Is the weak law of large numbers satisfied for the sequence of independent random variables $$ \xi_k, $$ so that $$P(\xi_k=\sqrt k)=P(\xi_k=-\sqrt k)=1/2 \sqrt k, $$$$P(\xi_k=0)=1-1/\sqrt k ?$$


1 Answer 1


Define $X_k= \frac{\sum_{i=1}^{k}{\xi_i}}{k}$.

Note that $E[X_k]=0$ holds for all $k$.

$Var(X_k)\leq\frac{1}{\sqrt{k}}$ which tends to $0$ as k tends to $\infty$

The above condition are sufficient to say that $X_k \to 0$ in probability.

But this is exactly what WLLN says. So the answer to your question is YES.

  • $\begingroup$ Why does $Var(X_k)=\frac{1}{k^{3/2}}$? $\endgroup$
    – daylegacy
    May 31, 2019 at 9:26
  • $\begingroup$ @daylegacy sorry made a silly mistake, corrected now. $\endgroup$ May 31, 2019 at 10:06

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