If ${(X_t)}$ is strictly stationary then $X_{t_1}+\ldots+X_{t_n} \overset{d}{=} X_{t_1+k}+\ldots+X_{t_n+k}$

Question

Let ${(X_t)}_{t \in \mathbb{Z}}$ be strictly stationary. I want to show that for all $t_1, \ldots, t_n \in \mathbb{Z}, k \in \mathbb{Z}$ it holds $$ X_{t_1}+\ldots+X_{t_n} \overset{d}{=} X_{t_1+k}+\ldots+X_{t_n+k} $$

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My reasoning:

Strict stationary means that the finite dimensional distributions are shift-invariant, i.e. for all $t_1, \ldots, t_n \in \mathbb{Z}, k \in \mathbb{Z}$ it holds

$$ (X_{t_1}, \ldots,X_{t_n}) \overset{d}{=} (X_{t_1+k}, \ldots X_{t_n+k}). $$

If the distribution of the vector $(X_{t_1}, \ldots,X_{t_n})$ is shift-invariant, then, it seems to me, so should be the distribution of $X_{t_1}+\ldots+X_{t_n}$.

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But how can one show it rigorously?

Answer 1

As it is mentioned by Did in the comments, you need to show that if $X\stackrel{d}{=}Y$, then for every function $g$, $g(X)\stackrel{d}{=}g(Y)$. I am tempted to give an information theoretic proof of this using data processing inequality for any kind of $f-$divergence, which includes KL-divergence. Data processing inequality writes as: $$ D(P_X||P_Y)\geq D(P_{g(X)}||P_{g(Y)}). $$ Note that KL-divergence is positive. Moreover $X\stackrel{d}{=}Y$ if and only if $D(P_X||P_Y)=0$. Hence: $$ D(P_{g(X)}||P_{g(Y)})=0\implies g(X)\stackrel{d}{=}g(Y). $$