If $\{X_t\}_{t \geq 0}$ is stationary, $E f(X_t) < \infty$ for a continuous function $f$, is $\{f(X_t)\}_{t\geq 0}$ stationary?

Question

It seems my lecturer thinks that if $\{X_t\}_{t\geq 0}$ is a stationary sequence of random variables, $E f(X_t) < \infty$ for all $t \geq 0$ and $f$ is continuous, then $\{ f(X_t)\}_{t \geq 0}$ is stationary. Is this true? If so, can you prove it or provide a reference?

Answer 1

I suppose you are talking about strict stationarity. Too show that $(f(X_{t_1}),f(X_{t_2}),...,f(X_{t_n}))$ has the same distribution as $(f(X_{t_1+s}),f(X_{t_2+s}),...,f(X_{t_n+s}))$ it is enough to show that $$P\{(f(X_{t_1}),f(X_{t_2}),...,f(X_{t_n}))\in A_1\times A_2\times ... \times A_n\}=\{(f(X_{t_1+s}),f(X_{t_2+s}),...,f(X_{t_n+s}))\in A_1\times A_2\times ... \times A_n\}$$ for all Borel sets $A_1,A_2,...,A_N$. This is same as $P\{(X_{t_1},X_{t_2},...,X_{t_n})\in B_1\times B_2\times ... \times B_n\}=\{(X_{t_1+s},X_{t_2+s},...,X_{t_n+s})\in B_1\times B_2\times ... \times B_n\}$ where $B_i=f^{-1}(A_i)$. This is true because $\{X_t\}$ is sattionary and $B_i$'s are Borel sets.