# What's the relationship between $(X_i)$ and $(X_i - \mathbb{E}[X_i | X_{<i}])$?

Let $$X_1,\cdots,X_n$$ be $$n$$ random variables on the same probability space $$(\Omega, F, P)$$, all with expectation $$0$$. Define $$Y_i=X_i - \mathbb{E}[X_i | X_1, \cdots, X_{i-1}]$$. Is it true that for any $$X = a_1 X_1 + \cdots + a_n X_n$$, there exist $$b_1,\cdots,b_n$$ such that $$X = b_1 Y_1 + \cdots + b_n Y_n$$, and vice versa?

If it is true, is there a textbook reference that I can cite? If it is not true, is there something similar that is true? I feel that $$X_i$$ and $$Y_i$$ must have some strong connections.

EDIT: The above statement is false. For example, let $$X_2=X_1^3$$, then $$Y_1 = X_1$$ and $$Y_2 = 0$$. In terms of the relationship between $$(X_i)$$ and $$(Y_i)$$, I only know that they generate the same $$\sigma$$-algebra. If someone can point out some other relationships, for example from an information theoretical perspective, then this would be really helpful.

• Do not make open ended questions. Mar 14 at 18:21

Take $$X_1=\cdots=X_n\neq0$$. Then $$Y_i=0$$. So if $$a_1+\cdots+a_n\neq0$$ then $$X=(a_1+\cdots+a_n)X_1\neq0=b_1Y_1+\cdots+b_nY_n$$.
EDIT: as the author pointed out it is not a counterexample as we would have $$Y_1=X_1$$. I leave it here for the moment until we get a proper answer...
• I think you mean to take $a_1+\dots+a_n \ne 0$ Mar 13 at 19:27
• But in this case $Y_1=X_1$ Mar 13 at 19:40
• @AlexOrtiz yes I meant $\neq0$ thank you.