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I am reading about the front door criterion and I want to make sure that I understand what is calculated in the second part of the formula, where the backdoor path between $Z$ and $Y$ is blocked by conditioning on $X.$

Graph:

Causal graph

Formula:

$$P(y|\operatorname{do}(X=x))=\sum_z P(z|x)\sum_{x'}P(y|x',z)\,P(x')$$

Using the graph and the formula above and assuming that all variables except $U$ are binary and $U$ is unobserved, if I am interested in the effect of $\operatorname{do}(X = 1)$ where $x$ set to $1$ on $y = 1,$ is the calculation below correct?:

  1. Calculate the sum of $P(z = 1 | x = 1)$ and $P(z = 0 | x = 1).$
  2. Calculate the sum of $P(y = 1 | z = 1, x = 1)$ and $P(y = 1 | z = 1, x = 0)$
  3. Multiply [result of step 1] * [result of step 2].

Is this right?

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  • $\begingroup$ I don't see any conditional probabilities in your calculations, but every probability in the front-door adjustment formula is conditional. That leads me to doubt seriously whether your calculations are correct. $\endgroup$
    – Adrian Keister
    Nov 25, 2020 at 16:27
  • $\begingroup$ I edited it Adrian, thank you. What I would like to understand is, if in the second part of forula, z is always set to 1 and we sum over the whole range of x ? $\endgroup$
    – Milo
    Nov 25, 2020 at 17:10
  • $\begingroup$ Please use the standard vertical pipe notation for conditionals: $P(z=1|x=1).$ $\endgroup$
    – Adrian Keister
    Nov 25, 2020 at 17:11

1 Answer 1

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It's not quite as simple as that: there is a $z$ in the inner sum so that the two sums are nested, not independent. Also, you need to be very careful about the order of operations. You really need to calculate the following: \begin{align*} \underbrace{P(y|\operatorname{do}(X=x))}_{y=1,\,x=1} &=\sum_z P(z|x)\sum_{x'}P(y|x',z)\,P(x')\\ &=P(z=0|x=1)\left[\sum_{x'}P(y=1|x',z=0)\,P(x')\right]\\ &\qquad+P(z=1|x=1)\left[\sum_{x'}P(y=1|x',z=1)\,P(x')\right] \end{align*}

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  • $\begingroup$ Thank you, Adrian. This is very helpful. I would like to ask for the final clarification of two things. First in every sqare bracket we will summarise probabilities over x' where x' includes x = 0 and x = 1. Second question is how do you know that the sums are not independent but nested? You expanded the expession with square brackets, so it is clear, but original expression does not have the square brackets. Is this because the z is present in the second part as well as in the first one? $\endgroup$
    – Milo
    Nov 25, 2020 at 17:39
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    $\begingroup$ Yes, that would be correct. You're very welcome! $\endgroup$
    – Adrian Keister
    Nov 25, 2020 at 17:40
  • $\begingroup$ Hi , I will come backwith the second question: how do you know that the sums are not independent but nested? You expanded the expession with square brackets, so it is clear, but original expression does not have the square brackets. Is this because the z is present in the second part as well as in the first one? $\endgroup$
    – Milo
    Nov 25, 2020 at 18:08
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    $\begingroup$ Yes, that's correct. According to the usual rules of writing sums, and which variables are bound and which are not, the $z$ in the second sum shows you that it has to be a nested sum, and not independent. $\endgroup$
    – Adrian Keister
    Nov 25, 2020 at 18:30

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