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Consider $X_n$ and $Y_n$ be two real-valued random sequences, if $$P(X_n \neq Y_n) \rightarrow 0 \text{ as $n \rightarrow \infty$}$$ is it equivalent to say that $X_n$ converges to $Y_n$ in probability?

Since the definition of convergence in probability says

for every $\varepsilon>0$, we have

$$\lim_{n \rightarrow \infty} P(|X_n - Y_n| \ge \varepsilon) = 0$$

hence, if we observe the event: $$\{|X_n - Y_n| \ge \varepsilon\} = \{X_n - Y_n \ge \varepsilon \text{ or } X_n - Y_n \le -\varepsilon\}$$ This seems to me I can write $\{X_n \neq Y_n\} = \{|X_n - Y_n| \ge \varepsilon\}$ for all $\varepsilon>0$

It seems to me I can say yes, but I'm not quite sure.

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  • $\begingroup$ IMO it seems correct. $\endgroup$ – Jimmy R. Oct 20 '14 at 20:48
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Consider the probability space $([0,1],\mathcal{B}([0,1]),ℙ=\text{Lebesgue})$ with random variables $$X_n(ω) := 1, \quad Y_n(ω) := 1-\frac{1}{n}$$

Then for all $n$, $ℙ(X_n ≠ Y_n) = ℙ([0,1]) = 1$, but for any $ε>0$, when $n>\frac{1}{\epsilon}$ we have $|X_n - Y_n| = \frac{1}{n} < ε$ (almost) surely so $ℙ(|X_n - Y_n| > ε) → 0$ as $n→ ∞$. Hence the conditions are not equivalent(although one implication does hold).

The reason this fails is that the sets are related as follows:

$$ [X_n ≠ Y_n] = \bigcup_{ε > 0} [ |X_n - Y_n| > ε ] $$

as opposed to an intersection(='for all') as you said.

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