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Let $X_n$ be a sequence of random variables such that $X_nc_n \rightarrow 0$ in probability for any sequence $c_n \to 0$. To show $\{X_n\}$ is a tight sequence.

DEFINITION : A family $\{X_n\}$ is tight if for any $\epsilon > 0$ we can find a bounded interval $K_\epsilon$ such that $\mathbf{P}(|X_n| < K_\epsilon) > 1 - \epsilon$.

The converse also holds: Let $\{X_n\}$ be a family of tight random variables. Then for any $c_n \rightarrow 0$ we have $X_nc_n \rightarrow 0$ in Probability.

Proof : Since $\{X_n\}$ is tight, we have for any $\epsilon > 0$ a $K_\epsilon$ s.t. $\mathbf{P}(|X_n| > K_\epsilon) < \epsilon$. Let $\{c_n\}$ be any sequence converging to $0$. We can find a $n_0$ s.t. $n>n_0 \implies |c_n| < \epsilon/K_\epsilon$.

Thus for all $n>n_0$ , $\mathbf{P}(|c_nX_n| > \epsilon) = \mathbf{P}(|c_nX_n| > \epsilon, |X_n| > K_\epsilon) + \mathbf{P}(|c_nX_n| > \epsilon, |X_n| < K_\epsilon)$ $$< \mathbf{P}(|X_n| > K_\epsilon) + \mathbf{P}(|X_n| > K_\epsilon , |X_n| < K_\epsilon) (\because |c_n| < \epsilon/K_\epsilon)$$

$$< \epsilon$$

So this is what convergence in probability asks us to find.

Any help is appreciated.

EDIT : So this is a direction that I am trying :

Since $\{X_nc_n\} \rightarrow 0$ in probability $\implies$ Every subsequence $\{X_{n_k}c_{n_k}\}$ has a further subsequence that converges a.s. to $0$ and hence in distribution. This implies $\{X_nc_n\}$ is tight. Now as $\{c_n\}$ converges it is bounded by say $C$. Thus for any given $\epsilon$ we had a $K_\epsilon$ s.t. $\mathbf{P}(|c_nX_n| > K_\epsilon) < \epsilon$ for all $n$ and using the bound of $C$, we have $\mathbf{P}(|X_n| > K_\epsilon/C) < \epsilon$ for all $n$.

Is this right? Any comments are appreciated.

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  • $\begingroup$ What do you mean, "the converse also holds", after quoting the definition? The converse of the definition?! As you were able to show that, show it, please, it may help to decide what you're talking about. Thank you! $\endgroup$
    – user436658
    Jan 14 '21 at 19:04
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Suppose $\{X_n\}$ is not tight. Then there exists $\epsilon > 0$ and a subsequence $(X_{n_k})$ such that $\mathbb{P}(|X_{n_k}| > k) \geq \epsilon$. Now define the sequence $(c_n)$ by $c_n = 1/k$ for all $n_{k-1} < n \leq n_k$. We have $c_n \to 0$ but $\mathbb{P}(|X_{n_k} c_{n_k}| > 1) = \mathbb{P}(|X_{n_k}| > k) \geq \epsilon$ for all $k$, and therefore $X_n c_n$ does not converge to $0$ in probability, a contradiction.

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  • $\begingroup$ Its "for all $c_n$ that converge to 0 $X_nc_n$ converges to 0" $\endgroup$
    – rostader
    Jan 15 '21 at 8:12
  • $\begingroup$ @db93 Edited to answer your intended question $\endgroup$
    – Adam
    Jan 15 '21 at 8:26
  • $\begingroup$ Yeah... it is more readable this way.. Thanks @Adam $\endgroup$
    – rostader
    Jan 15 '21 at 8:28
  • $\begingroup$ "Then there exists ϵ>0 and a subsequence (Xnk) such that P(|Xnk|>k)≥ϵ" Okay.. I kinda missed this subtelty. I was getting stuck thinking that there can be just one $X_{n*}$ for which this holds $\endgroup$
    – rostader
    Jan 15 '21 at 8:33

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