Let $U_1,U_2,...$ be a sequence of independent, identically uniformly distributed RVs on the interval $[0,1]$. Show that there is a constant $a \in \mathbb{R}$ so that $(U_1 U_2 \cdot\cdot\cdot U_n )^{\frac{1}{n}} \longrightarrow a$ for $n \longrightarrow \infty$. Here, the almost sure convergence is meant...
So, if I handle the $U_i$ as numbers between $0$ and $1$, I would suggest $X_n := (U_1 U_2 \cdot\cdot\cdot U_n )^{\frac{1}{n}}$ converges a.s. towards 0, because if you multiply infinitely many of them, there surely is a $0$ among them and then $X_n$ converges to $0$. If there is no $0$ among the factors, it would converge towards $1$, if I am correct.
But I don't really know how to show it for RVs, since we haven't done any similar tasks in class, so I hope someone could show me how to prove it using the right notation.
$$ (U_1U_2\cdots U_n)^{1/n} = \exp \left (\frac{1}{n}\sum_i \ln(U_i) \right) $$
$$ x = \ln(u) \to dx = 1/u\, du\to dx = e^{-x}du $$ $x\in (-\infty,0]$ so $-\ln(U_i) \in Exp(1)$ and the average converges a.s by the law of large numbers. $e^x$ is a continuous function hence the expression converges to $e^{-1}$ a.s.
Also we can do the expectation directly as this derivation confuses some people, $$ E[\ln(U)] = \int_0^1 \ln(x)\,dx = \Big [ x\ln(x) - x \Big ]_0^1 = -1. $$ Note that we used e.g. Lhopital's rule to see that $$ \lim_{x\to 0} x\ln(x) = 0. $$