I have seen two definitions of convergence in probability. The first one is, if $X_n \rightarrow c$ in probability,

$$\mathbb{P}(|X_n - c|>\epsilon) = 0$$ as $n\rightarrow\infty$.

But I have also seen the definition:

$$\mathbb{P}(|X_n - c| \ge \epsilon) = 0$$ as $n\rightarrow\infty$.

Is this basically saying that these two limits are equal?

  • 3
    $\begingroup$ These definitions are the same, since they both must hold for all $\epsilon>0$, and $$ P[|X_n-c|>\epsilon] \leq P[|X_n-c|\geq \epsilon] \leq P[|X_n-c|>\epsilon/2]$$ So, it is correct to say that $X_n$ converges to $c$ in probablity if for all $\epsilon>0$ we have $\lim_{n\rightarrow\infty} P[|X_n-c|>\epsilon]=0$. And it is also correct to replace this with "$\lim_{n\rightarrow\infty}P[|X_n-c|\geq \epsilon]=0$." $\endgroup$ – Michael Nov 16 '16 at 20:12
  • $\begingroup$ Thank you! This is very descriptive $\endgroup$ – Felicio Grande Nov 16 '16 at 20:16
  • $\begingroup$ Both "definitions" are wrong. What would be your source? $\endgroup$ – Did Nov 16 '16 at 20:48


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See the difference?

  • $\begingroup$ is this because I reversed the order of the statements? $\endgroup$ – Felicio Grande Nov 16 '16 at 22:03
  • 1
    $\begingroup$ @FelicioGrande : I think Did is just noticing that you wrote "$P[|X_n-c|\geq \epsilon]=0, n\rightarrow\infty$" rather than "$P[|X_n-c|\geq \epsilon]\rightarrow 0$ as $n\rightarrow\infty$." The former language could be ambiguous and might be interpreted as $P[|X_n-c|\geq \epsilon]$ is actually 0 for large $n$, which is not necessarily true. It converges to zero, but is not necessarily equal to zero for particular $n$ values. $\endgroup$ – Michael Nov 16 '16 at 22:18

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