How can I show that a random walk is not covariance stationary?

Question

How can I show that a random walk ($y$ follows a random walk) is not covariance stationary? I tried to work on the formula below (with no results) Could you give me just a hint on how to proceed please?

$$Cov(y_{t+h},y_t)=E(y_{t+h}\times y_t)-E(y_t)E(y_{t+h})$$

Is this approach right?

Important: $\epsilon_t$, the shock, is an iid sequence with mean $0$ and variance $\sigma^2_\epsilon$.

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If $y$ follows a RW we have $$y_t=y_{t-1}+\epsilon_t$$

then,

$$Var(y_t)=Var(y_{t-1}+\epsilon_t)=Var(y_{t-1})+\sigma_\epsilon+2Cov(\epsilon_t,y_{t-1})$$ Now I see that the variance of $y_t$ depends on the variance of $y_{t-1}$. This should suggest me that we lack covariance stationarity.

Answer 1

I think you're making life hard for yourself there. You just need to use a few elementary properties of variances and covariances.

Here's one approach:

• start with the algebraic definition of your random walk process.

• derive $\text{Var}(y_t)$ in terms of $\text{Var}(y_{t-1})$ and the variance of the error term

• show that $\text{Cov}(y_t,y_{t-1}) = Var(y_{t-1})$

• argue that $\text{Cov}(y_s,y_{s-1})\neq \text{Cov}(y_t,y_{t-1})$ if $s\neq t$.

... though, frankly, I think even just going to the second step (writing $\text{Var}(y_t)$ in terms of $\text{Var}(y_{t-1})$ and the variance of the error term) is sufficient to establish it's not covariance stationary.

Answer 2

For each integer $t$, $Y_t = \sum_{i=1}^t X_i$ where the $X_i$ are iid random variables. From the independence of the $X_i$, it follows that $\operatorname{var}(Y_t) = \sum_{i=1}^t \operatorname{var}(X_i) = t\sigma^2$. For integer $h$, let $W = \sum_{i=t+1}^{t+h} X_i$ and note that $W$ and $Y_t$ are independent random variables because they are functions (sums) of disjoint collections of independent random variables. Then, $$\begin{align} \operatorname{cov}(Y_{t+h},Y_t) &= \operatorname{cov}(Y_t+W,Y_t)\\ &= \operatorname{cov}(Y_t,Y_t) + \operatorname{cov}(W,Y_t) &{\scriptstyle\text{this step follows because the covariance operator is bilinear;}}\\ &= \operatorname{var}(Y_t) + 0 &\scriptstyle{\text{0 because independent RVs}~W~\text{and}~Y_t~\text{have zero covariance;}}\\ &= \operatorname{var}(Y_t)\\ &= t\sigma^2 \end{align}$$ and thus the covariance increases as a function of $t$ but is not a function of $h$ at all as is needed for covariance stationarity. More generally, you can show that $\operatorname{cov}(Y_t, Y_s) = \operatorname{var}\left(Y_{\min\{t,s\}}\right)$.

Answer 3

A usual way that we show this is by writing the random walk as

$$y_t = \sum_{i=1}^tu_t$$ and so

$$\operatorname{Var}(y_t) = \operatorname{Var}\left(\sum_{i=1}^tu_i\right) = t\sigma^2$$

and

$$\operatorname{Cov}(y_t, y_{t+k})= E\left(\sum_{i=1}^tu_i\right)\left(\sum_{i=1}^{t+k}u_i\right)-E\left(\sum_{i=1}^tu_i\right)E\left(\sum_{i=1}^{t+k}u_i\right)$$

$$=E\left(\sum_{i=1}^tu_i\right)\left(\sum_{i=1}^{t}u_i+ \sum_{i=t+1}^{t+k}u_i\right) -E\left(\sum_{i=1}^tu_i\right)E\left(\sum_{i=1}^{t}u_i+ \sum_{i=t+1}^{t+k}u_i\right)$$

$$=E\left(\sum_{i=1}^tu_i\right)^2 - \left[E\left(\sum_{i=1}^tu_i\right) \right]^2 +E\left(\sum_{i=1}^tu_i\right)\left(\sum_{i=t+1}^{t+k}u_i\right) -E\left(\sum_{i=1}^tu_i\right)E\left(\sum_{i=t+1}^{t+k}u_i\right)$$

$$=\operatorname{Var}\left(\sum_{i=1}^tu_i\right) + \operatorname{Cov}\left(\sum_{i=1}^tu_i, \sum_{i=t+1}^{t+k}u_i\right)$$

The two sums in the covariance term are independent since the white noises in the first do not appear in the second (different time-indices), so this covariance is zero, and we are left with

$$\operatorname{Cov}(y_t, y_{t+k}) = \operatorname{Var}\left(\sum_{i=1}^tu_i\right)=t\sigma^2$$