SlowGuess/NeurIPS_2023_OCR huggingface dataset:
where we dropped the superscript  $M$  for better readability. Now define  $q = \sum_{\mathbf{w} \in \mathcal{D}_{\mathbf{W}}^{s}} P^{M}(\mathbf{w}) = 1 - \sum_{\mathbf{w} \in \mathcal{D}_{\mathbf{W}}^{u}} P(\mathbf{w})$ , where we know  $q \geq 0.6$  since  $\mathbf{w}^{*} \in \mathcal{D}_{\mathbf{W}}^{s}$ . From the equation above we have
$$
P ^ {M} (Y = 1 \mid d o (\pi)) = \frac {1}{2} q + \pi (X = 0) (1 - q), \tag {13}
$$
where  $0.6 \leq q \leq 1$  and  $0 \leq \pi(X = 0) \leq 1$ . We claim that  $P(Y = 1|do(\pi)) < 0.7$ . Consider two cases: if  $\pi(X = 0) \leq \frac{1}{2}$ , then from Equation (13),
$$
P (Y = 1 | d o (\pi)) \leq \frac {1}{2} q + \frac {1}{2} (1 - q) = 0. 5 <   0. 7.
$$
Otherwise, assume  $\pi(X = 0) > \frac{1}{2}$ . Rewriting Equation (13),
$$
\begin{array}{l} P (Y = 1 | d o (\pi)) = q \left(\frac {1}{2} - \pi (X = 0)\right) + \pi (X = 0) <   0. 6 \left(\frac {1}{2} - \pi (X = 0)\right) + \pi (X = 0) \\ = 0. 3 + 0. 4 \pi (X = 0) \leq 0. 7. \\ \end{array}
$$
We showed that for any arbitrary policy  $\pi \in \Pi$
$$
P ^ {M} (Y = 1 \mid d o (\pi)) <   0. 7. \tag {14}
$$
Comparing this to Equaiton (12) completes the proof.
If part. Let  $I_X$  be the intervention vertex corresponding to  $X$ , i.e.,  $I_X = 0$  and  $I_X = 1$  indicate that  $X$  is passively observed (determined by its parents) and actively intervened upon (independent of its parents), respectively (refer to Figure 5). Assume that the 3-SAT instance is unsatisfiable. We claim that  $\pi(x) = P(x)$  is an imitating policy for  $P(y)$ . Since the 3-SAT is unsatisfiable, for any context  $\mathbf{w} \in \mathcal{D}_{\mathbf{W}}$ , there exists  $1 \leq i \leq m$  such that the edge  $S_{i-1} \to S_i$  is absent. Then, we have  $I_X \perp Y|X$ ,  $\mathbf{W} = \mathbf{w}$  in  $\mathcal{G}$  for every  $\mathbf{w} \in \mathcal{D}_{\mathbf{W}}$ , then  $I_X \perp Y|X$ ,  $\mathbf{W}$  in  $\mathcal{G}$ . Moreover, since the d-separation  $I_X \perp_{\mathcal{G}} \mathbf{W}$  holds, by contraction, we have  $I_X \perp Y$ ,  $\mathbf{W}|X$  in  $\mathcal{G}$ . As a result,  $I_X \perp Y|X$  in  $\mathcal{G}$ . Therefore,
$$
\begin{array}{l} P (y | d o (\pi)) = \sum_ {x \in \mathcal {D} _ {X}} P (y | d o (x)) \pi (x) = \sum_ {x \in \mathcal {D} _ {X}} P (y | x, I _ {X} = 1) P (x) = \sum_ {x \in \mathcal {D} _ {X}} P (y | x, I _ {X} = 0) P (x) \\ = \sum_ {x \in \mathcal {D} _ {X}} P (y | x) P (x) = \sum_ {x \in \mathcal {D} _ {X}} P (y | x) P (x) = P (y). \\ \end{array}
$$
Proposition 3.9. Given an  $LDAG\mathcal{G}^{\mathcal{L}}$  where  $\mathbf{Pa}(\mathbf{C}(\mathcal{L}))\subseteq \mathbf{C}(\mathcal{L})$  and a policy space  $\Pi$ , let  $Y$  be an arbitrary variable in  $\mathcal{G}^{\mathcal{L}}$ .  $P(y)$  is imitable w.r.t.  $\langle \mathcal{G}^{\mathcal{L}},\Pi\rangle$  if and only if  $P(y)$  is imitable w.r.t.  $\langle \mathcal{G}_{\mathbf{c}},\Pi\rangle$ , for every  $\mathbf{c}\in \mathcal{D}_{\mathbf{C}(\mathcal{L})}$ .
