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Is it true that if $$ 0 \leq x_n \leq y_n \xrightarrow{p} 0,$$ then $$x_n \xrightarrow{p} 0?$$ If this is true, I would be very grateful if you could give me some references where I can find the proof.

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    $\begingroup$ This follows directly from the definition of convergence in probability (and monotonicity of an arbitrary measure). $\endgroup$ – cardinal Nov 20 '13 at 19:47
  • $\begingroup$ Try to prove it yourself: Use reductio ad absurdum i.e. start by assuming that $x_n \xrightarrow{p} c,\; c>0$. And "translate" the symbol $\xrightarrow{p}$ into its full notation. $\endgroup$ – Alecos Papadopoulos Nov 20 '13 at 20:13
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    $\begingroup$ Hi, @Alecos. I fear I must be misinterpreting your comment. How would your argument help? At least conceivably, $X_n$ might not converge at all or might converge to a nondegenerate random variable, neither of which seem covered by that line of argument. I would work directly from the definition, exhibit a simple set inclusion, and call it a day. :-) $\endgroup$ – cardinal Nov 20 '13 at 20:52
  • $\begingroup$ Hi @cardinal. I guess I "read" into the OP's formulation that $x_n$ is by assumption always sandwiched between $0$ and $y_n$. Then one could start by assuming also these cases that you mention -although the day would last longer this way!:)... except if I am forgetting something basic here. $\endgroup$ – Alecos Papadopoulos Nov 20 '13 at 21:07
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By definition, $Y_n$ tends to $0$ in probability if $$ \lim_{n\to\infty} P(\omega:|Y_n(\omega)| <\epsilon) = 1 \, , $$ for every $\epsilon>0$. Since $0\leq X_n\leq Y_n$, by hypothesis, if $|Y_n|<\epsilon$, then $|X_n|<\epsilon$, yielding that $$ \{\omega:|Y_n(\omega)|<\epsilon\} \subset \{\omega:|X_n(\omega)|<\epsilon\} \, . $$ Please, make sure that the inclusion above is totally clear. Now, using the monotonicity of $P$, can you relate the probabilities of the two events above? What does this relation imply for $\lim_{n\to\infty} P(\omega:|X_n(\omega)|<\epsilon)$? This may help.

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