Let $\left\{X_t\right\}$ be a zero mean stationary process.
Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and consider the Hilbert space $L^2=L^2(\Omega,\mathcal{F},\mathbb{P})$. It has the inner product $\langle X,Y\rangle=E(XY)$. Moreover, consider $K_1=\overline{sp}\left\{X_2,\ldots,X_n\right\}$, $K_2=\overline{sp}\left\{X_1-P_{K_1}X_1\right\}$ and $K_n=\overline{sp}\left\{X_1,\ldots,X_n\right\}$. $K_1$ and $K_2$ are orthogonal subspaces of $K_n$.
For $Y\in L^2$, let $P_{K_1}Y$, $P_{K_2}Y$ and $P_{K_n}Y$ denote the orthogonal projection mappings of $L^2$ onto the subspaces $K_1, K_2, K_n$.
Show that for any $Y\in L^2$, we have $$ P_{K_n}Y=P_{K_1}Y+P_{K_2}Y. $$
My first idea was to write down the expressions of the projections as linear combinations.
Since $P_{K_n}Y\in K_n$, we can write it as some linear combination, i.e. $$ P_{K_n}Y=\sum_{i=1}^n a_i X_i. $$ By the projection theorem, this is unique.
On the other hand, we have that $P_{K_1}X_1=\sum_{i=2}^nd_iX_i$ so that $$ P_{K_1}Y+P_{K_2}Y=\sum_{i=2}^n b_iX_i+c\cdot (X_1-P_{K_1}X_1)\\ =\sum_{i=2}^n b_iX_i+cX_1-\sum_{i=2}^n cd_iX_i\\ =\sum_{i=2}^n (b_i-cd_i)X_i + cX_1. $$
This is a linear combination of $X_1,\ldots,X_n$, too.
How do I know now, that both expression do coincide? Do not see that.
If $M$ is a closed subspace of $L^{2}$, then $m$ is the orthogonal projection of $x$ onto $M$ iff $m \in M$ and $(x-m)\perp M$. That's the definition I'll use.
You have subspaces $K_1$, $K_2$, $K_n$ such that $$ K_n=K_1\oplus K_2,\\ L^{2}=K_{n}^{\perp}\oplus K_1\oplus K_2 $$ where the sums on the right are direct and orthogonal. You want to show that $$ P_{K_n}=P_{K_1}+P_{K_2}. $$ If $x \in L^2$, then $x=k_n^{\perp}+k_1+k_2$. Then $P_{K_1}x=k_1$ because $k_1 \in K_1$ and $(x-k_1)=k_n^{\perp}+k_2$ is orthogonal to $K_1$. So, $$ P_{K_1}x=k_1,\;\;P_{K_2}x=k_2, \\ \implies P_{K_n}x=k_1+k_2=P_{K_1}x+P_{K_2}x. $$