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Suponemos, sin aseveración repetida, que las variables aleatorias y funciones de vectores aleatorios son integrables, según sea necesario.
(CE1): Definición de condición. $$e(X) = E[g(Y)|X]$$a.s. iff$$E[I_M (X) g(Y)] = E[I_M (X) e(X)]$$ para cada conjunto de Borel$$M$$ en el codominio de$$X$$.
(CE1a): Si$$P(X \in M) > 0$$, entonces$$E[I_M(X) e(X)] = E[g(Y)|X \in M] P(X \in M)$$
(CE1b): Ley de probabilidad total. $$E[g(Y)] = E\{[g(Y)|X]\}$$
(CE2): Linealidad. Para cualquier constante$$a, b$$
$$E[ag(Y) + bh(Z)|X] = aE[g(Y)|X] + bE[h(Z)|X]$$ a.s.
(Se extiende a cualquier combinación lineal finita)
a.$$g(Y) \ge 0$$ a.s. implica$$E[g(Y)|X] \ge 0$$ a.s.
b.$$g(Y) \ge h(Z)$$ a.s. implica$$E[g(Y)|X] \ge E[h(Z)|X]$$ a.s.
(CE4): Convergencia monótona. $$Y_n \to Y$$a.s. implica monótonamente$$E[Y_n |X] \to E[Y|X]$$ a.s.
(CE5): Independencia. $$\{X, Y\}$$es un par independiente
a. iff$$E[g(Y)|X] = E[g(Y)]$$ a.s. para todas las funciones de Borel$$g$$
b. iff$$E[I_N (Y)|X] = E[I_N (Y)]$$ a.s. para todos los conjuntos de Borel$$N$$ en el codominio de$$Y$$
(CE6):$$e(X) = E[g(Y)|X]$$ a.s. iff$$E[h(X)g(Y)] = E[h(X)e(X)]$$ a.s. para cualquier Función Borel$$h$$
(CE7):$$E[h(X)|X] = h(X)$$ a.s. para cualquier función Borel$$h$$
(CE8):$$E[h(X)g(Y)|X] = h(X) E[g(Y)|X]$$ a.s. para cualquier función Borel$$h$$
(CE9): Si$$X = h(W)$$ y$$W = k(X)$$, con funciones$$h, k$$ Borel, entonces$$E[g(Y)|X] = E[g(Y)|W]$$ a.s.
(CE10): Si$$g$$ es una función de Borel tal que$$E[g(t, Y)]$$ es finita para todos$$t$$ en el rango de$$X$$ y$$E[g(X, Y)]$$ es finita, entonces
a.$$E[g(X, Y)|X = t] = E[g(t, Y)|X = t]$$ a.s.$$[P_X]$$
b. Si$$\{X, Y\}$$ es independiente, entonces$$E[g(X, Y)|X = t] = E[g(t, Y)]$$ a.s.$$[P_X]$$
(CE11): Supongamos que$$\{X(t): t \in T\}$$ es un proceso aleatorio medible de valor real cuyo conjunto de parámetros$$T$$ es un subconjunto de Borel de la línea real y$$S$$ es una variable aleatoria cuyo rango es un subconjunto de$$T$$, por lo que$$X(S)$$ es una variable aleatoria.
Si$$E[X(t)]$$ es finito para all$$t$$ in$$T$$ y$$E[X(S)]$$ es finito, entonces
a.\ 9E [X (S) |S = t] = E [X (t) |S = t]\) a.s$$[P_S]$$
b. Si, además,$$\{S, X_T\}$$ es independiente, entonces$$E[X(S)|S = t] = E[X(t)]$$ a.s.$$[P_S]$$
a. si$$Y$$ es integrable en$$A$$ y$$A = \bigvee_{n = 1}^{\infty} A_n$$.
entonces$$E[I_A Y|X] = \sum_{n = 1}^{\infty} E[I_A Y|X]$$ a.s.
b. Si$$\sum_{n = 1}^{\infty} E[|Y_n|] < \infty$$, thne$$E[\sum_{n = 1}^{\infty} Y_n|X]$$ a.s.
(CE13): Desigualdad triangular. $$|E[g(Y)|X]| \le E[|g(Y)||X]$$a.s.
(CE14): La desigualdad de Jensen. Si$$g$$ es una función convexa en un intervalo$$I$$ que contiene el rango de una variable aleatoria real$$Y$$, entonces$$g\{E[Y|X]\} \le E[g(Y)|X]$$ a.s.
(CE15): Supongamos$$E[|Y|^p] < \infty$$ y$$E[|Z|^p] < \infty$$ para$$1 \le p < \infty$$. Entonces$$E\{|E[Y|X] - E[Z|X]|^p\} \le E[|Y - Z|^p] < \infty$$