Consider a stochastic process $X_{t}$ defined by $X_{t}:=\cos(t+\phi)$, where $\phi\sim U[0,2\pi]$ the uniform distribution on $[0,2\pi]$.
Then, I want to show $\mathbb{P}(X_{t+h}\leq x)=\mathbb{P}(X_{t}\leq x).$ To do so, I took the expectation as follows: $$\mathbb{P}(X_{t+h}\leq x)=\mathbb{P}(\cos(t+h+\phi)\leq x)=\int_{0}^{2\pi}\dfrac{1}{2\pi}\mathbb{1}_{\{\cos(t+h+y)\leq x\}}\phi(dy),$$ where $\phi(dy)$ means we are integrating with respect to uniform measure.
Now, I made a change of variable, $\tilde{y}:=h+y$, so that $$\int_{0}^{2\pi}\dfrac{1}{2\pi}\mathbb{1}_{\{\cos(t+h+y)\leq x\}}\phi(dy)=\int_{h}^{2\pi+h}\dfrac{1}{2\pi}\mathbb{1}_{\{\cos(t+\tilde{y})\leq x\}}\phi(d\tilde{y}).$$
My question now is that is the following identity true, and if so, how could I prove it? $$\int_{h}^{2\pi+h}\dfrac{1}{2\pi}\mathbb{1}_{\{\cos(t+\tilde{y})\leq x\}}\phi(d\tilde{y})=\int_{0}^{2\pi}\dfrac{1}{2\pi}\mathbb{1}_{\{\cos(t+\tilde{y})\leq x\}}\phi(d\tilde{y}).$$
If this identity is true, is the following proof correct?
\begin{align*}
\mathbb{P}(X_{t+h}\leq x)&=\mathbb{P}(\cos(t+h+\phi)\leq x)\\
&=\int_{0}^{2\pi}\dfrac{1}{2\pi}\mathbb{1}_{\{\cos(t+h+y)\leq x\}}\phi(dy)\\
&=\int_{h}^{2\pi+h}\dfrac{1}{2\pi}\mathbb{1}_{\{\cos(t+\tilde{y})\leq x\}}\phi(d\tilde{y})\\
&=\int_{0}^{2\pi}\dfrac{1}{2\pi}\mathbb{1}_{\{\cos(t+\tilde{y})\leq x\}}\phi(d\tilde{y})\\
&=\mathbb{P}(\cos(t+\phi)\leq x)\\
&=\mathbb{P}(X_{t}\leq x).
\end{align*}
Thank you!
Best Answer
Okay, I get it. Thanks to the Herb Steinberg's comment:
In fact, I need to extend the result to finite dimensional distributions, that is, to show $$\mathbb{P}(X_{t_{1}+h}\leq x_{1},\cdots, X_{t_{n}+h}\leq x_{n})=\mathbb{P}(X_{t_{1}}\leq x_{1},\cdots, X_{t_{n}}\leq x_{n})$$ and thus only arguing the first dimension may not be sufficient.
Here is an alternative argument, \begin{align*} \mathbb{P}(X_{t_{1}+h}\leq x_{1},\cdots, X_{t_{n}+h}\leq x_{n})&=\mathbb{P}(\cos(t_{1}+h+\phi)\leq x_{1},\cdots, \cos(t_{n}+h+\phi)\leq x_{n})\\ &=\int_{0}^{2\pi}\dfrac{1}{2\pi}\mathbb{1}_{\cos(t_{1}+h+y)\leq x_{1}}\cdots\mathbb{1}_{\cos(t_{n}+h+y)\leq x_{n}}\phi(dy)\\ &=\int_{h}^{2\pi+h}\dfrac{1}{2\pi}\mathbb{1}_{\cos(t_{1}+\tilde{y})\leq x_{1}}\cdots\mathbb{1}_{\cos(t_{n}+\tilde{y})\leq x_{n}}\phi(d\tilde{y})\\ &=\int_{0}^{2\pi}\dfrac{1}{2\pi}\mathbb{1}_{\cos(t_{1}+\tilde{y})\leq x_{1}}\cdots\mathbb{1}_{\cos(t_{n}+\tilde{y})\leq x_{n}}\phi(d\tilde{y})\\ &=\mathbb{P}(X_{t_{1}}\leq x_{1},\cdots, X_{t_{n}}\leq x_{n}), \end{align*} where the third equality was obtained by a change of variable $\tilde{y}:=y+h$, and the fourth equality is obtained by the observation that the integrand is a $2\pi-$periodic function (since $\cos$ is ) and as such it has the same value of the integral over any interval of length $2\pi.$