# $X_1, X_2$, … independent variables with support in [0,1] , mean $\mu\in(0,1)$, $Y_n=X_1X_2…X_n,\;\;\;n\geq1$. Study convergence of $Y_n$ [closed]

$$X_1, X_2$$, ... 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, mrtaurhoApr 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.
• It is the infinite product $X_1X_2\cdots$. All the hypothesis about independence and mean or irrelevent. – Kavi Rama Murthy Apr 3 at 7:18