limiting distribution of $\frac{z_i}{z_1+...+z_i}$

Question

Let $z_i$ be i.i.d. random variables with finite mean. I would like to know the limiting distribution of $\frac{z_i}{z_1+...+z_i}$. I guess it is 0 almost surely, because the numerator is finite almost surely and the denominator tends to infinity. Is that correct and rigorous?

Answer 1

No, that is not rigorous. If $z_i\equiv 0$ the then ratio is not even defined. However, the limit is a.s. $0$ if we assume that $Ez_1 \neq 0$.

Divide numerator and denominator by $i$. Use SLLN to reduce the proof to showing that $z_i/i \to 0$ a.s. For this, apply Borel-Cantelli Lemma: $\sum P(|\frac {z_i} i|>\epsilon) =\sum P(|\frac {z_1} i|>\epsilon)=\sum P(|z_1| >n\epsilon) <\infty$ because $E|z_1|<\infty$. It follows that $z_i/i \to 0$ a.s.