<p>First, recall that a stochastic process <span class="math-container">$\{ Y_t \}$</span> is weakly stationary if :</p>
<p><span class="math-container">$i)$</span> The first moment is time independent and finite, i.e. <span class="math-container">$E(Y_t) \equiv \mu &lt; \infty$</span></p>
<p><span class="math-container">$ii)$</span> The Variance is time independent and finite, i.e. <span class="math-container">$Var(Y_t) = E[(Y_t - \mu)^2] \equiv \gamma_0 &lt; \infty$</span></p>
<p><span class="math-container">$iii)$</span> The Autocovariance only depends upon the order of lag <span class="math-container">$j$</span>, i.e., <span class="math-container">$Cov(Y_t,Y_{t-j}) = E[ (Y_t - \mu) (Y_{t-j} - \mu)] \equiv  \gamma_j$</span></p>
<p>Consider the following random walk process without the constant term:
<span class="math-container">\begin{equation}
Y_t = Y_{t-1} + \epsilon_t
\end{equation}</span></p>
<p>Now,consider <span class="math-container">$Y_{1}=Y_{0} + \epsilon_1$</span></p>
<p><span class="math-container">$Y_{2}=Y_{1} + \epsilon_2$</span></p>
<p><span class="math-container">$Y_{3}=Y_{2} + \epsilon_3$</span></p>
<p>Let's stop here, and using the expression of <span class="math-container">$Y_1$</span> and <span class="math-container">$Y_2$</span> into <span class="math-container">$Y_3$</span> and assuming that <span class="math-container">$Y_0=0$</span>, then</p>
<p><span class="math-container">$Y_3= \epsilon_1 + \epsilon_2 + \epsilon_3$</span></p>
<p>Generalizing the last equation</p>
<p><span class="math-container">\begin{equation}
Y_t = \sum_{j=1}^{t} \epsilon_j
\end{equation}</span></p>
<p>If you compute the moments, recalling that <span class="math-container">$\epsilon_t$</span> is a white noise process and so it has zero mean and constant variance, i.e., <span class="math-container">$\epsilon_t \sim i.i.d (0,\sigma^2)$</span> you have</p>
<p><span class="math-container">\begin{equation}
\mathbb{E}(Y_t) = \mathbb{E}(\sum_{j=1}^{t} \epsilon_j) =0
\end{equation}</span>
and the process is stationary in mean. It won't be stationary in variance. Indeed,
<span class="math-container">\begin{equation}
\mathbb{Var}(Y_t) = \sigma^2 \sum_{j=1}^{t} = t \sigma^2
\end{equation}</span>
As you can see, the variance increases with <span class="math-container">$t$</span>, making the variance infinitely large as <span class="math-container">$t \rightarrow \infty$</span>. Thus, <span class="math-container">${Y_t}$</span> is not stationary in variance. Thus, the process is not weakly stationary.</p>

CC BY-SA 4.0
2023-12-18T17:26:27.457Z
2023-12-22T08:39:30.043Z
2023-12-22T08:39:30.043Z
Tony
<p>Could you illustrate why a random walk process without a constant term exhibits stationarity in its first moment but not in the second?</p>

<econometrics><time-series><stationary>
Stationarity in Time Series
Dimitru
# Stationarity in Time Series

(Asked by: username_0 on 2023-12-18)

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

**Tags:** econometrics, time-series, stationary

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## 1 answers

### Answer by username_1 on 2023-12-18. Score: 10 | Accepted answer

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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