Let $\{ξ_i\}_{i=1}^n$ be i.i.d. random variable with $Eξ_i = 0$, $Dξ_i = σ^2 > 0$ and let $η_n = \frac{1}{\sigma\sqrt{n}}\sum_{i=1}^{n}ξ_i$. Prove/disprove existence of $(P) \lim_{n\to\infty} η_n$.
What is the meaning of $(P)$ in the above question?
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Dasheng Wang
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$(P) \lim $ stands for convergence in probability. Here we have convergence in distribution but not convergence in probability.
Kavi Rama Murthy
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Does it converge in distribution? I thought it should be $\eta_n-n\mu$ that converges to $N(0,1)$. – Dasheng Wang Apr 18 '20 at 12:47
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1@DashengWang It is given that the random variables have mean $0$. – Kavi Rama Murthy Apr 18 '20 at 12:49
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May I ask another question? When $n\to \infty$, does $\eta_n$ converge to $0$, since $\lim_{n\to\infty}\eta_n=\frac{\sqrt{n}\mu}{\sigma}$? – Dasheng Wang Apr 18 '20 at 16:10
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1@DashengWang No. If $a_n \to 0$ you cannot say $\sqrt n a_n \to 0$. – Kavi Rama Murthy Apr 18 '20 at 23:12
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There are different types of convergence of random variables in probability theory, for example convergence in probability, convergence in distribution, convergence almost surely.
In this case it seems to be convergence in probability.
SlavicDoomer
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