2
$\begingroup$

Let $\{X_n\}$ and $\{Y_n\}$ be sequences of variables and suppose that $Y_n$ converges in probability to some random variable $Y$, i.e. $Y_n\xrightarrow{p}Y$. Is it true then that:

$$\lim_{n\rightarrow\infty}\mathbb{P}[|X_n-Y_n|>\epsilon]=0 \text{ implies } X_n\xrightarrow{p}Y$$

If so, how can I show this?

$\endgroup$

1 Answer 1

1
$\begingroup$

Assume that (where I conveniently replaced Y with Z) $$\begin{split}X_n-Y_n&\overset p {\rightarrow} 0\\ Y_n&\overset p {\rightarrow} Z\end{split}$$

Fix $\epsilon.$ Notice that $|X_n-Y_n|\le\frac \epsilon 2$ and $|Y_n-Z|\le\frac \epsilon 2$ implies that $|X_n-Z|\le\epsilon$, by the triangle inequality. Reversing the logic, this means that $|X_n-Z|>\epsilon$ implies that $|X_n-Y_n|>\frac \epsilon 2$ (inclusive) or $|Y_n-Z|>\frac \epsilon 2$. In particular, if an event implies that at least one of two other events has occurred, this means that $A\subset B\cup C$, i.e. $P(A)\le P(B\cup C)$.

enter image description here

Now

$$\begin{split}P(|X_n-Z|>\epsilon)&\le P(|X_n-Y_n|>\frac \epsilon 2\cup|Y_n-Z|>\frac \epsilon 2)\text { what we just said}\\ &\le P(|X_n-Y_n|>\frac \epsilon 2)+P(|Y_n-Z|> \frac \epsilon 2)\text { definition of union} \end{split}$$

Take the limit to get $lim_{n\rightarrow\infty}P(|X_n-Z|>\epsilon)\le0$. Since probabilities are positive, it is 0.

$\endgroup$
4
  • $\begingroup$ Thank you - How does the first equality hold? Where $\epsilon/2$ first appears $\endgroup$
    – JDoe2
    Feb 4, 2021 at 22:35
  • $\begingroup$ @JDoe2 The first equality was actually not necessary, here is an updated proof. $\endgroup$
    – Vons
    Feb 5, 2021 at 1:20
  • $\begingroup$ Minor critique: The expression $$ X_n \rightarrow Y_n $$ does not really make sense; when we talk about limits, we do not want the RHS to depend on n. However, $$X_n - Y_n \rightarrow 0 $$ does make sense, and that is essentially what is being used. (Also, for OP, you if you know that $$ X_n + Y_n \rightarrow X + Y $$, you can use that to prove the claim as well, and the proof of this claim is also essentially the proof given to you in the answer above) $\endgroup$
    – E-A
    Feb 5, 2021 at 7:14
  • $\begingroup$ I see thank you both for your time! $\endgroup$
    – JDoe2
    Feb 5, 2021 at 15:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .