# Solved – Properties of the white noise process

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^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

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\$!

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