Exercise 6

  • Due No Due Date
  • Points None

6.1 Linear Gaussian SCM with  an open backdoor

Consider the DAG  and SCM from slide 9 on Lecture 6.

 gaussianDAG.pngLaTeX: \begin{align*} X_1 &:= N_1 \\ X_2 &:= 7X_1 + N_2 \\ X_3 &:= 4X_2 + N_3 \\ X_4 &:= 2X_1 + 3X_3 + N_4 \\ \end{align*}

where each LaTeX: N_i is Gaussian with mean 0 and variance 1.  Note that the causal effect of LaTeX: X_2 on LaTeX: X_4 is 12. Which of the following regressions give the correct value (i.e. the estimated constant before LaTeX: X_2 becomes 12 )?

a) LaTeX: X_4 \sim X_2

b)  LaTeX: X_4 \sim X_1 + X_2

c)  LaTeX: X_4 \sim X_2 + X_3

d) LaTeX: X_4 \sim X_1 + X_2 + X_3

Hint: Use the backdoor criterion and check your answer with the code ex8_1.ipynb

 

6.2 Linear Gaussian SCM - another example

Consider the SCM and DAG from Lecture 6 below. Write code, using the ols function performing ordinary least squares, similar to previous exercise, to check the statements made on slide 19 on the lecture.

ex8_2.png

 

6.3 Parental Adjustment

Explain why the theorem on parental adjustment (see slide 28 on Lecture 6) follows from the backdoor criterion.

 

6.4 The backdoor criterion in another Linear Gaussian SCM

Solve the exercise below using the backdoor criterion, and check your results using the code 8_4.ipynb

ex8_4.png

 

6.5 The frontdoor criterion and smoking

a) Use the frontdoor criterion and the information below to analytically determine

LaTeX: P(\textrm{cancer=1} | \mathbf{do}(\textrm{Smoke} = 0)) \\ P(\textrm{cancer=1} | \mathbf{do}(\textrm{Smoke}= 1))

b) Suggest  changes in the DAG that would mean that conditions (i), (ii) or (iii) in the frontdoor condition would be violated. Also describe what these changes could correspond to in real life.

frontdoor.png

 

6.6 Finding the causal structure - Myopia example

On Lecture 6 we described how statistical independence  tests could be used to determine that the causal structure explaining the myopia data is the following (figure to the left), ie that the use of night light (NL) does not cause child myopia (CM) even if these two variables are correlated.

myopia.png unique.png

But at the end of the lecture we claimed that the 3 structures in the figure to the right could not be separated purely from statistical independence tests. Why could we in this example rule out the two other possible structures ? (Hint: What would these alternative directions of the arrows imply in real life?)

 

Solution: solu6.pdf

Code solving 6.2 : exercise6_2.ipynb

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