0
$\begingroup$

Let $X_n$ be a nonnegative random variable such that $\mathbb{E}[X_n]\leq \frac{\alpha}{n}$ where $n \in \mathbb{N}$. Clearly, as $n \to \infty$, $\mathbb{E}[X_n] \to 0$. Is it possible to show that $X_n \to 0$ in probability?

My try:

To show $X_n \to 0$ in probability, one needs to show $\text{Prob}[|X_n-0|\geq \epsilon]$ for any $\epsilon>0$. Since $X_n$ is a nonnegative random variable, one can write Markov's inequality as the following:

$$ \text{Prob}[|X_n-0|\geq \epsilon]=\text{Prob}[X_n\geq \epsilon]\leq \frac{\mathbb{E}[X_n]}{\epsilon} \leq \frac{\alpha}{\epsilon n} $$

No matter how small $\epsilon$ is, one can get convergence in probability. Am I right? Is there any way to get convergence in mean square($\ell_2$).

$\endgroup$
3
  • 2
    $\begingroup$ Your proof of convergence in probability is correct. You don't even know that second moments exist so you cannot prove mean square convergence. $\endgroup$ Commented Mar 7, 2022 at 8:36
  • $\begingroup$ @Kavi Rama Murthy: What if we know $\mathbb{E}[|X_n-\mathbb{E}[X_n]|^2]\leq \sigma^2$ for some $\sigma>0$? $\endgroup$ Commented Mar 7, 2022 at 8:40
  • $\begingroup$ @Sepide You still cannot say it converges to zero in mean square sense. Recall mean square error is bias^2 + variance. You need the variance to go to zero. $\endgroup$ Commented Mar 7, 2022 at 9:22

1 Answer 1

Reset to default
1
$\begingroup$

Answer for the question in your comment above: Let $X_n$ take the values $0$ and $n$ with probabilities $1-\frac 1 {n^{2}}$ and $\frac 1 {n^{2}}$ respectively. Then $EX_n=\frac 1n$ and $E(X_n-EX_n)^{2}=1-\frac 1 {n^{2}}$ is bounded but $X_n$ does not converge in mean square.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.