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Let $\{A_n\}_{n \in \mathbb{N}}$ be a sequence of real numbers such that $\lim_{n \rightarrow \infty}A_n=0$. Does this imply that $plim_{n\rightarrow \infty}A_n=0$, where $plim$ is the probability limit?

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  • $\begingroup$ Are $A_n$ supposed to be random variables? What is a probability limit for real numbers? $\endgroup$ Oct 30 '14 at 17:53
  • $\begingroup$ $A_n$ are real numbers. $\endgroup$
    – TEX
    Oct 30 '14 at 17:57
  • $\begingroup$ A sequence of real numbers is a sequence of random numbers with trivial distribution. $\endgroup$ Oct 30 '14 at 18:31
  • $\begingroup$ 1) It is not a good idea ask another question in comments. 2) Where have you find the notation $plim$ probability limits? $\endgroup$ Oct 30 '14 at 18:50
  • $\begingroup$ in the Econometric book Hayashi $\endgroup$
    – TEX
    Oct 30 '14 at 18:53
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Using Markov's inequality:

$$ P( |A_n-0|> \epsilon) \leq \frac{E(|A_n|)}{\epsilon}=\frac{|A_n|}{\epsilon} $$ Letting $n\rightarrow \infty$ on both sides proves convergence in probability.

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Defenition for probability limit:

$X_n\to X$ in probability iff $\forall \epsilon>0$ $\lim\limits_{n\to\infty}P\{|X_n-X|>\epsilon\} = 0$.

We should concern sequence of random numbers $X_n$ such that $P\{X_n=A_n\}=1$. As $\lim\limits_{n\to\infty}A_n=0$ then for any $\epsilon>0$ there is $n_\epsilon$ such that $|A_n|<\epsilon$ for any $n>n_\epsilon$ and the same for $X_n$. It implies that $P\{|X_n-0|>\epsilon\}=0$ for $n>n_\epsilon$.

So we have that for any $\epsilon>0$ $\lim\limits_{n\to\infty}P\{|X_n-0|>\epsilon\} = 0$.

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