0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.