Lemma on conditioning
Matija Vidmar
November 4, 2012
Proposition 1 (Lemma on conditioning). Let (Ω, F, P) be a probability space; Y : (Ω, F) → (S, S)
and Z : (Ω, F) → (T, T ) be two random elements; and G any sub-σ-algebra of F, such that σ(Y ) ⊂ G
and σ(Z) ⊥ G (i.e. σ(Z) independent of G). Let f be any S ⊗ T /B([−∞, ∞])-measurable mapping,
for which E[f + ◦ (Y, Z)] < ∞ (resp. E[f − ◦ (Y, Z)] < ∞). Then for any y ∈ S, f ◦ (y, Z) is
F/B([−∞, ∞])-measurable; a.s.-PY in y ∈ S, E[f + ◦ (y, Z)] < ∞ (resp. E[f − ◦ (y, Z)] < ∞); on
the PY -almost sure set A ∈ S, where it is defined, (y 7→ E[f ◦(y, Z)]) is S|A /B([−∞, ∞])-measurable
and, on a P|G -almost sure set,
E[f ◦ (Y, Z)|G] = (y 7→ E[f ◦ (y, Z)]) ◦ Y.
(1)
(Remark: If f is bounded from below or from above, then A = S.)
Proof. Fix everything but f ; let H be the set of functions f for which the statement of the theorem
holds true. It is easy to see that H contains the indicator functions of the elements of the π-system
{s × t : (s, t) ∈ S × T } generating S ⊗ T . Since the class of sets X ∈ S ⊗ T , whose indicators belong
to H is clearly a monotone class (by linearity and monotonicity), it follows by the monotone class
theorem that H contains all indicators of sets in S ⊗ T . By linearity H contains simple functions
over the latter σ-algebra. By monotonicity and approximation, it contains all nonnegative and
nonpositive S ⊗ T -measurable functions. In particular, the final remark obtains. Finally, if f is a
S ⊗ T /B([−∞, ∞])-measurable mapping, for which E[f + ◦ (Y, Z)] < ∞ (resp. E[f − ◦ (Y, Z)] < ∞),
then f + and f − of course belong to H by what we have just proven. Further, since the expectation
of the LHS in (1) is finite for f + (resp. f − ) by assumption, the same is true of the RHS, and in
particular a.s.-PY in y ∈ S, E[f + ◦ (y, Z)] < ∞ (resp. E[f − ◦ (y, Z)] < ∞). Then, by linearity (1)
obtains, and it follows that f ∈ H, which is what we wanted.
References
[1] Cinlar, Erhan: Probability and Stochastics, Springer, 2011.
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