Express an I(2) process as a sum of telescoping cumulative sums Suppose $z_t$ is  zero mean I(2) process
Show that $z_t$ can be written as $$z_t = \sum_{i=1}^t e_i +\sum_{i=1}^{t-1}e_i +\sum_{i=1}^{t-2}e_i +...+e_1 $$ where $e_i$ is white noise process.
 A: Hi: If $z_t$ is I(2) with zero mean then that means that
$(z_{t} - z_{t-1})  =  (z_{t-1} - z_{t-2}) + \epsilon_{t}$.
So, if you difference the differences, that gives an I(0) process.
$(z_{t} - z_{t-1})  -  (z_{t-1} - z_{t-2}) = \epsilon_{t}$
The above was for $z_t$. Keep doing the same thing for $z_{t+1},z_{t+2},\ldots, z_{t+n}$. Then put everything except $z_{t+n}$ on the RHS and you should see the pattern.
I'm putting more material in ( still not solving it but it should work out ) Using the same idea as above, but for $t+n$, gives
$(z_{t+n} - z_{t+n-1})  =  (z_{t+n-1} - z_{t+n-2}) + \epsilon_{t}$.
So, if you difference the differences, that gives an I(0) process.
($z_{t+n} - z_{t+n-1})  -  (z_{t+n-1} + z_{t+n-2}) = \epsilon_{t}$
Now, re-write the above as: $z_{t+n} = 2 z_{t+n-1} - z_{t+n-2}$
Now, try to write $z_{t+n-1}$ and $z_{t+n-2}$ in terms of the
$\epsilon_{t}$ by going backwards until you get back to the beginning.
A: Hint 1: random walk is a cumulative sum of i.i.d. increments and is I(1). A cumulative sum of a random walk is a cumulative sum of an I(1) process and thus I(2). Your case is where i.i.d. is replaced by something more general, namely, an I(0) process.

I will write $z_t=a_0+a_1z_{t−1}+a_2z_{t−2}+e_t$. Is this true?

Hint 2: It is more like $z_t=y_t+y_{t−1}+y_{t−2}+\dots+y_1$ where $y_j$ is a cumulative sum of $e$s from $1$ to $j$. The $y_j$s are like random walks, but they may be more general if $e$s are not strictly i.i.d. but merely I(0).
