Showing $\int |f_n - g_n| dx \to 0$ when normal $X_n \overset{p}{\to} \mu, Y_n \overset{p}{\to} \mu $

Question

I'd like to show the following.

Let $X_n, Y_n$ be sequences of normal random variables. If $$ X_n \overset{p}{\to} \mu, Y_n \overset{p}{\to} \mu $$ then $\int |f_n - g_n| dx \to 0$, with $f_n, g_n$ : the PDF's of $X_n, Y_n$

But I'm stuck.

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My attempt

What I can only think of is, the definition of convergence in probability, $\forall \epsilon >0,$

$$P(|X_n-\mu| \ge \epsilon) \xrightarrow{n\to\infty} 0\\ P(|Y_n-\mu| \ge \epsilon) \xrightarrow{n\to\infty} 0$$

and this seems something related to the uniform convergence of the distribution functions. But I cannot relate the density to the convergence of random variables.

Answer 1

Without any additional assumptions (e.g. on the speed of convergence), the assertion is wrong.

(Counter)Example Let $Z \sim N(0,1)$ be a standard Gaussian random variable. If we set $$X_n := \frac{1}{n} Z \qquad Y_n:= \frac{1}{\sqrt{n}} Z$$ then clearly $X_n \to 0$, $Y_n \to 0$ in probability. The density $p$ of $Z$,

$$p(y) = \frac{1}{\sqrt{2\pi}} \exp \left( - \frac{y^2}{2} \right)$$

is continuous and satisfies $p(0)=1/\sqrt{2\pi}$, and therefore it follows that we can choose $r>0$ such that

$$p(y) \geq \frac{3}{4} \frac{1}{\sqrt{2\pi}} \quad \text{for all $|y| \leq r$}.$$

This implies that the transition density $f_n$ of $X_n$ satisfies

$$f_n(y) = \sqrt{n} p \left( \sqrt{n} y \right) \geq \frac{3}{4} \frac{\sqrt{n}}{\sqrt{2\pi}} \quad \text{for all $|y| \leq r/\sqrt{n}$.}$$

Since

$$\int_{\mathbb{R}} |f_n(y)-g_n(y)| \, dy \geq \int_{|y| \leq r/\sqrt{n}} (f_n(y))-g_n(y)) \, dy$$

and $\|g_n\|_{\infty} \leq n^{1/4}/\sqrt{2\pi}$, we get

$$\int_{\mathbb{R}} |f_n(y)-g_n(y)| \, dy \geq \left( \frac{3}{4} \frac{\sqrt{n}}{\sqrt{2\pi}} - \frac{n^{1/4}}{\sqrt{2 \pi}} \right) \frac{2r}{\sqrt{n}} \xrightarrow[]{n \to \infty} \frac{3}{2} \frac{r}{\sqrt{2\pi}}>0$$

and so

$$\int_{\mathbb{R}} |f_n(y)-g_n(y)| \, dy \to 0$$

does not hold true.

Remark For random variables $X$ and $Y$ with density $f$ and $g$, respectively, the total variation distance is defined as

$$d_{\text{TV}}(X,Y) = \frac{1}{2} \int |f-g|.$$

It is possible to show that

$$d_{\text{TV}}(X,Y) = \sup_{A \in \mathcal{B}(\mathbb{R})} |\mathbb{P}(X \in A)-\mathbb{P}(Y \in A)|.$$

The above (counter)example shows that $X_n \to \mu$, $Y_n \to \mu$ in probability does, in general, not imply $d_{\text{TV}}(X_n,Y_n) \to 0$.