Convergence in probability Can anyone tell me how they got the regions $0&lt;\epsilon&lt;\theta$ and $\epsilon &gt;0 $.

Also to clarify, is the last step where it says $\lim_{n \to \infty} P(|Y_n-\theta|&gt;\epsilon)=0$
\begin{align}
\mathbb{P}(|Y_n - \theta | \geq \epsilon) &amp;= 1- \mathbb{P}(-\epsilon \leq Y_n - \theta \leq \epsilon) \\
&amp;= 1 - \mathbb{P}(Y_n \leq \theta + \epsilon) + \mathbb{P}(Y_n \leq \theta - \epsilon) \\
&amp;= 1 - F_{Y_n}(\theta + \epsilon) + F_{Y_n}(\theta-\epsilon). \\
&amp;= F_{Y_n}(\theta- \epsilon). 
\end{align}
Note that the last equality follows from the fact that $\theta + \epsilon \geq \theta$ for any $\epsilon&gt;0$ and consequently $F_{Y_n}(\theta+ \epsilon) = 1$. According to the definition of the distribution function of $Y_n$, there are two cases to be considered:

(1) Assume $0&lt;\epsilon&lt;\theta$, then $0 &lt; \theta-\epsilon &lt; \theta$ and consequently
$$\mathbb{P}(|Y_n - \theta| \geq \epsilon) = \left(\frac{\theta-\epsilon}{\theta}\right)^n \to 0,$$ 
as $n \to \infty$. 

(2) Assume $\epsilon &gt; \theta$, then $\theta - \epsilon &lt; 0$ and consequently $F_{Y_n}(\theta- \epsilon) = 0, \forall n \geq 1$. Therefore, the result holds in the limit.
Imagine we have $X$ representing the toss of a fair die.

Then $\text{Range}(X) = \{1,2,\cdots,6\}$

$$P(X = x) = \frac16$$

So:

$$P(X \le 1) = \frac16$$
$$P(X \le 2) = \frac26$$
$$P(X \le 3) = \frac36$$
$$P(X \le 4) = \frac46$$
$$P(X \le 5) = \frac56$$
$$P(X \le 6) = \frac66 = 1$$

Also:

$$P(X \le 7) = \frac66 = 1$$
$$P(X \le 0) = \frac06$$
$$P(X \le 0.5) = \frac06$$
$$P(X \le -2000) = \frac06$$
$$P(X \le 4.5) = \frac46$$
$$P(X &gt; 7) = \frac06$$

Thus:

The '$\varepsilon &gt; \theta$' (I'll assume '$\varepsilon &gt; 0$' is a typo) is like '$X &gt; 7$'

The '$0 &lt; \varepsilon &lt; \theta$' (I'll assume '$\varepsilon &gt; 0$' is a typo) is like '$1 &lt; X &lt; 6$'



Also

Show that $Y_n$ converges in probability