I've got to prove, when $X_n \overset{p}{\to} X $ and $Y_n \overset{p}{\to} Y$ then $h(X_n,Y_n) \overset{p}{\to} h(X,Y)$, $(\overset{p}{\to}$:convergence in probability). $h:\mathbb{R} \times \mathbb{R} {\to} \mathbb{R} $ is a continous function.

My strategy would be the following: I want to prove $h(X_n,Y) \overset{p}{\to} h(X,Y)$ and the same thing for the other random variable , i.e. $h(X,Y_n) \overset{p}{\to} h(X,Y)$. Is this enought to prove the assertion?


This is not sufficient. Consider for example $h(x, y) = xy$, $X_n = Y_n = n$ and $X = Y = 0$. Then $h(X_n, Y) \to h(X, Y)$ and also $h(X, Y_n) \to h(X, Y)$, but clearly not $h(X_n, Y_n) \to XY$.

Instead, use the fact that $Z_n \xrightarrow{P} Z$ if and only if each subsequence of $Z_n$ has a further subsequence that converges to $Z$ almost surely.

  • $\begingroup$ thanks, so the logic is that each of $X_n$ and $Y_n$ has a further subsequence that converges almost surely($X_n$ to$ X$, $Y_n$ to $Y$) and that under a continous map it won't change, i.e. $h(X_n,Y_n) \overset{a.s.}{\to} h(X,Y)$ ,where the almost sure convergence then implies convergence in probability, right? $\endgroup$ – Thesinus Dec 18 '16 at 17:21
  • $\begingroup$ Not quite. For each subsequence of $h(X_n, Y_n)$ you can find a further subsequence along both $X_n$ and $Y_n$ converge almost surely. This yields a subsubsequence of $h(X_n, Y_n)$ that converges to $h(X, Y)$ almost surely (you need the continuity of $h$ in the last step). $\endgroup$ – Dominik Dec 18 '16 at 17:25

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