# Convergence in parameter implies convergence in distribution

I'm interested in the following question. It seems pretty elementary but I don't know where to actually find reference on it.

Suppose we have a scale (one parameter) distribution family $$\{\mathcal{F}_{\sigma}, \sigma>0\}$$. And further assume $$\{X_i\}$$ with $$i\ge 1$$ are random variables from this family, such that $$X_i\sim \mathcal{F}_{\sigma_i}$$ (with scale parameter $$\sigma_i$$).

If the paramter $$\sigma_n\xrightarrow{n\to\infty}\sigma$$, do we have $$X_n\xrightarrow[weak]{n\to\infty} X$$ (convergence in distribution), where $$X$$ follows $$\mathcal{F}_{\sigma}$$.

Note that the commonly seen distribution family: $$N(0,\sigma)$$, $$Laplace(location=0,\ scale=\sigma)$$ or discrete distribution $$Geom(p)$$, etc. all have this satisfied.

Yes, this is correct. By convention, in probability "$$\mathcal{F}$$" is reserved to denote $$\sigma$$-fields rather than distribution functions. For this reason, I will replace $$\mathcal{F}$$ in your post with $$F$$ below.
Recall that by definition, $$\{F_\tau: \tau > 0\}$$ is a scale family if $$F_\tau(x) = F_1(\tau^{-1}x)$$ for all $$x \in \mathbb{R}$$, where $$F_1$$ is a fixed distribution function. Let $$x_0$$ be a continuity point of $$F_\sigma$$, i.e., $$F_\sigma(y) \to F_\sigma(x_0)$$ as $$y \to x_0$$. It then follows by $$\sigma_n \to \sigma$$ as $$n \to \infty$$ that $$\sigma\sigma_n^{-1}x_0 \to x_0$$, whence \begin{align} F_{\sigma_n}(x_0) = F_1(\sigma_n^{-1}x_0) = F_1(\sigma^{-1}\sigma\sigma_n^{-1}x_0) = F_\sigma(\sigma\sigma_n^{-1}x_0) \to F_\sigma(x_0), \end{align} which shows $$F_{\sigma_n} \to_d F_\sigma$$ as $$n \to \infty$$. This completes the proof.
• That looks nice. Thanks for the proof. I’m wondering if the family $\mathcal{F}$ is an arbitrary family, we generally don’t have any conclusion like this, right? Commented Apr 24, 2023 at 22:50