If $X_{n}\xrightarrow{\mathcal{P}}X$ and $X_{n}\xrightarrow{\mathcal{P}}Y$, does $X = Y$ almost surely?

Question

Suppose that $Y,X,X_{1},X_{2},X_{3},\ldots$ are random variables such that $X_{n}\xrightarrow{\mathcal{P}}X$ and $X_{n}\xrightarrow{\mathcal{P}}Y$. Does $X = Y$ almost surely?

Here is my attempt.

Due to the definitions involved, the following limits hold for every $\varepsilon > 0$: \begin{align*} \lim_{n\to\infty}\mathbb{P}(\{\omega\in\Omega : |X_{n}(\omega) - X(\omega)| \geq \varepsilon\}) = \lim_{n\to\infty}\mathbb{P}(\{\omega\in\Omega : |X_{n}(\omega) - Y(\omega)| \geq \varepsilon\}) = 0 \end{align*}

We want to prove that: \begin{align*} \mathbb{P}(\{\omega\in\Omega : X(\omega) = Y(\omega)\}) = 1 & \Longleftrightarrow \mathbb{P}(\{\omega\in\Omega : |X(\omega) - Y(\omega)| = 0\}) = 1 \end{align*}

But then I get stuck. The hint says to apply the triangle inequality.

Can someone help me with this?

Answer 1

A common way of proving $$ \mathbb{P}(\{ \omega \in \Omega : |X(\omega)-Y(\omega)| = 0 \}) = 1 $$ is to show that $$ \mathbb{P}(\{ \omega \in \Omega : |X(\omega)-Y(\omega)| \ge \varepsilon \}) = 0 $$ for any arbitrary $\varepsilon > 0$. So, fix $\varepsilon > 0$ arbitrary and suppose $\omega \in \Omega$ is such that $$ |X(\omega)-Y(\omega)| \ge \varepsilon. $$ It follows $$ \varepsilon \le |X(\omega)-Y(\omega)| \le |X(\omega)-X_n(\omega)| + |X_n(\omega)-Y(\omega)| $$ by the triangle inequality. Assuming for contradiction that we also have $|X(\omega)-X_n(\omega)| < \varepsilon/2$ and $|X_n(\omega)-Y(\omega)| < \varepsilon/2$, we obtain $$ \varepsilon \le |X(\omega)-X_n(\omega)| + |X_n(\omega)-Y(\omega)| < \varepsilon/2 + \varepsilon/2 = \varepsilon, $$ which is impossible (this contradiction argument is what @SeverinSchraven is referring to). So, we must have that at least one of the following inequalities holds true: $|X(\omega)-X_n(\omega)| \ge \varepsilon/2$ and / or $|X_n(\omega)-Y(\omega)| \ge \varepsilon/2$. Since we have been working with an arbitrary $\omega \in \Omega$, we have proven the set inclusion $$ \{ \omega \in \Omega : |X(\omega)-Y(\omega)| \ge \varepsilon \} \subseteq \{ \omega \in \Omega : |X(\omega)-X_n(\omega)| \ge \varepsilon/2 \} \cup \{ \omega \in \Omega : |X_n(\omega)-Y(\omega)| \ge \varepsilon/2 \}. $$ Now, taking $\mathbb{P}$ of both sides above, applying limits, and using the relevant standard properties, try to show that $$ \mathbb{P}(\{ \omega \in \Omega : |X(\omega)-Y(\omega)| \ge \varepsilon \}) = 0. $$ Since we fixed an arbitrary $\varepsilon > 0$, the proof is complete from there. Feel free to show your work and thoughts. If you continue to have some trouble, we can comment / edit more detail into our answers. Best of luck with your studies!

Answer 2

$X_n \to X$ in probability implies that some subsequence of $(X_n)$ converges to $X$ almost surely. Then since this subsequence converges to $Y$ in probability, we obtain a subsequence of this subsequence that converges to $Y$ almost surely. Thus we obtained a subsequence of $(X_n)$ that converges to $X$ almost surely and converges to $Y$ almost surely. Hence $X = Y$ almost surely.