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

Question

$X_1, X_2$, ... 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.

Answer 1

$(Y_n)$ 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.