Show that two white noise definitions are equivalent Given $(\varepsilon_t)\sim WN(0,\sigma^2)$ a white noise. By definition
$$E(\varepsilon_t)=0,\,\, E(\varepsilon_t^2)=\sigma^2 \quad \forall t$$
and
$$E(\varepsilon_t \varepsilon_s) = 0, \quad s\neq t$$
I want to show that this equivalent to
$$E(\varepsilon_t |\varepsilon_s)=0$$
The intuition of one direction is ok. If $\varepsilon_t$ and $\varepsilon_s$ are ortogonal,i.e. $E(\varepsilon_t \varepsilon_s) = 0$, then the projection $E(\varepsilon_t |\varepsilon_s)=0$. But I don't know how to formally prove it.
The reciprocal or another direction I can't prove either.
Help
 A: This is not true in general. As a counterexample, let $(u_t)_{t\,\in\,\mathbb{Z}}$ be i.i.d. random variables with $u_t \sim \mathcal{N}(0,1)$. Define the stochastic process $(a_t)_{t\,\in\,\mathbb{Z}}$ as follows:
$$a_t =
\left\{
 \begin{array}{ll}
  u_t  & \mbox{if } t \equiv0 \mod 2 \\
  \frac{1}{\sqrt{2}}(u_{t-1}^2 - 1) & \mbox{if } t \equiv 1 \mod 2
 \end{array}
\right. \quad.
$$
It is straightforward to show that $\mathbb{E}[a_t] = 0$ and $Var(a_t) = 1$ for all $t$ (use $E[u_t^4] = 3$ for odd $t$). Moreover,
$$\mathbb{E}[a_{2t}a_{2t+1}] = \frac{1}{\sqrt{2}}\mathbb{E}[u_{2t}^3 - u_{2t}] = 0 \quad.$$
For the other values of $t,s \, \in \, \mathbb{Z}$ with $t \neq s$, you can show that $\mathbb{E}[a_{t}a_{s}] = 0$ just by verifying that it is the expected product different $u_t$, which is $0$. Therefore, $\mathbb{E}[a_{t}a_{s}] = 0$ for all $t,s \, \in \, \mathbb{Z}$ with $t \neq s$.
Finally, notice that $a_{2r} = u_{2r}$ and $a_{2r+1} = \frac{1}{\sqrt{2}}(u_{2r}^2 - 1)$ for any $r\,\in\,\mathbb{Z}$, so
$$\mathbb{E}[a_{2r+1}|a_{2r}] = \frac{1}{\sqrt{2}}(a_{2r}^2 - 1) \quad, $$
which concludes the argument.
