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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?

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  • $\begingroup$ 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]. $\endgroup$ – Kavi Rama Murthy Jul 12 at 9:55
  • $\begingroup$ @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? $\endgroup$ – Lundborg Jul 12 at 9:59
  • $\begingroup$ No, uniform convergence is also not good enough. $\endgroup$ – Kavi Rama Murthy Jul 12 at 10:14

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