Could you illustrate why a random walk process without a constant term exhibits stationarity in its first moment but not in the second?


1 Answer 1


First, recall that a stochastic process $\{ Y_t \}$ is weakly stationary if :

$i)$ The first moment is time independent and finite, i.e. $E(Y_t) \equiv \mu < \infty$

$ii)$ The Variance is time independent and finite, i.e. $Var(Y_t) = E[(Y_t - \mu)^2] \equiv \gamma_0 < \infty$

$iii)$ The Autocovariance only depends upon the order of lag $j$, i.e., $Cov(Y_t,Y_{t-j}) = E[ (Y_t - \mu) (Y_{t-j} - \mu)] \equiv \gamma_j$

Consider the following random walk process without the constant term: \begin{equation} Y_t = Y_{t-1} + \epsilon_t \end{equation}

Now,consider $Y_{1}=Y_{0} + \epsilon_1$

$Y_{2}=Y_{1} + \epsilon_2$

$Y_{3}=Y_{2} + \epsilon_3$

Let's stop here, and using the expression of $Y_1$ and $Y_2$ into $Y_3$ and assuming that $Y_0=0$, then

$Y_3= \epsilon_1 + \epsilon_2 + \epsilon_3$

Generalizing the last equation

\begin{equation} Y_t = \sum_{j=1}^{t} \epsilon_j \end{equation}

If you compute the moments, recalling that $\epsilon_t$ is a white noise process and so it has zero mean and constant variance, i.e., $\epsilon_t \sim i.i.d (0,\sigma^2)$ you have

\begin{equation} \mathbb{E}(Y_t) = \mathbb{E}(\sum_{j=1}^{t} \epsilon_j) =0 \end{equation} and the process is stationary in mean. It won't be stationary in variance. Indeed, \begin{equation} \mathbb{Var}(Y_t) = \sigma^2 \sum_{j=1}^{t} = t \sigma^2 \end{equation} As you can see, the variance increases with $t$, making the variance infinitely large as $t \rightarrow \infty$. Thus, ${Y_t}$ is not stationary in variance. Thus, the process is not weakly stationary.


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