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ColorStatistics
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There is a more straightforward way to show thatderive the total causal effect in this example canto be computed using $P(y|do(x))=\sum\limits_{z_1,z_3}P(Z_1=z_1)P(Z_3=z_3|Z_1=z_1)P(Y=y|Z_1=z_1,Z_3=z_3,X=x)$

Because the set {$Z_1,Z_3$} satisfies the backdoor criterion, we can get the causal effect of $X$ on $Y$ by conditionalizing the conditional probability $P(Y=y|X=x)$ on the set {$Z_1,Z_3$} (in the book "Causal Inference in Statistics: A Primer" he calls this the adjustment rule): $$P(Y=y|do(X=x))=\sum\limits_{z_1,z_3}P(Y=y|X=x,Z_1=z_1,Z_3=z_3)P(Z_1=z_1,Z_3=z_3)$$ As a last step, we use the causal graph to express $P(Z_1=z_1,Z_3=z_3)$ as $P(Z_1=z_1)P(Z_3=z_3|Z_1=z_1)$ and we arrive at the desired expression:

$P(y|do(x))=\sum\limits_{z_1,z_3}P(Z_1=z_1)P(Z_3=z_3|Z_1=z_1)P(Y=y|Z_1=z_1,Z_3=z_3,X=x)$

There is a more straightforward way to show that the total causal effect in this example can be computed using $P(y|do(x))=\sum\limits_{z_1,z_3}P(Z_1=z_1)P(Z_3=z_3|Z_1=z_1)P(Y=y|Z_1=z_1,Z_3=z_3,X=x)$

Because the set {$Z_1,Z_3$} satisfies the backdoor criterion, we can get the causal effect of $X$ on $Y$ by conditionalizing the conditional probability $P(Y=y|X=x)$ on the set {$Z_1,Z_3$} (in the book "Causal Inference in Statistics: A Primer" he calls this the adjustment rule): $$P(Y=y|do(X=x))=\sum\limits_{z_1,z_3}P(Y=y|X=x,Z_1=z_1,Z_3=z_3)P(Z_1=z_1,Z_3=z_3)$$ As a last step, we use the causal graph to express $P(Z_1=z_1,Z_3=z_3)$ as $P(Z_1=z_1)P(Z_3=z_3|Z_1=z_1)$ and we arrive at the desired expression:

$P(y|do(x))=\sum\limits_{z_1,z_3}P(Z_1=z_1)P(Z_3=z_3|Z_1=z_1)P(Y=y|Z_1=z_1,Z_3=z_3,X=x)$

There is a more straightforward way to derive the total causal effect to be $P(y|do(x))=\sum\limits_{z_1,z_3}P(Z_1=z_1)P(Z_3=z_3|Z_1=z_1)P(Y=y|Z_1=z_1,Z_3=z_3,X=x)$

Because the set {$Z_1,Z_3$} satisfies the backdoor criterion, we can get the causal effect of $X$ on $Y$ by conditionalizing the conditional probability $P(Y=y|X=x)$ on the set {$Z_1,Z_3$} (in the book "Causal Inference in Statistics: A Primer" he calls this the adjustment rule): $$P(Y=y|do(X=x))=\sum\limits_{z_1,z_3}P(Y=y|X=x,Z_1=z_1,Z_3=z_3)P(Z_1=z_1,Z_3=z_3)$$ As a last step, we use the causal graph to express $P(Z_1=z_1,Z_3=z_3)$ as $P(Z_1=z_1)P(Z_3=z_3|Z_1=z_1)$ and we arrive at the desired expression:

$P(y|do(x))=\sum\limits_{z_1,z_3}P(Z_1=z_1)P(Z_3=z_3|Z_1=z_1)P(Y=y|Z_1=z_1,Z_3=z_3,X=x)$

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ColorStatistics
ColorStatistics
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There is a more straightforward way to show that the total causal effect in this example can be computed using $P(y|do(x))=\sum\limits_{z_1,z_3}P(Z_1=z_1)P(Z_3=z_3|Z_1=z_1)P(Y=y|Z_1=z_1,Z_3=z_3,X=x)$

