Bandits.ArrayModel.indepFun_cond_comp
indepFun_cond_comp🔗
Lemma
Bandits.ArrayModel.indepFun_cond_compNo docstring.
theorem
Bandits.ArrayModel.indepFun_cond_comp.{u_3, u_4, u_5, u_6} {𝓐 : Type u_3} {β : Type u_4} {γ : Type u_5} {δ : Type u_6} {m𝓐 : MeasurableSpace 𝓐} {mβ : MeasurableSpace β} {mγ : MeasurableSpace γ} {mδ : MeasurableSpace δ} [MeasurableSingletonClass δ] {μ : MeasureTheory.Measure 𝓐} {X : 𝓐 → β} {Y : 𝓐 → γ} (hXY : ProbabilityTheory.IndepFun X Y μ) (hY : Measurable Y) {Z : γ → δ} (hZ : Measurable Z) (z : δ) : ProbabilityTheory.IndepFun X Y μ[|Z ∘ Y ⁻¹' {z}]Bandits.ArrayModel.indepFun_cond_comp.{u_3, u_4, u_5, u_6} {𝓐 : Type u_3} {β : Type u_4} {γ : Type u_5} {δ : Type u_6} {m𝓐 : MeasurableSpace 𝓐} {mβ : MeasurableSpace β} {mγ : MeasurableSpace γ} {mδ : MeasurableSpace δ} [MeasurableSingletonClass δ] {μ : MeasureTheory.Measure 𝓐} {X : 𝓐 → β} {Y : 𝓐 → γ} (hXY : ProbabilityTheory.IndepFun X Y μ) (hY : Measurable Y) {Z : γ → δ} (hZ : Measurable Z) (z : δ) : ProbabilityTheory.IndepFun X Y μ[|Z ∘ Y ⁻¹' {z}]
Code
lemma indepFun_cond_comp {𝓐 β γ δ : Type*} {m𝓐 : MeasurableSpace 𝓐} {mβ : MeasurableSpace β}
{mγ : MeasurableSpace γ} {mδ : MeasurableSpace δ} [MeasurableSingletonClass δ] {μ : Measure 𝓐}
{X : 𝓐 → β} {Y : 𝓐 → γ} (hXY : X ⟂ᵢ[μ] Y) (hY : Measurable Y)
{Z : γ → δ} (hZ : Measurable Z) (z : δ) :
X ⟂ᵢ[μ[|(Z ∘ Y) ⁻¹' {z}]] YBody uses (1)
Used by (1)
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Proof
by
have h_preim : (Z ∘ Y) ⁻¹' {z} = Y ⁻¹' (Z ⁻¹' {z}) := by grind
simp_rw [h_preim]
exact indepFun_cond_of_indepFun hXY hY (hZ (measurableSet_singleton z))