# Martingale Difference Sequence

I saw that in the link on page 3 it is said $$Y_t = e_t\cdot e_{t-1}$$ is martingale difference sequence and dependent where $$e_t$$ is i.i.d with $$N(0,\sigma^2)$$ Could you provide me with the proof of it?

Apparently, $$E[Y_t]=E[e_te_{t-1}]=E[e_t]E[e_{t-1}]=0$$; and with some algebraical manipulations we can reach the conclusion $$E[Y_t|Y_{t-1},...,Y_1]=0$$: \begin{align}E[Y_t|Y_{t-1},...,Y_1]=E[e_te_{t-1}|e_{t-1}e_{t-2},...,e_1e_0]\end{align} Now let $$\mathbf{e}_t=[e_t,...,e_0]$$, so the given side in the above expectation is a function of $$\mathbf{e}_{t-1}$$,i.e. a function $$f$$ such that $$f: \mathbb{R}^{t}\rightarrow \mathbb{R}^{t-1}$$; therefore we need to find $$E[e_te_{t-1}|f(\mathbf{e}_{t-1})]$$. Using Law of Iterated Expectations, we have $$E[e_te_{t-1}|f(\mathbf{e}_{t-1})]=E[E[e_te_{t-1}|f(\mathbf{e}_{t-1}),\mathbf{e}_{t-1}]]$$ Since, $$f(\mathbf{e}_{t-1})$$ is a deterministic function of $$\mathbf{e}_{t-1}$$, we have $$E[e_te_{t-1}|f(\mathbf{e}_{t-1}),\mathbf{e}_{t-1}]=E[e_te_{t-1}|\mathbf{e}_{t-1}]$$ which is equal to $$E[e_te_{t-1}|e_{t-1},e_{t-2},...,e_0]=e_{t-1}E[e_t|e_{t-1},...,e_0]=e_{t-1}E[e_t]=0$$. So, the expected value inside the outer expectation is $$0$$, i.e. $$E[e_te_{t-1}|f(\mathbf{e}_{t-1}),\mathbf{e}_{t-1}]=0$$, which means the expectation is $$0$$, i.e. $$E[e_te_{t-1}|f(\mathbf{e}_{t-1})]=0$$, which also means the expectation we query in the first place for $$Y_t$$ is $$0$$, leading to $$Y_t$$ being MDS.