# Stationarity in Time Series

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

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: $$$$Y_t = Y_{t-1} + \epsilon_t$$$$

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

$$$$Y_t = \sum_{j=1}^{t} \epsilon_j$$$$

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

$$$$\mathbb{E}(Y_t) = \mathbb{E}(\sum_{j=1}^{t} \epsilon_j) =0$$$$ and the process is stationary in mean. It won't be stationary in variance. Indeed, $$$$\mathbb{Var}(Y_t) = \sigma^2 \sum_{j=1}^{t} = t \sigma^2$$$$ As you can see, the variance increases with $$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.