \Phi(t)$ for all $t$ at which $\Phi$ is continuous - the entire real line, in this case.

Goint through the steps to show this requires a little bit more work even though the result is obvious. The easiest way to do this is to use the identities, $ϕ(x)=Z(n)(p_n (x)+q_n (x))$, $F_{X_n} (t)=1-F_{Y_n} (-t)$ and $Φ(t)=1-Φ(-t)$.

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