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Let $U \sim \mathrm{Uniform}[0,1]$.

$$ X_n= \begin{cases} 3 & U \leq \frac{2}{3}-\frac{1}{n} \\ 8 & \text{otherwise} \end{cases} $$

$$ Y = \begin{cases} 3 & U \leq \frac{2}{3} \\ 8 & \text{otherwise} \end{cases} $$

We have for $\epsilon > 0:$

$$ X_n = Y \text{ unless } \frac{2}{3}-\frac{1}{n} <U \leq\frac{2}{3}$$

$$P(\mid X_n-Y\mid \;\geq \epsilon) \leq P(X_n \neq Y)$$

$$= P(\frac{2}{3}-\frac{1}{n} < U \leq \frac{2}{3})$$ $$=\frac{1}{n}$$

Hence, $P(|X_n-Y| \geq \epsilon) \to 0$ as $n \to \infty$ for all $\epsilon > 0$
and the sequence $\{X_n\}$ converges in probability to $Y.$

I don't understand why: $P(|X_n-Y| \geq \epsilon) \leq P(X_n \neq Y)$
Could someone explain?

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  • $\begingroup$ To have $|X_n-Y| \geq \epsilon$ you certainly must have $X_n \neq Y$. When $\epsilon<5$ these are actually the same in this case (why?) $\endgroup$ – Ian May 26 '16 at 2:19
  • $\begingroup$ @Ian, $P(X_n \neq Y) = P(\mid X_n - Y \mid \; \geq \; \epsilon) + P(0 \;<\; \mid X_n - Y \mid \; < \; \epsilon)$. Thanks! $\endgroup$ – user137481 May 26 '16 at 14:35
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This is because both $X_n$ and $Y$ only take values from a set of two points $\{3, 8\}$, which means given any positive $\epsilon \in (0, 5]$, the set \begin{align} \{\omega: |X_n(\omega) - Y(\omega)| \geq \epsilon\} & = \{\omega: X_n(\omega) = 8, Y(\omega) = 3\} \cup \{\omega: X_n(\omega) = 3, Y(\omega) = 8\} \\ & = \{\omega: X_n(\omega) \neq Y(\omega)\}. \end{align}

And if $\epsilon > 5$, then $$\{\omega: |X_n(\omega) - Y(\omega)| \geq \epsilon\} = \varnothing.$$

So in either case, we have $P[|X_n - Y| \geq \epsilon] \leq P[X_n \neq Y]$.

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