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.
Best Answer
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.