# Convergence in probability for two sequences of random variables

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?

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. $$P(A)\le P(B\cup C)$$.

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.

• Thank you - How does the first equality hold? Where $\epsilon/2$ first appears Feb 4, 2021 at 22:35
• @JDoe2 The first equality was actually not necessary, here is an updated proof.
– Vons
Feb 5, 2021 at 1:20
• 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)
– E-A
Feb 5, 2021 at 7:14
• I see thank you both for your time! Feb 5, 2021 at 15:33