1
$\begingroup$

The Question is,

Give an example of a sequence $\{Y_n\}$ of jointly defined random variables, s.t. as $n \to \infty$

(i) ${Y_n \over n}$ converges to 0 in probability.

(ii) ${Y_n \over n^2}$ converges to 0 with probability 1. (almost sure convergence)

(iii) ${Y_n \over n}$ does not converge to 0 with probability 1.

Does anyone know of an example this? I found some examples of sequences that converge in probability and not almost surely converge, but I'm not sure those satisfy the condition (ii). Also, I wonder how to satisfy "jointly defined random variables" condition.


(I edited to add the example I thought.)

My example is, Let $\{Y_n\}$ be independent and $P(Y_n=1)={1 \over n}$ and $P(Y_n=0)=1-{1 \over n}$.

Then $P({Y_n \over n}=1)={1 \over n}$ and $P({Y_n \over n}=0)=1-{1 \over n}$,

${Y_n \over n}$ converges to $0$ in probability not almost surely. But I'm not sure whether ${Y_n \over n^2}$ converges to 0 with probability 1 or not.

Thanks for any help.

$\endgroup$
  • $\begingroup$ You could start by mentioning the examples you've fond, and if they comply condition 2 or not (or if you are not sure, or if have no idea). $\endgroup$ – leonbloy Oct 18 '17 at 3:07
  • $\begingroup$ I checked on my examples again, and I found that my examples were independent sequences, so it does not satisfy the jointly defined random variables. I have no idea... $\endgroup$ – Siri Oct 18 '17 at 3:40
  • 3
    $\begingroup$ Independent random variables are jointly defined. Jointly defined means that you know how the random variables interact, namely to know what $P(Y_1 \in A_1, \cdots , Y_n \in A_n)$ is for all $n$ and for all choices of subsets $A_i$. $\endgroup$ – Kore-N Oct 18 '17 at 5:44
  • $\begingroup$ Oh Thank you!!! I didn't know that. $\endgroup$ – Siri Oct 18 '17 at 6:00
  • $\begingroup$ Your example is odd since you chose $(Y_n)$ such that $|Y_n|\leqslant1$ almost surely, which implies that $Y_n/n^a\to0$ almost surely, for every $a>0$. $\endgroup$ – Did Oct 18 '17 at 7:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.