Conditional independence of A and C given D if A and (B, C) are conditionally independent given D

1,799

$\frac{P(ABCD)}{P(D)}=\frac{P(AD)}{P(D)}\frac{P(BCD)}{P(D)}$ (by definition of conditional probability) $\Rightarrow P(ABCD)P(D)=P(AD)P(BCD)$ $\Rightarrow \frac{P(ABCD)}{P(BCD)}=\frac{P(AD)}{P(D)}$ $\Rightarrow P(A|BCD)=P(A|D)=P(A|BD)$, since $A$ and $B$ are independent in the conditional universe given $D$, we have $P(A|BD) = P(A|D)$

Share:
1,799

Related videos on Youtube

Dorik
Author by

Dorik

Just a student

Updated on September 21, 2020

Comments

  • Dorik
    Dorik about 3 years

    Suppose that $A$ is conditionally independent of $(B,C)$ given $D$ (that is $P(A, B, C | D) = P(A|D)P(B,C|D))$.

    I want to show the following is true:

    $P(A | B, C, D) = P(A | B, D)$

    I am not sure how to show this. I have tried beginning with transformations like the product rule and Bayes' Rule, but I am not sure how I can remove $C$ from the equations.

  • Dorik
    Dorik about 7 years
    How does $P(A,B,C,D)P(D) = P(A,D)P(B,C,D)$ follow from $\frac{A,B,C,D}{P(D)} = \frac{A}{P(D)}\frac{B,C,D}{P(D)}$ ?
  • Sandipan Dey
    Sandipan Dey about 7 years
    Assuming $P(D) \neq 0$ just multiply both sides by $(P(D))^2$
  • Dorik
    Dorik about 7 years
    Shouldn't there be a $P(D)$ on the right hand side then? That is, shouldn't it be $P(AD)P(BCD)P(D)$?
  • Sandipan Dey
    Sandipan Dey about 7 years
    $\frac{P(ABCD)}{P(D)}=\frac{P(AD)}{P(D)}\frac{P(BCD)}{P(D)}$ $\Rightarrow \frac{P(ABCD)}{P(D)}.P(D).P(D)=\frac{P(AD)}{P(D)}\frac{P(BCD)}{P(D)}.P(D).P(D)$ $\Rightarrow P(ABCD).P(D)=P(AD)P(BCD)$
  • Dorik
    Dorik about 7 years
    Ah, I see now. I am very foolish. I kept thinking the common denominator could be combined, but that only applies in additive operations.