Suppose that a sequence of random variables is given by $X_n = Z + \epsilon_n, n \geq 1$, such that $\epsilon_n$ and $Z$ are independent, and $E[{\epsilon^2}_n]\leq 1$ for all $n\geq 1$ (Note that Z does not depend on n). I want to show that $X_n = O_P(1)$ (i.e. $X_n$ is bounded in probability).

The formal definition of $O_P(1)$ convergence I use is:

$X_n = O_P(1)$ if for each $\epsilon > 0$, $\exists$ an $M_\epsilon$ such that $\limsup_{n\to\infty} P\{||X_n|| > M_\epsilon\} < \epsilon$

I have an intuitive understanding, but am unable to start the proof. If we think of $n$ as the sample size, then as it grows, $X_n$ will vary with $\epsilon_n$. Since $E[{\epsilon^2}_n]$ is bounded above by 1, then $X_n'$'s probability mass should be bounded as well (i.e. not escape to the tails).

How can I construct a formal proof?


We have that $$ P(|X_n|>\delta)=P(|Z+\varepsilon_n|>\delta)\le P(|Z|>\delta/2)+P(|\varepsilon_n|>\delta/2) $$ since $\{|Z+\epsilon_n|>\delta\}\subset\{|Z|>\delta/2\}\cup\{|\varepsilon_n|>\delta/2\}$.

$P(|Z|>\delta/2)\to0$ as $\delta\to\infty$ using the properties of the cumulative distribution functions. By Markov's inequality, $$ P(|\varepsilon_n|>\delta/2)\le\frac{4\operatorname E|\varepsilon_n|^2}{\delta^2}\le\frac{4}{\delta^2}. $$ Hence, by choosing large enough $\delta$, we can make the $P(|X_n|>\delta)$ as small as we want. This shows that $X_n=O_p(1)$.

  • $\begingroup$ The hypothesis that $Z$ is square integrable is not needed. Start with $K$ large enough for $P(|Z|\geqslant K)\leqslant\epsilon$ and essentially proceed as before. $\endgroup$ – Did Nov 22 '17 at 9:20
  • $\begingroup$ @Did Thank you for your comment. I rewrote my answer. $\endgroup$ – Cm7F7Bb Nov 22 '17 at 9:25
  • $\begingroup$ @Cm7F7Bb Thank you. Would it be correct to say that the first line follows from the triangle inequality for vector noms? Or, is it necessary to state the union condition in the second line? $\endgroup$ – elhalconloco Nov 22 '17 at 23:23
  • $\begingroup$ @elhalconloco I don't think that the triangle inequality is enough. By the triangle inequality, $P(|X+\varepsilon_n|>\delta)\le P(|X|+|\varepsilon_n|>\delta)$. Then we have that $$\{|X|+|\varepsilon_n|>\delta\}\subset\{|X|>\delta/2\}\cup\{|\varepsilon|>\delta/2\}$$ since $$\{|X|\le\delta/2\}\cap\{|\varepsilon|\le\delta/2\}\subset\{|X|+|\varepsilon_n|\le\delta\}.$$ Here we use the fact $A\subset B$ if and only if $B^c\subset A^c$ and $(A\cup B)^c=A^c\cap B^c$. $\endgroup$ – Cm7F7Bb Nov 23 '17 at 8:56

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