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How do you show that the white noise process is strictly stationary?

Let's consider the i.i.d. white noise process $a_t$: \begin{align} E[a_t] &= 0\\ Var[a_t] &= \sigma_a^2 \end{align}

The weak stationarity is obvious. But I don't know how to show the strict stationarity. Basically I would like to show that for all $t_1, \dots, t_n, k \in \mathbb{Z}$, the joint distribution of $a_{t_1},\dots, a_{t_n}$ is the same as $a_{t_1 -k},\dots, a_{t_n-k}$.

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  • $\begingroup$ Hi: the joint distribution of the $n$ $X_{i}$ random variables at time $t_{n}$, ( the $n$ element vector ) is multivariate normal with mean zero and covariance matrix with diagonal elements $\sigma_{a}^2$ and zero everywhere else. So, you just need to show that the same thing is true for the $n$ random variables at time $t_{k+n}$ (i.e: the $n$ vector $X_{t+k+1}, X_{t+k+2}, \ldots, X_{t+k+n}$. ). I hope that helps. $\endgroup$ – mlofton Apr 16 at 2:20
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That cryptic initialism i.i.d. is the key; it stands for independent and identically distributed which says to the cognoscenti that

  • Every random variable in the process/time series has the same distribution, and so for any given distinct integers $t_1, t_2, \ldots, t_n$ and any $k \in \mathbb Z$, the $2n$ random variables $A_{t_1}, A_{t_2}, \ldots, A_{t_n}, A_{t_1-k}, A_{t_2-k}, \ldots, A_{t_n-k}$ all have the same distribution.
  • The random variables $A_{t_1}, A_{t_2}, \ldots, A_{t_n}$ are independent, and so their joint distribution is just the product of their $n$ identical marginal distributions. Similarly, $A_{t_1-k}, A_{t_2-k}, \ldots, A_{t_n-k}$ are also independent and since the have the same marginal distributions as the $A_{t_1}, A_{t_2}, \ldots, A_{t_n}$, the joint distribution of $A_{t_1-k}, A_{t_2-k}, \ldots, A_{t_n-k}$, which is the product of their marginal distributions, is the same as the joint distribution of $A_{t_1}, A_{t_2}, \ldots, A_{t_n}$.
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