The question popped up when I was reading these lecture notes online. Consider the MA$(1)$ process given by $X_t=W_t+bW_{t-1}$ where $W_t$ is white noise distributed with constant variance $\sigma_W^2.$

They then proceed to calculate autocovariance function by

\begin{align} \gamma(k) &=E[X_tX_{t-k}]=E[(W_t+bW_{t-1})(W_{t-k}+bW_{t-k-1})]\\ &=E[W_tW_{t-k}]+E[bW_tW_{t-k-1}]+E[bW_{t-1}W_{t-k}]+E[b^2W_{t-1}W_{t-k-1}]. \end{align}

For $k=0$ we get the variance by $\gamma(0)=\text{Cov}[X_t,X_t]=\text{Var}[X_t]=\sigma_X^2,$ which expressed in terms of $\sigma_W^2$ becomes

\begin{align} \gamma(0) &=E[W_t^2]+bE[W_tW_{t-1}]+E[bW_{t-1}W_{t}]+b^2E[W_{t-1}^2]\\ &=\sigma^2_W+0+0+b^2\sigma^2_W = (1+b^2)\sigma_W^2. \end{align}

But the above only holds if the mean of the process $W$ is equal to zero.

Question 1: Why is $E[W_tW_{t-1}]=0?$

Question 2: What if $W\sim\text{WN}(\mu,\sigma_W^2)?$ That is, how would the calculations change?

Here is my attempt: If $\mu \neq 0$ then $E[W_t^2]=\text{Var}[W_t]+E[W_t]^2=\sigma_W^2+\mu^2$. If it's still true that $E[W_tW_{t-1}]=0$ then we obtain $\gamma(0)=\sigma^2_W+\mu^2+b^2(\sigma^2_W+\mu^2)=(\sigma^2_W+\mu^2)(1+b^2).$

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    $\begingroup$ E[W(t) W(t-1)] is zero because white noise terms are zero mean and uncorrelated. If they are no longer zero mean, then this does not hold any more, it will be non-zero. $\endgroup$
    – nbbo2
    Apr 20, 2021 at 15:27
  • $\begingroup$ @noob2 - Thank you. But do you know how to calculate $E[W_tW_{t-1}]$ to know exactly which non-zero value it will be? $\endgroup$
    – Parseval
    Apr 20, 2021 at 15:31
  • $\begingroup$ By a reasoning very similar to what you already did, I believe it will be $\mu^2$, but check the math, I may be wrong. $\endgroup$
    – nbbo2
    Apr 20, 2021 at 16:13
  • 1
    $\begingroup$ Writing $W_t = \mu + Z_t$ where $Z_t$ are zero mean, independent Gaussians, then using 'FOIL', you should get an expected value of $\mu^2$. $\endgroup$ Apr 20, 2021 at 16:22

2 Answers 2


I feel that question 1 has already been answered in the comments, so I will provide a brief answer and insight for question 2 (which is also your title question).

Question 2: What happens when $W \sim N(\mu, \sigma^2)$?

You will get a moving average with a specific intercept-term. The intercept does not have any impact on the autocovariances and thus they will stay the same as the original moving average. To see this, let $\bar{W} \sim N(\mu,\sigma^2)$ be a white-noise process with constant drift. Then, we can quickly observe that:

\begin{align*} X_t &= \bar{W}_t + b \bar{W}_{t-1}\\ &= (\mu + W_t) + b \cdot (\mu + W_{t-1})\\ &= \mu + b \mu + W_t + bW_{t-1}\\ &= \nu + W_t + bW_{t-1}, \end{align*} with $\nu = \mu + b\mu$ and $W_t \sim N(0, \sigma^2)$ be an uncorrelated white-noise process. Evidently, we also get that $\mathbb{E}\left[X_t\right]=\nu$. Now, we have a MA(1) process with constant drift, and therefore we cannot use the equation on page 3 in the slides (as seen in your question), since the author assumes $X_t$ is a zero-mean process. Instead, taken directly from the definition of autocovariance, it is observed that the mean does not have any impact on the autocovariances:

\begin{align} \gamma(h)&=\mathbb{C}ov\left(X_t,X_{t-h}\right)\\ &=\mathbb{C}ov\left(\nu + W_t + bW_{t-1}, \: \nu + W_{t-h} + bW_{t-h-1}\right)\\ &=\mathbb{C}ov\left(W_t,W_{t-h}\right) + b\mathbb{C}ov\left(W_t,W_{t-h-1}\right) + b\mathbb{C}ov\left(W_{t-1},W_{t-h}\right) + b^2\mathbb{C}ov\left(W_{t-1},W_{t-h-1}\right),\\ \end{align}

where we have used covariance properties of linear combinations. Therefore, we get the same results found in the slides for all $h$:

\begin{equation} \gamma(h) = \begin{cases} (1+b^2)\sigma^2 & \text{if } h = 0 \\ b\sigma^2 & \text{if } h=1\\ 0 & \text{if } h\geq 2 % \end{cases} . \end{equation}

If you want to verify this yourself, then you could proceed as done in the slides, however, the formula on page 3 for $X_t$ being a process with non-zero mean, can be written as (directly from the definition of (auto-)covariance):

\begin{equation} \gamma(h)=\mathbb{C}ov\left(X_t,X_{t-h}\right) = \mathbb{E}\left[X_t X_{t-h}\right] - \mathbb{E}\left[X_t\right] \mathbb{E}\left[X_{t-h}\right]. \end{equation}

Simulating the results:

As a quick conclusion, we can verify the results by simulating a MA(1) process with drift (this is done in R) and see whether the mean of the process is equal to $\nu$ and whether the autocorrelations (for $h\geq 2$) converge towards zero. Let, $b=0.5, \: \mu=10$ and $\sigma^2 = 1$, then $\nu=15$ and can be verified by running the following code:

MA_process_sim <- function(b, mu, sigma, Nmesh){

  X <- numeric()

  W <- rnorm(Nmesh+1, mu, sigma)

  X[1] <- 0

  for(i in 1:Nmesh){

    X[i+1] <- W[i+1] + b*W[i]



test <- MA_process_sim(0.5, 10, 1, 425000)

acfs <- acf(test, plot = F)

with mean(test) giving us 14.9979. The acfs for $h=1$ results in $0.4$, which is the same as the theoretical result, $\rho(1)=\frac{\gamma(1)}{\gamma(0)}= \frac{b\sigma^2}{(1+b^2)\sigma^2}= 0.4$. Furthermore, for $h \geq 2$ the acf-values are very close to zero. I hope this provide some help and insight.

Additional picture of simulation results:



(Q2) Adding a constant $\mu$ to the white noise, we have

$$Y_t := (\mu+W_t) + b(\mu + W_{t-1}) = \mu(1+b) + X_t, $$

which is MA(1) with drift:

$$E[Y_t] = \mu(1+b) $$


$$ {\rm Cov}(Y_t, Y_{t-k}) = {\rm Cov}(\mu(1+b) + X_t, \mu(1+b) + X_{t-k}) $$ $$ = {\rm Cov}(X_t, X_{t-k})$$

(last equality coming from general covariance properties).


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