Prove that $X_n = Z + \epsilon_n$ is Op(1)

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 $X_n=O_p(1)$.
• 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. – Did Nov 22 '17 at 9:20
• @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 $(A\cup B)^c=A^c\cap B^c$. – Cm7F7Bb Nov 23 '17 at 8:56