Let $\phi$ be a continuous bounded function. Show that if $(X_n)_{n \geq 1}$ converges in probability to $c \in \mathbb{R}$ then $\mathbb{E}[\phi(X_n)]$ converges to $\phi(c)$.

It seems to me that it is slightly modified version of Portmanteau's theorem, but I can't manage to see how the boundedness is used in the proof.


1 Answer 1


A sequence of random variable $(X_n)_{n \geq 1}$ which converges in probability to some random variable $X$, also converges in distribution to $X$. In particular, if $X_n$ convergence in probability to a constant $c$, then $X_n$ converges to $c$ in distribution (in this case, the converse is also true). Now, since $\phi$ is a bounded continuous function, we may apply Portmontau's theorem to get

$$ \lim_{n \to \infty} \mathbb{E}(\phi(X_n)) = \mathbb{E}(\phi(c)) = \phi(c). $$

  • $\begingroup$ I see. And how would you prove Portmanteau's theorem? All the proofs that I see use measure theory, which I haven't seen yet. $\endgroup$
    – RFTexas
    Commented Feb 9, 2021 at 15:48
  • 1
    $\begingroup$ For the proof of Portmontau, you can look here: (1) bit.ly/3cV0iVU or (2) bit.ly/3p3ywZX (Theorem 7.5). If you have not seen measuer theory, the first link will serve you better. $\endgroup$
    – ferhenk
    Commented Feb 9, 2021 at 15:52

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