Properties of the white noise process The stochastic process $\{u_t\}$ is a white noise process if and only if


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*$Eu_t=0$ for all integers $t$; and

*$E(u_t u_{t+k})=\sigma^2\textbf{1}\{k=0\}$ for all integers $t$ and $k$, where $\sigma>0$ and $\textbf{1}\{k=0\}$ is equal to $1$ if and only if $k=0$, and equal to $0$ if and only if $k\neq 0$.


I have heard from my lecturer that a white noise process satisfies $E_tu_{t+1}=0$, where $E_t$ is expectation given information about $\{u_k\}_{k\leq t}$. 
Question. Is it true that a white noise process $\{u_t\}$ satisfies $E_tu_{t+1}=0$? If 'yes', then why is it true and how do I derive that conclusion? If 'not', then what is a counterexample, and is it true under some reasonable assumption (e.g., assuming that the random variables in the stochastic process $\{u_t\}$ are independent)?
Attempt 1. By (2), $E(u_tu_{t+1})=0$. From this it follows by the law of total expectations that $E(E_t(u_tu_{t+1}))=0$. Since we are conditioning on information about the white noise process for the time periods $k\leq t$, it follows that $E(u_tE_tu_{t+1})=0$. Now, $E_tu_{t+1}=0$ is consistent with the last expression, but I do not see how it follows deductively (if it does). (Since I view $E_tu_{t+1}$ as a given real number, I think the last expression simplifies to $Eu_t\cdot E_tu_{t+1}=0$, which is satisfied whether or not $E_tu_{t+1}=0$ because $Eu_t=0$ by (1).)
Attempt 2. After considering Alexey's comment to my question, I tried to write an answer.
To begin with, if we assume independence in the sense that $u(t)$ is a stochastic variable independent of its history before time period $t$, then the distribution of $u_{t+1}|I_t$ coincide with the distribution of $u_{t+1}$, where $I_t$ is the information set up to time period $t$. Thus, in this case we have $E_tu_{t+1}=Eu_{t+1}=0$. 
After this I tried to find a counterexample to the conclusion that $E_tu_{t+1}=0$ for any white noise process $\{u_t\}$. I found a dependent white noise process, but not one that satisfied $E_tu_{t+1}\neq 0$. The example is the following.
Let $\{v_t\}$ be an i.i.d. process such that $P(v_t=-1)=P(v_t=1)=1/2$ for all integers $t$. Define a new stochastic process by $$u_t=v_t(1-v_{t-1}).$$ First, let me check that it is a white noise process. Firstly,\begin{align}Eu_t&=E(v_t(1-v_{t-1}))\\ &=Ev_tE(1-v_{t-1})\\ &=0\cdot 0 =0\end{align} where the second equality follows by independence and the third equality from the fact that $Ev_t=1/2-1/2=0$. 
Secondly, for any integer $t$,  \begin{align}Eu_tu_{t+k}&=E(v_t(1-v_{t-1})v_{t+k}(1-v_{t+k-1}))\\
&=E(v_t(1-v_{t-1})(1-v_{t+k-1}))E(v_{t+k})\\
&=E(v_t(1-v_{t-1})(1-v_{t+k-1}))\cdot 0\\
&=0\end{align} if $k\neq 0$, and, if $k=0$, \begin{align}Eu_t^2 &=Ev_t^2E(1-v_{t-1})^2\\ &=((-1)^2/2+1^2/2)+(2^2/2+0^2/2)\\
&=2\end{align} which is finite.
Thus, $\{u_t\}$ is a white noise process. Is the process dependent? Yes, since e.g. $u_t=2$ implies $v_t=1$ and $v_{t-1}=-1$ and thus $u_{t+1}=v_{t+1}(1-v_t)=0$. This means that $$P(u_t=2,u_{t+1}=2)=0.$$ However, $$P(u_t=2)P(u_{t+1}=2)=1/4\cdot 1/4=1/16,$$ and hence $$P(u_t=2,u_{t+1}=2)\neq P(u_t=2)P(u_{t+1}=2).$$
From here on, I have tried to construct an information set $I_t$ such that $E_tu_{t+1}\neq 0$,  but without success. I have also tried to somehow change the definition of $v_t$ or $u_t$. Maybe it would work if $u_t$ was a product of two distinct stochastic processes.
 A: Consider the i.i.d. process $\{x_t\}$ where $P(x_t=0)=P(x_t=2)=1/2$ for each integer $t$. This sequence satisfies $E(x_t)=1$ for each integer $t$. Now, construct the process $\{u_t\}=\{x_t^2(1-x_{t-1})\}$. For each integer $t$, then, we have \begin{align}E(u_t)&=E(x^2_t(1-x_{t-1}))\\ &=E(x_t^2)(1-E(x_{t-1}))\\ &=E(x_t^2)(1-1)\\ &=0\end{align} where the second inequality follows from independence and the linearity of expectation. 
Furthermore, \begin{align}Eu_t^2&=Ex_t^4E(1-x_{t-1})^2\\
&=2^4/2\cdot [1^2/2+(-1)^2/2]\\ &=8,\end{align} which is finite and independent of $t$. For $k\neq - 1$ we also have \begin{align}Eu_tu_{t+k}&=E(x_t^2(1-x_{t-1})x_{t+k}^2(1-x_{t+k-1}))\\ &=E(1-x_{t-1})E(x_t^2x_{t+k}^2(1-x_{t+k-1}))\\ &= 0\cdot E(x_t^2x_{t+k}^2(1-x_{t+k-1}))\\ &=0, \end{align} and if $k=-1$ we may do as follows: \begin{align}Eu_tu_{t-1}&=E(x_t^2(1-x_{t-1})x_{t-1}^2(1-x_{t-2}))\\ &=E(1-x_{t-2})E(x_t^2(1-x_{t-1})x_{t-1}^2)\\ &= 0\cdot E(x_t^2(1-x_{t-1})x_{t-1}^2)\\ &=0. \end{align}
In other words, $\{u_t\}$ is a white noise process. 
To show that $E_tu_{t+1}\neq 0$, consider the information set $I_t$ up until time period $t$ which says that $u_t=0$. Then $x_t^2(1-x_{t-1})=0$. By construction this can only be the case if $x_t=0$. Hence, $u_{t+1}=x_{t+1}^2(1-0)=x_{t+1}^2$ if $u_t=0$ is given. (Note that this suggests that $u_{t+1}$ depends on information in time periods $k\leq t$.) Hence, as the information set $I_t$ says nothing about the value of $x_{t+1}$, the distribution of $u_{t+1}|I_t$ is equivalent to the distribution of $x_{t+1}^2$. Thus, \begin{align}E_tu_{t+1} &=Ex_{t+1}^2\\ &=0^2/2+2^2/2\\ &=2\\ &\neq 0.\end{align}
Thus, we have found a white noise process not satisfying $E_tu_{t+1}= 0$!
