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?
Best Answer
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.