Because the set {$Z_1,Z_3$} satisfies the backdoor criterion, we can get the causal effect of $X$ on $Y$ by conditionalizing the conditional probability $P(Y=y|X=x)$ on the set {$Z_1,Z_3$} (in the book "Causal Inference in Statistics: A Primer" he calls this the adjustment rule): $$P(Y=y|do(X=x))=\sum\limits_{z_1,z_3}P(Y=y|X=x,Z_1=z_1,Z_3=z_3)P(Z_1=z_1,Z_3=z_3)$$ As a last step, we use the causal graph to express $P(Z_1=z_1,Z_3=z_3)$ as $P(Z_1=z_1)P(Z_3=z_3|Z_1=z_1)$ and we arrive at the desired expression:

$P(y|do(x))=\sum\limits_{z_1,z_3}P(Z_1=z_1)P(Z_3=z_3|Z_1=z_1)P(Y=y|Z_1=z_1,Z_3=z_3,X=x)$

There is a more straightforward way to show that the total causal effect in this example can be computed using $P(y|do(x))=\sum\limits_{z_1,z_3}P(Z_1=z_1)P(Z_3=z_3|Z_1=z_1)P(Y=y|Z_1=z_1,Z_3=z_3,X=x)$

Because the set {$Z_1,Z_3$} satisfies the backdoor criterion, we can get the causal effect of $X$ on $Y$ by conditionalizing the conditional probability $P(Y=y|X=x)$ on the set {$Z_1,Z_3$}: $$P(Y=y|do(X=x))=\sum\limits_{z_1,z_3}P(Y=y|X=x,Z_1=z_1,Z_3=z_3)P(Z_1=z_1,Z_3=z_3)$$ As a last step, we use the causal graph to express $P(Z_1=z_1,Z_3=z_3)$ as $P(Z_1=z_1)P(Z_3=z_3|Z_1=z_1)$ and we arrive at the desired expression:

$P(y|do(x))=\sum\limits_{z_1,z_3}P(Z_1=z_1)P(Z_3=z_3|Z_1=z_1)P(Y=y|Z_1=z_1,Z_3=z_3,X=x)$

There is a more straightforward way to show that the total causal effect in this example can be computed using $P(y|do(x))=\sum\limits_{z_1,z_3}P(Z_1=z_1)P(Z_3=z_3|Z_1=z_1)P(Y=y|Z_1=z_1,Z_3=z_3,X=x)$

Because the set {$Z_1,Z_3$} satisfies the backdoor criterion, we can get the causal effect of $X$ on $Y$ by conditionalizing the conditional probability $P(Y=y|X=x)$ on the set {$Z_1,Z_3$} (in the book "Causal Inference in Statistics: A Primer" he calls this the adjustment rule): $$P(Y=y|do(X=x))=\sum\limits_{z_1,z_3}P(Y=y|X=x,Z_1=z_1,Z_3=z_3)P(Z_1=z_1,Z_3=z_3)$$ As a last step, we use the causal graph to express $P(Z_1=z_1,Z_3=z_3)$ as $P(Z_1=z_1)P(Z_3=z_3|Z_1=z_1)$ and we arrive at the desired expression:

$P(y|do(x))=\sum\limits_{z_1,z_3}P(Z_1=z_1)P(Z_3=z_3|Z_1=z_1)P(Y=y|Z_1=z_1,Z_3=z_3,X=x)$

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ColorStatistics
ColorStatistics
  • 4.8k
  • 1
  • 19
  • 54

There is a more straightforward way to show that the total causal effect in this example can be computed using $P(y|do(x))=\sum\limits_{z_1,z_3}P(Z_1=z_1)P(Z_3=z_3|Z_1=z_1)P(Y=y|Z_1=z_1,Z_3=z_3,X=x)$

Because the set {$Z_1,Z_3$} satisfies the backdoor criterion, we can get the causal effect of $X$ on $Y$ by conditionalizing the conditional probability $P(Y=y|X=x)$ on the set {$Z_1,Z_3$}: $$P(Y=y|do(X=x))=\sum\limits_{z_1,z_3}P(Y=y|X=x,Z_1=z_1,Z_3=z_3)P(Z_1=z_1,Z_3=z_3)$$ As a last step, we use the causal graph to express $P(Z_1=z_1,Z_3=z_3)$ as $P(Z_1=z_1)P(Z_3=z_3|Z_1=z_1)$ and we arrive at the desired expression:

$P(y|do(x))=\sum\limits_{z_1,z_3}P(Z_1=z_1)P(Z_3=z_3|Z_1=z_1)P(Y=y|Z_1=z_1,Z_3=z_3,X=x)$