# Sum of two parts of a continuous stochastic process

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.

Consider the Fourier basis of $L^2(0,2)$: $a_n(t)=cos(n\pi t/2)$ ($n\in \mathbb N$) and $b_n(t)=sin(n\pi t/2)$ ($n\in \mathbb N, n\geq 1$).
If you compute $a_n(t)+a_n(t+1)$ and $b_n(t)+b_n(t+1)$, depending on whether $n$ is even or odd, some of them vanish. So no, they're not a basis.