Apêndice 6.B Seção 2.5 (Modelo Causal (Causal Model))

Prova do Lema 2.27.

Realizaremos a demonstração por indução. Para tal, defina 𝒱(1)={V∈𝒱:P⁢a⁢(V)=∅︀}{\mathcal{V}}^{(1)}=\{V\in{\mathcal{V}}:Pa(V)=\emptyset\}, e para cada i>2i>2, 𝒱(i)={V∈𝒱:P⁢a⁢(V)⊆𝒱(i−1)}{\mathcal{V}}^{(i)}=\{V\in{\mathcal{V}}:Pa(V)\subseteq{\mathcal{V}}^{(i-1)}\}.

Se Y∈𝒱(1)Y\in{\mathcal{V}}^{(1)}, então por construção 𝔼⁢[Y]=μY{\mathbb{E}}[Y]=\mu_{Y}. Também, como P⁢a⁢(Y)=∅︀Pa(Y)=\emptyset, ℂV,Y=∅︀\mathbb{C}_{V,Y}=\emptyset, para todo V≠YV\neq Y, e ℂY,Y\mathbb{C}_{Y,Y} tem apenas o caminho unitário, C∗=(Y)C^{*}=(Y). Assim,

∑V∈𝒱∑C∈ℂV,YμV⋅∏i=1|C|−1βCi+1,Ci\displaystyle\sum_{V\in\mathbb{{\mathcal{V}}}}\sum_{C\in\mathbb{C}_{V,Y}}\mu_{% V}\cdot\prod_{i=1}^{|C|-1}\beta_{C_{i+1},C_{i}} =μY⋅∏i=1|C∗|−1βCi+1∗,Ci∗\displaystyle=\mu_{Y}\cdot\prod_{i=1}^{|C^{*}|-1}\beta_{C^{*}_{i+1},C^{*}_{i}}
=μY\displaystyle=\mu_{Y}

A seguir, suponha que para todo W∈𝒱(i−1)W\in{\mathcal{V}}^{(i-1)}, 𝔼⁢[W]=∑V∈𝒱∑C∈ℂV,WμV⋅∏i=1|C|−1βCi+1,Ci{\mathbb{E}}[W]=\sum_{V\in\mathbb{{\mathcal{V}}}}\sum_{C\in\mathbb{C}_{V,W}}% \mu_{V}\cdot\prod_{i=1}^{|C|-1}\beta_{C_{i+1},C_{i}} e tome Y∈𝒱(i)Y\in{\mathcal{V}}^{(i)}. Como todo caminho direcionado que chega em YY a partir de V≠YV\neq Y tem como penúltimo elemento um pai de YY, podemos escrever

ℂV,Y\displaystyle\mathbb{C}_{V,Y} =∪W∈P⁢a⁢(Y){(C,Y):C∈ℂV,W}\displaystyle=\cup_{W\in Pa(Y)}\{(C,Y):C\in\mathbb{C}_{V,W}\} (7)

Portanto, obtemos:

∑V∈𝒱∑C∈ℂV,YμV⋅∏i=1|C|−1βCi+1,Ci\displaystyle\sum_{V\in\mathbb{{\mathcal{V}}}}\sum_{C\in\mathbb{C}_{V,Y}}\mu_{% V}\cdot\prod_{i=1}^{|C|-1}\beta_{C_{i+1},C_{i}}
=\displaystyle= μY+∑V≠Y∑∪W∈P⁢a⁢(Y){C∗=(C,Y):C∈ℂV,W}μV⋅(∏i=1|C∗|−1βCi+1∗,Ci∗)\displaystyle\mu_{Y}+\sum_{V\neq Y}\sum_{\cup_{W\in Pa(Y)}\{C^{*}=(C,Y):C\in% \mathbb{C}_{V,W}\}}\mu_{V}\cdot\left(\prod_{i=1}^{|C^{*}|-1}\beta_{C^{*}_{i+1}% ,C^{*}_{i}}\right)
=\displaystyle= μY+∑V≠Y∑W∈P⁢a⁢(Y)∑C∈ℂV,WμV⋅(∏i=1|C|−1βCi+1,Ci)⋅βY,W\displaystyle\mu_{Y}+\sum_{V\neq Y}\sum_{W\in Pa(Y)}\sum_{C\in\mathbb{C}_{V,W}% }\mu_{V}\cdot\left(\prod_{i=1}^{|C|-1}\beta_{C_{i+1},C_{i}}\right)\cdot\beta_{% Y,W}
=\displaystyle= μY+∑W∈P⁢a⁢(Y)βY,W⁢∑V∈𝒱∑C∈ℂV,WμV⋅(∏i=1|C|−1βCi+1,Ci)\displaystyle\mu_{Y}+\sum_{W\in Pa(Y)}\beta_{Y,W}\sum_{V\in{\mathcal{V}}}\sum_% {C\in\mathbb{C}_{V,W}}\mu_{V}\cdot\left(\prod_{i=1}^{|C|-1}\beta_{C_{i+1},C_{i% }}\right)
=\displaystyle= μY+∑W∈P⁢a⁢(Y)βY,W⁢𝔼⁢[W]\displaystyle\mu_{Y}+\sum_{W\in Pa(Y)}\beta_{Y,W}{\mathbb{E}}[W] W∈𝒱(i−1)\displaystyle W\in{\mathcal{V}}^{(i-1)}
=\displaystyle= 𝔼⁢[μY+∑W∈P⁢a⁢(Y)βY,W⁢W]\displaystyle{\mathbb{E}}\left[\mu_{Y}+\sum_{W\in Pa(Y)}\beta_{Y,W}W\right]
=\displaystyle= 𝔼⁢[𝔼⁢[Y|P⁢a⁢(Y)]]=𝔼⁢[Y]\displaystyle{\mathbb{E}}[{\mathbb{E}}[Y|Pa(Y)]]={\mathbb{E}}[Y]

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