Why this sequence converges to $0$ and not to $1$ in probability?

Question

I didn't understand why this sequence bellow converges to $0$ in probability?

Shouldn't this sequence converge to $1$ in probability?

This is the definition of convergence in probability:

Answer 1

Write $X_n=I(A_n)$ where $I$ is the indicator function and $A_n$ are the sequence of sets given. Given $0<\varepsilon<1$ $$ P(|X_n|>\varepsilon)=\lambda(A_n)\to 0 $$ as $n\to \infty$. If $\varepsilon\geq 1$, then trivially $P(|X_n|>\varepsilon)=0$.

Answer 2

The definition of '$1_{A_n}$ converges in probability to $0$' is 'For all $\varepsilon > 0$, $\lim P(|1_{A_n} -0| >\varepsilon) = 0$'

The $\varepsilon$'s in question actually depend on whether or not they are greater than $1$.

Let us consider each case, and show $\lim P(|1_{A_n} -0| >\varepsilon) = 0$ in each case.

Each case begins as follows: $\lim P(|1_{A_n} -0| >\varepsilon) = \lim P(|1_{A_n}| >\varepsilon)$

$ = \lim P(1_{A_n} >\varepsilon) = \lim [P(1_{A_n} >\varepsilon)] = \lim [1- P(1_{A_n} \le \varepsilon)] = 1- \lim [P(1_{A_n} \le \varepsilon)]$

1. Case 1: $0<\varepsilon<1$

   • Observe that for $0 < \varepsilon < 1$, we have that $\{1_{A_n} \le \varepsilon\} = \{1_{A_n} = 0\}$. Then

$$1- \lim P(1_{A_n} \le \varepsilon) = 1- \lim P(1_{A_n} = 0) = 1- \lim P(1_{A_n^C} = 1) = 1- \lim P(A_n^C)$$

$$= 1- \lim [P(A_n^C)] = \lim [1- P(A_n^C)] = \lim [P(A_n)] = 0$$

2. Case 2: $\varepsilon=1$

   • Observe that for $\varepsilon = 1$, we have that $\{1_{A_n} \le 1\} = \{1_{A_n} \le \varepsilon \}$. Observe also that $\{1_{A_n} \le 1\} = [1_{A_n} = 1] \cup [1_{A_n} = 0] = \Omega$ Then

$$1- \lim [P(1_{A_n} \le \varepsilon)] = 1- \lim P(1_{A_n} \le 1) = 1- \lim P([1_{A_n} = 1] \cup [1_{A_n} = 0]) = 1- \lim P(A_n \cup A_n^C) = 1- \lim P(\Omega) = 1 - 1 = 0$$

$$= 1- \lim [P(A_n^C)] = \lim [1- P(A_n^C)] = \lim [P(A_n)] = 0$$

3. Case 3: $\varepsilon>1$

   • Observe that we still have for $\varepsilon > 1$ that $\{1_{A_n} \le 1\} = \{1_{A_n} \le \varepsilon \}$. Observe also that we still have that $\{1_{A_n} \le \varepsilon \} = \{1_{A_n} \le 1\} = [1_{A_n} = 1] \cup [1_{A_n} = 0]$ Then

$$1- \lim [P(1_{A_n} \le \varepsilon)] = 1- \lim P(1_{A_n} \le 1) = 1- \lim P([1_{A_n} = 1] \cup [1_{A_n} = 0]) = 1- \lim P(A_n \cup A_n^C) = 1- \lim P(\Omega) = 1 - 1 = 0$$

Therefore, regardless of the $\varepsilon$, we end up with $$\lim P(|1_{A_n} -0| >\varepsilon) =0$$