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If $X_n\rightarrow X$ almost surely and $Y_n\rightarrow Y$ a.s., then is $X_n/Y_n\rightarrow X/Y$ almost surely true?

Is there a theorem for this or is this not correct?

lightfish
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  • I can get $1/Y_n\rightarrow 1/Y$ a.s. from continuous mapping theorem, but can I multiply that with $X_n\rightarrow X$a.s. to get what I need? – lightfish Nov 23 '13 at 21:12

1 Answers1

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Claim: Let $f : S \subset \mathbb{R}^2 \to \mathbb{R}$ be a continuous map, $\{X_1, X_2, ..., Y_1, Y_2, ..., X, Y\}$ be random variables on $(\Omega,P)$ taking values in $S$ such that, $X_n \to X$ a.s. and $Y_n \to Y$ a.s., then $f(X_n,Y_n) \to f(X,Y)$ a.s.

Proof: Let $N = \{X_n \not\to X\}$ and $M = \{Y_n \not\to Y\}$. Notice that $P(N \cup M) = 0$. By assumption of continuity, $f(X_n,Y_n) \to f(X,Y)$ on $(N \cup M)^c$ and $P\big((N \cup M)^c\big) = 1$.

Caution: You do need the continuity of $f$ here. So, if you are dividing by zero, say, you can't apply this claim.

Tom
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  • S open would avoid pathologies. – Did Nov 23 '13 at 22:14
  • @Did I should have made the assumption that $S$ was a (Borel) measurable subset of $\mathbb{R}^2$. Do you have an example off the top of your head where $S$ begin open rather than simply measurable (endowed with the induced topology) could cause problems? Thanks! – Tom Nov 23 '13 at 22:26