White noise stricly stationary proof

Question

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 $a_{t_1 -k},\dots, a_{t_n-k}$.

Answer 1

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 $A_{t_1}, A_{t_2}, \ldots, A_{t_n}$.

Answer 2

IID is a stronger condition than strict stationarity, and it implies the latter. If you have a sequence of IID values then the sequence is exchangeable, which implies strict stationarity.