Let $X$ be a centered continuous stochastic process which is square integrable on $[0,2]\times \Omega$ and the basis of $L^2(0,2)$ is $\{e_i\}$. By using Karhunen-Leove Theorem one can write for all $t\in[0,2]$, $$X_t=\sum_{i\geq 1}\left(\int_0^2X_te_i(t)dt\right) e_i(t), in\; L^2(\Omega).$$

Let us define another stochastic process $Y$ on $[0,1]\times \Omega'$, which is defined as $$Y_t=X_t+X_{t+1},\; \; \forall t\in [0,1].$$

Then using the above expansion of $X$, we can write for all $t\in[0,1], $$$Y_t=\sum_{i\geq 1}\left(\int_0^2X_te_i(t)dt\right) \left(e_i(t)+e_i(t+1)\right).$$

I am trying addres the following issues:

Firstly: Whether $\left\{e_i(t)+e_i(t+1)\right\}$ is a basis for $L^2[0,1]$ and

Secondly: Is the last expression gives the Karhunen-Leove expansion of the process $Y$ in $[0,1]$?

It may be true that if the answer to the first question is yes, then the answer to the second also be yes. But I cann't satisfy myself for the first question.

Could anyone guide me: how to check whether $\left\{e_i(t)+e_i(t+1)\right\}$ is a basis for $L^2[0,1] or not. Thanks.

Let $X$ be a centered continuous stochastic process which is square integrable on $[0,2]\times \Omega$ and the basis of $L^2(0,2)$ is $\{e_i\}$. By using Karhunen-Leove Theorem one can write for all $t\in[0,2]$, $$X_t=\sum_{i\geq 1}\left(\int_0^2X_te_i(t)dt\right) e_i(t), in\; L^2(\Omega).$$

Let us define another stochastic process $Y$ on $[0,1]\times \Omega'$, which is defined as $$Y_t=X_t+X_{t+1},\; \; \forall t\in [0,1].$$

Then using the above expansion of $X$, we can easily write for all $t\in[0,1], $$$Y_t=\sum_{i\geq 1}\left(\int_0^2X_te_i(t)dt\right) \left(e_i(t)+e_i(t+1)\right).$$

This expression have a similar type of expression as that of Karhunen-Leove expansion of $Y$. But to show that, this is actually K-L expansion, we need to show that $\left\{e_i(t)+e_i(t+1)\right\}$ is a basis for $L^2[0,1]$.

Here is I am getting stuck: Could anyone please guide me: how to check whether $\left\{e_i(t)+e_i(t+1)\right\}$ is a basis for $L^2[0,1] or not. Thanks.