Proof. It suffices to show that for an arbitrary  $M \in \mathcal{M}_{\langle \mathcal{G}^{\mathcal{L}} \rangle}$ , there exists a policy  $\pi \in \Pi$  uniquely computable from  $P(\mathbf{O})$  such that  $P^{M}(y|do(\pi)) = P^{M}(y)$ . For ease of notation, let  $\mathbf{C} \coloneqq \mathbf{C}(\mathcal{L})$ . For every  $\mathbf{c} \in \mathcal{D}_{\mathbf{C}}$ , construct an SCM  $M_{\mathbf{c}}$  over  $\mathbf{V}$  by setting  $\mathbf{C} = \mathbf{c}$ , and replacing every occurrence of variables  $\mathbf{C}$  by the constant value  $\mathbf{c}$  in the equations of  $M$ . Therefore,  $M_{\mathbf{c}}$  is compatible with  $\mathcal{G}_{\mathbf{c}}$  and
$$
P ^ {M _ {\mathbf {c}}} (\mathbf {O}) = P ^ {M} (\mathbf {O} | d o (\mathbf {c})).
$$
By assumption, we know that for every  $\mathbf{c} \in \mathcal{D}_{\mathcal{C}}$ , there exists a policy  $\pi_{\mathbf{c}} \in \Pi$  uniquely computable from  $P(\mathbf{O})$  such that
$$
P ^ {M _ {\mathbf {c}}} (y \mid d o (\pi_ {\mathbf {c}})) = P ^ {M _ {\mathbf {c}}} (y). \tag {15}
$$
Suppose  $\pi_c$  relies on the values of  $\mathbf{Z}_{\mathbf{c}} \subseteq \mathbf{P}\mathbf{a}^{\Pi}$ . We can write
$$
\begin{array}{l} P ^ {M _ {\mathbf {c}}} (y \mid d o (\pi_ {\mathbf {c}})) = P ^ {M} (y \mid d o (\pi_ {\mathbf {c}}), d o (\mathbf {c})) \\ = \sum_ {x \in \mathcal {D} _ {X}, \mathbf {z} _ {\mathbf {c}} \in \mathcal {D} _ {\mathbf {Z} _ {\mathbf {c}}}} P ^ {M} (y | d o (x), d o (\mathbf {c}), \mathbf {z} _ {\mathbf {c}}) P ^ {M} (x | d o (\pi_ {\mathbf {c}}), d o (\mathbf {c}), \mathbf {z} _ {\mathbf {c}}) P ^ {M} (\mathbf {z} _ {\mathbf {c}} | d o (\mathbf {c})) \\ = \sum_ {x \in \mathcal {D} _ {X}, \mathbf {z} _ {\mathbf {c}} \in \mathcal {D} _ {\mathbf {Z} _ {\mathbf {c}}}} P ^ {M} (y | d o (x), \mathbf {c}, \mathbf {z} _ {\mathbf {c}}) P ^ {M} (x | d o (\pi_ {\mathbf {c}}), \mathbf {c}, \mathbf {z} _ {\mathbf {c}}) P ^ {M} (\mathbf {z} _ {\mathbf {c}} | \mathbf {c}) \\ = P ^ {M} (y \mid d o (\pi_ {\mathbf {c}}), \mathbf {c}), \tag {16} \\ \end{array}
$$
where the first and third parts of (a) hold since  $Y \perp \mathbf{C}|X$ ,  $\mathbf{Z}_{\mathbf{c}}$  in  $\mathcal{G}_{\overline{X}\underline{\mathbf{C}}}$  and  $\mathbf{Pa}(\mathbf{C}) \subseteq \mathbf{C}$ . For the second part, let  $M_{\pi_{\mathbf{c}}}$  be the model obtained from substituting the model of  $X$  given its parents by  $\pi_{\mathbf{c}}$  in the equations of  $M$ . Then, we get
$$
P ^ {M} (x \mid d o (\pi_ {\mathbf {c}}), d o (\mathbf {c}), \mathbf {z} _ {\mathbf {c}}) = P ^ {M _ {\pi_ {\mathbf {c}}}} (x \mid d o (\mathbf {c}), \mathbf {z} _ {\mathbf {c}}) = P ^ {M _ {\pi_ {\mathbf {c}}}} (x \mid \mathbf {c}, \mathbf {z} _ {\mathbf {c}}) = P ^ {M} (x \mid d o (\pi_ {\mathbf {c}}), \mathbf {c}, \mathbf {z} _ {\mathbf {c}}).
