= Fractional moment decay of independent mean-one products
{title2=$\mathbb E[Y_n^r]=(\mathbb E[X_1^r])^n,\quad 0<r<1$}
Suppose $(X_n)$ are identically distributed <independent random variables> with $X_n\geq0$, $\mathbb EX_n=1$, and $X_1$ not constant <almost surely>. For each $0<r<1$, strict <concavity> of $t^r$ gives
$$
0<q:=\mathbb E[X_1^r]<1.
$$
The <moment> is finite because $t^r\leq1+t$ on $[0,\infty)$. Let $Y_n=\prod_{k=1}^nX_k$. <Independence> and <Markov's inequality> give
$$
\mathbb E[Y_n^r]=q^n,\qquad
\mathbb P(Y_n>\varepsilon)\leq\varepsilon^{-r}q^n.
$$
The <series> of the latter <probabilities> is finite. The <First Borel-Cantelli lemma>, applied for the countable choices $\varepsilon=1/m$, proves $Y_n\to0$ <almost surely>. This proof requires neither existence nor <integrability> of $\log X_1$.
Although this nonnegative <martingale> has $\mathbb EY_n=1$ at every time, its terminal <expected value> is zero. It is therefore not <uniformly integrable>. The contrast is useful: bounded <expected values> ensure <almost sure convergence> here, but do not allow preservation of the <expected value> through the limit.
Back to article page
