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A sequence $X_n$ converges in probability to $X$ if

$$\lim_{n\rightarrow \infty}P(|X_n-X|>\epsilon)=0$$ when $n \rightarrow \infty$ for all $\epsilon >0$.

How do you formulate the antithesis of this in order to display "does not converge in probability"?

Is it

$$\lim_{n\rightarrow \infty}P(|X_n-X|<\epsilon)=0$$ or

$$\lim_{n\rightarrow \infty}P(|X_n-X|>\epsilon)=c>0$$

or perhaps both?

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    $\begingroup$ you're using the word antithesis wrong. $\endgroup$ Dec 11 '15 at 15:29
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In your formulation you missed the existential quantifier "for all $\epsilon >0$:" so the definition must read:

A sequence $X_n$ of random variables converges in probability to the random variable $X$ if for all $\epsilon >0$ we have: $$ \lim_{n\rightarrow \infty}P(|X_n-X|>\epsilon)=0 $$ So to say that $X_n$ do not converge in probability to $X$ you need that to show there exists some (only one is enough!) positive real number $\epsilon$ such that $$ P(|X_n-X|>\epsilon) $$ does not converge to zero, it is larger than some positive number $\delta$ for arbitrarily large $n$.

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    $\begingroup$ "...larger than $\delta$ infinitely often" is another way to put it. $\endgroup$
    – Adrian
    Dec 11 '15 at 12:59
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There is a general rule of predicate logic called the negation law, which says that when you take the negation ("antithesis") of a universal statement, this is logically equivalent to an existential statement of the negation of the condition. In logical symbols, for any statement $P(x)$ about a quantifier $x$ the negation law says:

$$\sim (\forall x) \text{ } P(x) \equiv (\exists x) \sim P(x).$$

The negation of the condition for convergence in probability is:

$$\sim (\forall \epsilon > 0) \text{ } \lim \limits_{n \rightarrow \infty} \mathbb{P}(|X_n - X| > \epsilon) = 0.$$

Applying the negation law, this is logically equivalent to:

$$(\exists \epsilon > 0) \text{ } \lim \limits_{n \rightarrow \infty} \mathbb{P}(|X_n - X| > \epsilon) \neq 0.$$

In words, this says that if you want to show the convergence in probability does not hold, you must show that there exists a value $\epsilon>0$ for which the limiting probability is not zero. Now, if you want to go further, you can break down the limit statement into its logical definition. The logical definition of the limit statement $\lim \limits_{n \rightarrow \infty} \mathbb{P}(|X_n - X| > \epsilon) = 0$ is:

$$(\forall \varepsilon >0) (\exists N \in \mathbb{N}) (\forall n \in \mathbb{N}_{N+}) \text{ } \text{ } \mathbb{P}(|X_n - X| > \epsilon) < \varepsilon.$$

Hence, the negation of convergence in probability can be written as:

$$(\exists \epsilon > 0) (\exists \varepsilon >0) (\forall N \in \mathbb{N}) (\exists n \in \mathbb{N}_{N+}) \text{ } \text{ } \mathbb{P}(|X_n - X| > \epsilon) \geqslant \varepsilon.$$

In words, this says that if you want to show that convergence in probability does not hold, you need to show that there exist values $\epsilon>0$ and $\varepsilon>0$ such that, for any positive integer $N$, there is a corresponding integer $n \geqslant N$ such that $\mathbb{P}(|X_n - X| > \epsilon) \geqslant \varepsilon$.

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