The first is the definition of conditional expectation:
The random variable X here is a measurable function from the probability space Ω to the Euclidean space R n, and its conditional expectation is E[X|HH] (I use HH to represent the flower H). Firstly, it is a measurable function of HH, and the integral on any H is equal to the integral of X on H. The conditional expectation obtained from these two conditions is unique (in the sense of almost everywhere equality), but what is the conditional expectation defined in this way?
If HH={empty，\Omega}。
That is, HH is the smallest Borel algebra on \Omega, with only two elements, empty set and full space. E[X|HH] satisfies two conditions, one is measurable on HH, and the other is that the integral on h is equal to the integral of x on h, so look at the first one first. The measurable function on HH={empty, \Omega} is only a constant function, so the reduction to absurdity can be considered. If there are two different points in the range, first find the open set V 1 and V2 to separate these two points, then the original images of V 1 and V2 are two non-empty measurable set, which is impossible. So E[X|HH] is a constant function. Let's look at the second condition, and examine the integral on H. H has only two choices, and the integral on the empty set is meaningless, so only the integral on the whole space is left. The second condition is equivalent to that the integral of a constant in the whole space is equal to the integral of x in the whole space, so the constant function E[X|HH] is the expected E[X].
If HH = {empty, a, b, \Omega}. (b is the complement of a)
At this time, apart from the empty set and the whole space, HH also has the complement set of A and A. Then first, the measurable function on HH can be written as a* 1_A+b* 1_B, which is a linear combination of the characteristic functions on A and B, and the proof method is similar to the above. First, it can be proved that there are at most two points in the image set, and the same reduction to absurdity is used. Condition 2 examines the integral on measurable set in HH, that is, H can take A, B and the whole space, while on A and B, E[X|HH] is a constant, and H=A can get a*P(A)=\int_A X dP, that is, a=(\int_A X dP)/P(A), that is, X.
If HH is a finite set, or more generally, there is a finite measurable set H 1, ..., HN makes them disjoint, they are not in the whole space, and there is no measurable set smaller than each Hi.
At this time, the measurable function on HH can be written as a1*1_ h1+...+an *1_ HN. First of all, all simple functions on HH have this form, and it is these coefficients ai that change, so a finite-dimensional space is formed. Measurable functions can be approximated point by point with simple functions, but the closure of the finite-dimensional space is still its own, so it is proved. In the second condition, let H=Hi respectively, and we can get the probability that ai is the integral of x to Hi divided by Hi, that is, ai=(\int_Hi X dP)/P(Hi), which is similar to the well-known discrete case.
Of course, the HH discussed above also has some "limitations". For the general HH, the expression is more complicated and even impossible to write, but I hope the above discussion can help you have a certain feeling and better understanding.