The stochastic process ${u_t}$ is a white noise process if and only if

- $Eu_t=0$ for all integers $t$; and
- $E(u_t u_{t+k})=sigma^2textbf{1}{k=0}$ 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 $kneq 0$.

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 ${u_k}_{kleq t}$.

*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 $kleq t$, 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 $Eu_tcdot E_tu_{t+1}=0$, 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})\ &=0cdot 0 =0end{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\

&=0end{align} if $kneq 0$, 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)\

&=2end{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, $$P(u_t=2)P(u_{t+1}=2)=1/4cdot 1/4=1/16,$$ 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.

**Contents**hide

#### Best Answer

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)\ &=0end{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\ &=2^4/2cdot [1^2/2+(-1)^2/2]\ &=8,end{align} which is finite and independent of $t$. For $kneq – 1$ 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}))\ &= 0cdot 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)\ &= 0cdot 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 $kleq t$.) 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 $E_tu_{t+1}= 0$!