# Joint uniform convergence in distribution of random variables and constant

by Lundborg   Last Updated July 12, 2019 10:20 AM - source

Let $$(X_{n, \theta})_{n \in \mathbb{N}, \theta \in \Theta}$$ be a sequence of parameter dependent real-valued random variables where $$\Theta$$ is some parameter space.

Assume that $$X_{n, \theta}$$ converges uniformly to $$X_\theta$$, i.e. for any continuous and bounded $$f: \mathbb{R} \to \mathbb{R}$$ $$\sup_{\theta} \left|E(f(X_{n, \theta})) - E(f(X_\theta)) \right| \to 0$$ as $$n \to \infty$$.

Let $$(y_\theta)_{\theta \in \Theta}$$ be some family of real numbers. Does then $$(X_{n, \theta}, y_\theta)$$ converge uniformly to $$(X_\theta, y_\theta)$$, i.e. for any continuous and bounded $$f: \mathbb{R}^2 \to \mathbb{R}$$ $$\sup_{\theta} \left|E(f(X_{n, \theta}, y_\theta)) - E(f(X_\theta, y_\theta)) \right| \to 0$$ as $$n \to \infty$$.

Intuitively I find it crazy that adding a constant that does nothing would change this convergence but perhaps I need some assumptions like boundedness of $$y_\theta$$ (which would be fine) but I just cant figure out a way to show it.

Usually arguments like this will be of the form: note that $$g(x) = f(x, y_\theta)$$ is continuous and then we're done but $$g$$ is now $$\theta$$-dependent and therefore I don't think the argument works. Any ideas?

Tags :