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Let $(X_n)_{n \geq 1}$ and $(Y_n)_{n \geq 1}$ be two sequences of random variables s.t. $X_n$ converges to X and $Y_n$ to $Y$ both in probability. Furthemore, $X$ = $Y$ a.s.

How can I prove that, for every $\epsilon > 0$, $lim_{n \rightarrow \infty} P(|X_n - Y_n| > \epsilon) = 0 $ ?

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Given $\epsilon > 0$ and given $\delta$,

$X_n$ converges to $X$ in probability $\implies$ there is $N_1 \in \mathbb{N}$ such that $n \geq N_1 \implies P(|X_n - X| > \epsilon) < \delta/2$

$Y_n$ converges to $Y = X$ in probability $\implies$ there is $N_2 \in \mathbb{N}$ such that $n \geq N_2 \implies P(|Y_n - X| > \epsilon) < \delta/2$

Let $N = \max \{N_1, N_2\}$.

Then $P(|X_n - Y_n| > \epsilon) = P(|X_n - X + X - Y_n| > \epsilon) \leq P(|X_n - X| + |X - Y_n| > \epsilon)$

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  • $\begingroup$ Thanks! Just another question: $X_n$ converges to $X$ in probability implies that if there is N such that $n \geq N$ then $P(|X_n - X|>\epsilon) < \delta$, $\delta>0$ ? Or $P(|X_n - X|>\epsilon) = 0$? $\endgroup$ – user0102910 May 3 '16 at 2:37
  • $\begingroup$ You are right! Already edited my answer! Thanks $\endgroup$ – Luísa Borsato May 3 '16 at 11:03
  • $\begingroup$ Thanks very much indeed, Luisa =) $\endgroup$ – user0102910 May 4 '16 at 0:15

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