6
$\begingroup$

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

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

$\endgroup$
  • 1
    $\begingroup$ Under the introduced definition it is not true. You should think on counterexample, and in this case it is not hard to construct. You should take into account that there is no independence assumption (actually, the common definition of the white noise process is different from yours, see wiki for an example) $\endgroup$ – Alexey Zaytsev Dec 14 '16 at 12:49
  • $\begingroup$ Thanks for the input, I will try to come up with a counterexample, I was using lectures.quantecon.org/jl/arma.html and the definition given there seems to be the same as the definition given in en.wikipedia.org/wiki/White_noise#White_noise_vector. I.i.d. noise is a special case of a white noise process, does the conclusion follow if the process is an i.i.d. noise process? $\endgroup$ – Elias Dec 14 '16 at 12:54
  • $\begingroup$ As you can see from wiki usually we need an independence property for $u_t$ (also an alternative "weak" definition is possible). For i.i.d. process your conclusion follows from the fact that the conditional distribution is similar to the initial distribution, as the values of a realization of the process at different times are i.i.d. $\endgroup$ – Alexey Zaytsev Dec 14 '16 at 13:07
  • $\begingroup$ Yes, I see that. Hmm, okay, in essence, as I understand you, independence implies that the distribution of $u_{t+1}|I_t$ is equal to the distribution of $u_{t+1}$, where $I_t$ is the information set up until period $t$? $\endgroup$ – Elias Dec 14 '16 at 13:21
  • $\begingroup$ Yes, it implies. $\endgroup$ – Alexey Zaytsev Dec 14 '16 at 13:23
3
$\begingroup$

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 $E_tu_{t+1}= 0$!

$\endgroup$

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.