# Joint uniform convergence in distribution of random variables and constant

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?

• In the hypothesis you have a fixed function $f$. What you want is to vary the function itself and get uniform convergence over all those functions. Surely this is not true without further hypothesis. [ $f(x,\theta)$ is a family of bounded continuous functions]. – Kavi Rama Murthy Jul 12 at 9:55
• @KaviRamaMurthy It just seems absurd to me that simply considering a constant together with the random variable breaks the convergence. Maybe my intuition is just poor. What if we assumed that $f$ was uniformly continuous, would that help anything? – Lundborg Jul 12 at 9:59
• No, uniform convergence is also not good enough. – Kavi Rama Murthy Jul 12 at 10:14