# Question about slides in lecture note: What if we can't assume $\mu=0?$

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 $$\gamma(0)=\sigma^2_W+\mu^2+b^2(\sigma^2_W+\mu^2)=(\sigma^2_W+\mu^2)(1+b^2).$$

• 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. – noob2 Apr 20 at 15:27
• @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? – Parseval Apr 20 at 15:31
• 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. – noob2 Apr 20 at 16:13
• Writing $W_t = \mu + Z_t$ where $Z_t$ are zero mean, independent Gaussians, then using 'FOIL', you should get an expected value of $\mu^2$. – steveo'america Apr 20 at 16:22

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 $$h$$:

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

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):

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

### 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]

}

return(X[-1])

}

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

mean(test)
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.

(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:

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

and

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

(last equality coming from general covariance properties).