Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 30K / b / Solution

If $Y_1,Y_2,\ldots$ are independent integrable random variables with $\mathbb E[Y_i]=1$, then
$$
M_n=\prod_{i=1}^nY_i
$$
is a martingale for $\mathcal F_n=\sigma(Y_1,\ldots,Y_n)$, because independence gives $\mathbb E[M_{n+1}\mid\mathcal F_n]=M_n\mathbb E[Y_{n+1}]=M_n$.


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 Incoming links: Product martingale from independent mean-one factors