Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 4 / 28J / a

The product $z_n=z_0{\prod }_{j=0}^{n-1}{X}_j$ uses only past multipliers, so all $X_n,\dots ,X_{N-1}$ are independent of it. With $m=\mathbb{E}{X}_0$, independence givesThe formula is finite when $m<\infty$; since the variables are positive, it also has the extended value $+\infty$ for $m=\infty$ and $n<N$. At $n=N$, it simply reads $z_N$.