2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.