September 22, 2025

• Introductions
• Emails
• Vibe Check
• Homework 0
• Homework 1
• The GOAT's Primer Book 💯
• Bareinboim book examples
• Conditional, Bayesian Decomp, Marginalization

Resources

• Ancestors, Parents, Children, Descendants (do this!)
• Covariates in DAGs (do this!)
• Graph Editor - Old
• Graph Editor - New 🤩
• Sample Github Project for study/practice.

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If X and Y are independent

$$P(X|Y) = P(X) \space \text{and} \space P(Y|X) = P(Y)$$

The independent event we're conditioning on $X$ really doesn't matter: $P(X|\text{Apple falls from tree}) = P(X)$. We can use that to establish a case where $(X \perp Y)$ regardless of some other Random Variable $Z$.

I.e., if $(X \perp Y | Z)$, we will need to prove that:

$$\begin{align*} P(X=x, Y=y | Z=z) &= P(X=x,Y=y) \\ &= P(X)P(Y) \\ &= P(X|Z)P(Y|Z) \\ \end{align*}$$