I am stuck with a problem:

$X_n $ converges in probability to $X$ is equivalent to $\inf \{ \epsilon>0: P(|X_n-X|>\epsilon)<\epsilon \} \rightarrow 0$

See that $X_n$ converges to $X$ in probability $\iff$ $\exists N $ such that $\forall n \geq N$, $P[|X_n-X| > \epsilon]<\epsilon \iff \lim_{n \rightarrow \infty} \inf \{ \epsilon: P(|X_n-X|>\epsilon)<\epsilon \}=0$

I am not sure whether I can write the last step. If I can then it's done.

Can anyone help?


1 Answer 1


Your proof is missing details.

Let $X_n \to X$ in probability. Let $\eta >0$. Then there exists $n_0$ such that $P(|X_n-X| >\eta) <\eta$ for $n \geq n_0$. Hence $\inf \{\epsilon >0: P(|X_n-X| >\epsilon) <\epsilon\} \leq \eta$ for $n \geq n_0$. This proves that $\inf \{\epsilon >0: P(|X_n-X| >\epsilon) <\epsilon\} \to 0$ as $n \to \infty$.

Conversely suppose $\inf \{\epsilon >0: P(|X_n-X| >\epsilon) <\epsilon\} \to 0$ as $n \to \infty$. Let $\eta >0$. Then $\inf \{\epsilon >0: P(|X_n-X| >\epsilon) <\epsilon\} <\eta$ for $n$ sufficiently large, say $n \geq n_0$. This implies that there exists $\epsilon \in (0, \eta) $ such that $P(|X_n-X| >\epsilon) <\epsilon\} $. But then $P(|X_n-X| >\eta) <\eta\} $ for $n \geq n_0$. [Verify this!]. This proves that $X_n \to X$ in probability.


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