Ilya Shpitser
CS262Z
Final exam.

Problem 1.

Assume, without loss of generality, that we are dealing with a zero-mean
linear model.

Since we are dealing with linear models, most general function for Y would be:

(i) Y = CX + I + U

where C is the Wright-rule sum of the path costs originating from X and
ending in Y, I represents the terms containing other variables in the model,
and U representing the unmeasurable variable terms.
(Another way to view C is as dE(Y|do(x))/dx, see p.161)

Note that there is always a way to bring the function determining Y into
the form (i) by merely recursively substituting the rhs functions for
all variables which have X as an ancestor, and grouping all X terms together.

If Y=y' and X=x' are observed, we have:

E(Y|Y=y',X=x') = y' = Cx' + E(U|Y=y',X=x') (since we are dealing with
a zero-mean model, all observable variable terms not related to actual
observations vanish).

Thus,

(ii) E(U|Y=y',X=x') = y' - Cx'

Since no U is a child of X, E(U|do(x)) = E(U) = 0, by assumption.
Since do(x) substitutes x for the function determining X, we have:

(iii) E(Y|do(x)) = Cx + E(U|do(x)) = Cx

To answer our counterfactual query, we must first perform abduction on U,
given the evidence:

E(Y |Y=y',X=x') = Cx + E(U|Y=y',X=x') = Cx + y' - Cx' (by (ii))
   x

This simplifies to:

E(Y |Y=y',X=x') = y' + E(Y|do(x)) - E(Y|do(x')) (by (iii)),
   x

which is our result.

Problem 2.

Example 1.

All variables in this model are assumed to be binary.

A: Fred has an accident.
H: Fred is at the hospital.
L: Larry lurks outside Fred's workplace.
C: Larry cuts off Fred's finger.
F: Condition of Fred's finger.

The functions are:
L = 0
A = 1
H = A = 1
C = L AND NOT H = 0
F = NOT C AND (NOT A OR H) = 1

The causal structure is therefore:

L	A--.
|	|  |
V	V  |
C<------H  |
|	|  |
|	V  |
`------>F<-'

(a)  The events in question are L=0, F=1.

Both are true in the story, which satisfies (AC1).
Let W = {A}, w = {0} (the contingency is that the factory machines were turned
off).

Then F       = 1, while F       = 0
      A=0,L=0            A=0,L=1

The choices of Z are {}, {C}, {H}, {C,H}

If Z is {}, F       = 1
             A=0,L=0
If Z is {C}, F           = 1
              A=0,L=0,C=0
If Z is {H}, F           = 1
              A=0,L=0,H=1
If Z is {C,H}, F               = 1
                A=0,L=0,H=1,C=0
Thus, (AC2) is satisfied.
If W = {}, F   = 1, and F   = 1, thus W is minimal, and (AC3) is satisfied.
            L=0          L=1
Thus, L=0 is the actual cause of F=1.
This makes sense, since on most days, Larry's failure to show up would
have spared Fred's finger.

(b)  The events in question are A=1, F=1.

Both are true in the story, which satisfies (AC1).
Let W = {L}, w = {1} (the contingency is that the loan shark is made to show up)

Then F       = 1, while F       = 0
      L=1,A=1            L=1,A=0

The choices of Z are {}, {C}, {H}, {C,H}

If Z is {}, F       = 1
             L=1,A=1
If Z is {C}, F           = 1
              L=1,A=1,C=0
If Z is {H}, F           = 1
              L=1,A=1,H=1
If Z is {C,H}, F               = 1
                L=1,A=1,C=0,H=1

Thus, (AC2) is satisfied.
If W = {}, F   = 1, and F   = 1, thus W is minimal, and (AC3) is satisfied.
            A=1          A=0
Thus, A=1 is the actual cause of F=1.
This also makes sense, since the accident produced the disturbance which
caused Fred to miss running into the loan shark.

(c) The events in question are A=1, H=1, F=1.

All (minimal) actual causes are singleton variables, in this case A=1.
Thus, the answer is no, since AC3 fails.

Example 2.

All variables in this model are assumed to be binary.

B: there is a bomb under Suzie's chair
E: Suzie's chair explodes
F: Suzie notices the bomb and flees
D: Suzie dies
H: Suzie is pronounced healthy

The functions are:
B = 1
E = B = 1
F = B = 1
D = E AND NOT F = 0
H = NOT D = 1

B-------.
|	|
V	|
E	|
|	|
V	V
D<------F
|
V
H

(a)  The events in question are E=1, H=1.

Both events are true in the story, which satisfies (AC1)

The choices for W are {}, {B}, {F}, {D}, {B,F}, {D,F}, {B,D}, {B,D,F}.

Since D completely determines H, we can exclude any set W which includes D.

If W is {}, H   = 1, and H   = 1
             E=1          E=0
If W is {B}, H       = 0, while H       = 1
              E=1,B=0            E=0,B=0
which doesn't work since the values are flipped.
While H       = 1, and H       = 1, which also doesn't work.
       E=1,B=1          E=0,B=1
If W is {F}, H       = 0, while H       = 1
              E=1,F=0            E=0,F=0
which again doesn't work since the values are flipped.
While H       = 1, and H       = 1, which also doesn't work.
       E=1,F=1          E=0,F=1
If W is {B,F}, we construct a table for various values of the intervention:

		H			H
do(B)	do(F)    E=1,do(B),do(F)         E=0,do(B),do(F)
0	0	0			1
0	1	0			1
1	0	0			1
1	1	1			1

Since, going from forcing an explosion to forcing no explosion can never
make Suzie go from healthy to not healthy, this W does not work also.

Since there is no set W which satisfies (AC2), the chair exploding is not
the actual cause of Suzie being pronounced healthy.

(b) The events in question are F=1, H=1.

Both events are true in the story, which satisfies (AC1).

Let W = {}.

Then H   = 1, while H   = 0
      F=1            F=0

The possible choices of Z are {}, {B}, {E}, {D}, {B,E}, {B,D}, {D,E}, {B,D,E}.

Since D(u) = 0, and D completely determines H = 1, the condition holds for
any set Z which includes D.

If Z is {}, H   = 1
             F=1
If Z is {B}, H       = 1
              F=1,B=1
If Z is {E}, H       = 1
              F=1,E=1
if Z is {B, E), H           = 1
                 F=1,B=1,E=1
Thus, (AC2) is satisfied.
Since W is the empty set, (AC3) is also trivially satisfied.

Thus, F=1 is the actual cause of H=1.

(c)  The events in question are B=1, H=1.

Both events are true in the story, which satisfies (AC1).

Let W = {E}, w = {1}, (the contingency is that the chair exploded
because it had nitroglycerin in the lining)

Then H       = 1, while H       = 0
      E=1,B=1            E=1,B=0

The possible choices of Z are {}, {F}, {D}, {F,D}.

Since D(u) = 0, and D completely determines H = 1, the condition holds for
any set Z which includes D.

If Z is {}, H       = 1
             E=1,B=1
If Z is {F}, H           = 1
              E=1,B=1,F=1
Thus, (AC2) is satisfied.
If W = {}, H   = 1, and H   = 1, thus W is minimal, and (AC3) is satisfied.
            B=1          B=0

Thus, B=1 is the actual cause of H=1.