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I see this page Ordinary least square and random projection, and I am thinking that how $L^2$ integrable random variables be regarded as projections over a defined filtration sequence $\mathcal{F_n}$ ?

It seems that a vector space of square integrable random variables defined on the same probability space is not enough? Is there existing literature addressing a formalism of this projection view? Thank you!

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No. Vector space structure is not enough, we actually need a compatible lattice structure to make things work. To apply the conditional expectation operator $E(\bullet\mid Y)$ onto the Hilbert space consisting of $L^2(\mathcal{X})$ random variables defined on the probability space $\mathcal{X}$, it requires us to consider following construction. A probability Hilbert space $H$(of $L^2(\mathcal{X})$ with $<x,y>=E(x-Ex)(y-Ey)$) if it satisfies following conditions

(1) it has unitary element $\boldsymbol{1}_{\mathcal{X}}$.

(2) it adopts a lattice structure compatible with the vector space structure $(H,\ge)$ such that $x\vee\boldsymbol{1}=\boldsymbol{0}$ iff $x=\boldsymbol{0}$

(3) For $x\ge \boldsymbol{0}$ and $y\ge \boldsymbol{0}$ $<x,y>\geq 0$. equality holds iff $x\wedge y=\boldsymbol{0}$

A closed subspace $G$ of a probability Hilbert subspace $H$ is a closed probability space such that it contains $\boldsymbol{1}_G$ and closed under $\vee \wedge$ operations of the lattice. The probability span of $y\in H$ is defined to be the intersection of all probability Hilbert subspaces that contains ${y}$(which is like the way we define ideals/spectrum). Then $E(\bullet\mid y)$ is a projection operator onto the probability span of $y$.

[Small&McLeish] contains more details of construction of probability spans.

[Small&McLeish]Small, Christopher G., and Don L. McLeish. Hilbert space methods in probability and statistical inference. Vol. 920. John Wiley & Sons, 2011.

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  • $\begingroup$ What is a lattice that is compatible with $H$? $\endgroup$
    – lsn
    Commented Aug 4, 2017 at 2:22
  • $\begingroup$ @lsn_rice That lattice is distributive under scalar and vector addition. $\endgroup$
    – Henry.L
    Commented Aug 4, 2017 at 2:24
  • $\begingroup$ What if x,y are not L^2? This still works? $\endgroup$
    – lsn
    Commented Aug 4, 2017 at 2:25
  • $\begingroup$ @lsn_rice No, the Hilbert structure is lost and we need more consideration. $\endgroup$
    – Henry.L
    Commented Aug 4, 2017 at 2:25

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