Let $(X_n)_n$ i.i.d with distribution function $F_{X_i}(x)=(1-x^{-\lambda})1_{x>1}, \lambda >0$. I calculated the expected value $(\frac{- \lambda}{1-\lambda})\,\textrm for \lambda \in (1,\infty)$ and undefined for $\lambda =1$ I also calculated the distribution from $X:=log(X_1)$. Now I want to show that $(Y_n)_n :=((X_1X_2 ... X_n)^\frac{1}{n})_{n\in \mathbb N} $ is convergent in Probability and calculate the limes.

My attempt:

I tried to use the law of large numbers to get almost sure convergence (and this implies convergenc in probability),i.e. $lim_{n\to \infty}\frac{1}{n}\sum_{j=1}^nY_j=\mathbb E(Y_1)$ and $\mathbb E(Y_1) $ is the same like $\mathbb E(X_1)$ but I do not see how this could help me in any way. Any help is much appreciated!


$$ \ln{Y_n}=\frac{1}{n}\sum_{i=1}^n\ln{X_i}\xrightarrow{p}\mathsf{E}\ln{X}_1=\lambda^{-1}. $$

Thus, $Y_n\xrightarrow{p}e^{1/\lambda}$.

$F_{\ln{X_1}}(x)=\mathsf{P}(\ln{X_1}\le x)=(1-e^{-\lambda x})1\{x\ge 0\}$ and $\mathsf{E}\ln{X_1}=\lambda^{-1}$.

  • $\begingroup$ nice answer! I was missing this trick with log. One last question remaining, why is here $1_{0 \le x}$ and not $1_{1 \le x} $ $\endgroup$ – user409387 Jun 7 '17 at 18:12
  • $\begingroup$ @JohnDoe If $x\ge 1$, then $\ln(x)\ge 0$... $\endgroup$ – d.k.o. Jun 7 '17 at 18:15
  • $\begingroup$ I do not really understand it .:( $\endgroup$ – user409387 Jun 7 '17 at 18:20
  • $\begingroup$ $\mathsf{P}(\ln{X_1}\le x)=\mathsf{P}({X_1}\le e^x)=(1-e^{-\lambda x})1\{e^x\ge 1\}=(1-e^{-\lambda x})1\{x\ge 0\}$ $\endgroup$ – d.k.o. Jun 7 '17 at 18:22
  • $\begingroup$ oh lol ok got it, sorry for stupid question $\endgroup$ – user409387 Jun 7 '17 at 18:25

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