Show that $X_n \to 0$ in probability under given condition.

Question

Let $k > 0$. Suppose that

$$\forall \epsilon > 0: \exists N: \forall n \geq N: P(|X_n| \geq \epsilon) \leq \epsilon k$$

Show that $X_n \xrightarrow{P}{ 0}$.

Attempt:

We have to show:

$$\forall \tilde{\epsilon} > 0: \lim_n P(|X_n| \geq \tilde{\epsilon}) = 0$$

This is equivalent with showing that

$$\forall \tilde{\epsilon} > 0: \forall \epsilon \in ]0, \tilde{\epsilon}[: \exists N: \forall n \geq N : P(|X_n| \geq \tilde{\epsilon}) \leq \epsilon k$$

So, let's show this.

Let $\tilde{\epsilon} > 0$ and take $\epsilon \in ]0 , \tilde{\epsilon}[$. Choose $N$ such that $P(|X_n| \geq \epsilon) \leq \epsilon k$ whenever $n \geq N$. Then, if $n \geq N$, we have

$$P(|X_n| \geq \tilde{\epsilon}) \leq P(|X_n| \geq \epsilon) \leq \epsilon k$$

and the claim follows.

Is this correct?

Answer 1

One has to show that $\lim_{n\rightarrow\infty}\mathbb{P}[|X_n|>\delta]=0$ for all $\delta>0$. Fix $\delta>0$. Given $\varepsilon>0$, choose $N_\varepsilon$ such that $\mathbb{P}[|X_n|>\frac{\varepsilon\wedge \delta}{k}]<\varepsilon\wedge\delta$ whenever $n\geq N$ (This can be done by the conditions in the hypothesis). Consequently, if $n\geq N_\varepsilon$ $$ \mathbb{P}[|X_n|>\delta]\leq \mathbb{P}[|X_n|>\frac{\varepsilon\wedge\delta}{k}]<\varepsilon\wedge\delta\leq\varepsilon$$ This shows that $\lim_{n\rightarrow\infty}\mathbb{P}[|X_n|>\delta]=0$ for any $\delta>0$.