0
$\begingroup$

My statistics professor says:

Let $X_n \xrightarrow{d} X$, $Y_n \xrightarrow{d} y$, where $y$ is some constant. Suppose for generality that $X \in \mathbb{R}^n$, and $y \in \mathbb{R}^m$. Then $$\begin{pmatrix}X_n \\ Y_n \end{pmatrix} \xrightarrow{d} \begin{pmatrix}X\\y\end{pmatrix}$$

I saw here that this isn't generally the case. Why do we have this when $Y_n$ converges in distribution to a constant? My Professor also said that this doesn't have to do with independence and I'm thinking about it the wrong way, but I'm unsure about this. A clarification on this particular point would be helpful.

(I also don't fully understand the proof from Rao (1973) page 122/123. Why do those limits imply the convergence in distribution result, and what does it mean for that limit to be indeterminable?)

For context, the result from Rao (1973):

If $y > c$, then

$$ \begin{multline} P(X_n < x) \geq P(X_n < x, Y_n \leq y) = P(X_n < x) - P(X_n<x, Y_n > y) \geq \\ P(X_n < x) - P(Y_n > y), \end{multline} $$

so in the limit we have

$$ \lim_{n \rightarrow \infty} P(X_n < x, Y_n < y) = P(X < x), $$

and $0$ if $y<c$.

I get this, but I just don't understand what is meant by

When $y=c$, the limit is indeterminable and $(x, c)$ is a discontinuity point for any $x$.

Improve this question
$\endgroup$
5
  • $\begingroup$ In general, $X_n$ and $Y_n$ does not need to be defined on the same probability space, if so, the notation "$(X_n, Y_n)$" is meaningless. $\endgroup$ Commented Apr 20 at 16:25
  • $\begingroup$ @Zhanxiong: But convergence in distribution do not need definition on the same probability space ... $\endgroup$ Commented Apr 20 at 17:11
  • $\begingroup$ @kjetilbhalvorsen That's exactly what I meant. Saying "$X_n \to_d X$ and $Y_n \to_d y$" is perfectly meaningful. However, it does not guarantee the existence of $(X_n, Y_n)$ as a random vector. $\endgroup$ Commented Apr 20 at 17:29
  • $\begingroup$ Add the assumption that $X_n$ and $Y_n$ are defined on the same probability space (which you can always achieve by extending the probability spaces) and try to use that $Y_n$ converges in probability (hence in distribution) to some constant. Then try to use that any random variable is independent of any constant (easy exercise to prove). $\endgroup$ Commented Apr 20 at 17:46
  • $\begingroup$ If $y= c$ creates a discontinuity, then the limit must be different for $y<c$ and $y> c;$ is this the case here? Check. $\endgroup$ Commented Apr 20 at 18:35

0

Reset to default

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.