where P(F= Prob(Y E F). Then the conditional average of f(X) given F is (T(F),f)=/f(a)(F)(dx) //m(0y(h P(F) (r(y), f)dy P(F Note that the last equality uses the invariance property(101). If A is a possible set of values of X(i.e. X Eo(X)), with f= IA, then(106)reduces to the familiar elementary formula prob(X∈AY∈F)=Prob(X∈A,Y∈ F)/Prob(Y∈F) Example 3. 3 Consider the(finite)discrete set Q2=1, 2, 3. Let F be the standard discrete a-algebra, namely all subsets F=0, 0, (1, 23, 3, 1, 21.1, 3, 2, 31, and let P be the uniform distribution, P=(3,3,3). Let X=(2, 4, 9), i.e. X(1)=2, X(2)=4,X(3)=9, and let y=(1,1,4). Then y=0(Y)={0,9,{1,2},{3}. Using the property(101 )with F=1, 2) we find that E[(w)=3 for w E (1, 2) while F=3 gives E[X|y](u)=9forw∈{3} The random variable X has been averaged over the atoms (1, 2, 3 of ]. The random variables Y and E(XID] are constant on these atoms( measurability). The conditional expectation E[X] can be viewed as a function of the values y E 1, 4 of Y. Indeed, let 9(1)=3andg(4)=9.Then EX|y](u)=9(Y(u) We write simply EX(y) for g(y)(a slight abuse of notation) 3.1. 3 Stochastic Processes Heuristics. A stochastic process is a random function of time. e. g. Noise, Figure 3 defined on(s, F, P). For each.stXnlno is a sequence of random variables Xn Definition 3.4 A stochastic proces. X(u)={X0(u),X1(u),…} denotes a sample pathwhere p(F) = Prob(Y ∈ F). Then the conditional average of f(X) given F is hπ(F), fi = Z f(x)π(F)(dx) = Z Z F pX,Y (x, y)f(x)dydx p(F) = Z F hπ(y), fidy p(F) (106) Note that the last equality uses the invariance property (101). If A is a possible set of values of X (i.e. X ∈ σ(X)), with f = IA, then (106) reduces to the familiar elementary formula Prob(X ∈ A|Y ∈ F) = Prob(X ∈ A, Y ∈ F)/Prob(Y ∈ F). Example 3.3 Consider the (finite) discrete set Ω = {1, 2, 3}. Let F be the standard discrete σ-algebra, namely all subsets F = {∅, Ω, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}}, and let P be the uniform distribution, P = ( 1 3 , 1 3 , 1 3 ). Let X = (2, 4, 9), i.e. X(1) = 2, X(2) = 4, X(3) = 9, and let Y = (1, 1, 4). Then Y = σ(Y ) = {∅, Ω, {1, 2}, {3}}. Using the property (101) with F = {1, 2} we find that E[X|Y](ω) = 3 for ω ∈ {1, 2} while F = {3} gives E[X|Y](ω) = 9 for ω ∈ {3}. The random variable X has been averaged over the atoms {1, 2}, {3} of Y. The random variables Y and E[X|Y] are constant on these atoms (Y measurability). The conditional expectation E[X|Y] can be viewed as a function of the values y ∈ {1, 4} of Y . Indeed, let g(1) = 3 and g(4) = 9. Then E[X|Y](ω) = g(Y (ω)). We write simply E[X|Y](y) for g(y) (a slight abuse of notation). 3.1.3 Stochastic Processes Heuristics. A stochastic process is a random function of time. e.g. Noise, Figure 3. Definition 3.4 A stochastic process {Xn} ∞ n=0 is a sequence of random variables Xn defined on (Ω, F, P). For each ω ∈ Ω, X(ω) = {X0(ω), X1(ω), . . .} denotes a sample path. 25