$X_1, X_2$, ... independent variables with support in the interval [0,1] and equal mean $\mu\in(0,1)$, but not necessarily identically distributed. Study the convergence in quadratic mead and convergence in probability that $Y_n=X_1X_2...X_n,\;\;\;n\geq1$

Convergence in probability:

If for every $\epsilon>0$, $\lim\limits_{n \to\infty} P(|X_n-X|>\epsilon)=0$, then $X_n$ converges in probability to X.

Convergence in quadratic mean:

If $\lim\limits_{n \to\infty} E((X_n-X)^2)=0$, then $X_n$ converges in quadratic mean to X.


closed as off-topic by saz, RRL, NCh, Cesareo, mrtaurho Apr 6 at 9:46

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$(Y_n)$ is non-negative and decreasing so $Y =\lim Y_n$ exists almost surely, hence in probability. By Bounded Convergence Theorem it also converges in quadratic mean.

  • $\begingroup$ (Yn) is non-negative and decreasing implies Y=limYn exists almost surely? What would be the value of Y? $\endgroup$ – Maria Apr 3 at 6:48
  • $\begingroup$ It is the infinite product $X_1X_2\cdots$. All the hypothesis about independence and mean or irrelevent. $\endgroup$ – Kavi Rama Murthy Apr 3 at 7:18
  • $\begingroup$ Why (Yn) non-negative and decreasing implies Y=limYn exists almost surely?? $\endgroup$ – MJ_FG-CCS Apr 11 at 17:20
  • $\begingroup$ If you have a positive an decreasing succession and Yn converges in probability to Y, then Yn converges almost sure to Y. Thatis true or maybe I was wrong? $\endgroup$ – MJ_FG-CCS Apr 11 at 17:23
  • $\begingroup$ That is true. The sequence converges a.s. and the limit in probability has to be the same as a.s. limit. $\endgroup$ – Kavi Rama Murthy Apr 11 at 23:05

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