$$
Note that, since  $\mathbf{Pa}(\mathbf{C})\subseteq \mathbf{C}$ $P^{M_{\pi_{\mathbf{c}}}}(x|do(\mathbf{c}),\mathbf{z}_{\mathbf{c}}) = P^{M_{\pi_{\mathbf{c}}}}(x|\mathbf{c},\mathbf{z}_{\mathbf{c}}).$
$$
P ^ {M} _ {\mathbf {c}} (y) = P ^ {M} (y \mid d o (\mathbf {c})) = P ^ {M} (y \mid \mathbf {c}). \tag {17}
$$
Combining Equations (16), (17) and (15) we get the following for every  $\mathbf{c} \in \mathcal{D}_{\mathbf{C}}$
$$
P ^ {M} (y | d o (\pi_ {\mathbf {c}}), \mathbf {c}) = P ^ {M} (y | \mathbf {c}). \tag {18}
$$
Next, define a policy  $\pi^{*}(x|\mathbf{Pa}^{\Pi})\coloneqq \sum_{\mathbf{c}\in \mathcal{D}_{\mathbf{C}}}\mathbb{1}_{\{(\mathbf{Pa})_{\mathbf{C}}^{\Pi} = \mathbf{c}\}}\pi_{\mathbf{c}}$ , where  $\mathbb{1}_{\{(\mathbf{Pa})_{\mathbf{C}}^{\Pi} = \mathbf{c}\}}$  is an indicator function such that  $\mathbb{1}_{\mathbf{c}} = 1$  when  $\mathbf{C} = \mathbf{c}$ , and is equal to zero otherwise. Now, we can write
$$
\begin{array}{l} P ^ {M} (y | d o (\pi^ {*})) \\ = \sum_ {\mathbf {c} \in \mathcal {D} _ {\mathbf {C}}} P ^ {M} (y | d o (\pi^ {*}), \mathbf {c}) P ^ {M} (\mathbf {c} | d o (\pi^ {*})) \\ = \sum_ {\mathbf {c} \in \mathcal {D} _ {\mathbf {C}}} P ^ {M} (y | d o (\pi_ {\mathbf {c}}), \mathbf {c}) P ^ {M} (\mathbf {c}) \\ = \sum_ {\mathbf {c} \in \mathcal {D} _ {\mathbf {C}}} P ^ {M} (y | \mathbf {c}) P ^ {M} (\mathbf {c}) = P ^ {M} (y), \\ \end{array}
$$
where the last line holds due to Equation (18) and the fact that  $\mathbf{Pa}(\mathbf{C})\subseteq \mathbf{C}$ . We proved that there exists a policy  $\pi^{*}\in \Pi$  uniquely computable from  $P(\mathbf{O})$  such that  $P^{M}(y|do(\pi^{*})) = P^{M}(y)$ . Hence,  $P(y)$  is imitable w.r.t.  $\langle \mathcal{G}^{\mathcal{L}},\Pi ,\rangle$ .
Proposition 4.2. Given an  $LDAG\mathcal{G}^{\mathcal{L}}$  and a policy space  $\Pi$ , let  $\mathbf{C}$  be a subset of  $\mathbf{Pa}^{\Pi}\cap \mathbf{C}(\mathcal{L})$ . If for every realization  $\mathbf{c}\in \mathcal{D}_{\mathbf{C}}$  at least one of the following holds, then  $P(y)$  is imitable w.r.t.  $\langle \mathcal{G}^{\mathcal{L}},\Pi ,P(\mathbf{O})\rangle$ .
- There exists a subset  $\mathbf{S}_{\mathbf{c}}$  such that  $X \perp Y|\{\mathbf{S}_{\mathbf{c}}, \mathbf{C}\}$  in  $\mathcal{G}^{\mathcal{L}}(\mathbf{c}) \cup \Pi$ , and  $P(\mathbf{S}_{\mathbf{c}}|do(\pi), \mathbf{C} = \mathbf{c}) = P(\mathbf{S}_{\mathbf{c}}|\mathbf{C} = \mathbf{c})$  has a solution  $\pi_{\mathbf{c}} \in \Pi$ .
- There exists  $\mathbf{Z}_c \subseteq \mathbf{P}\mathbf{a}^{\Pi}$  such that  $X \perp Y|\{\mathbf{Z}_{\mathbf{c}}, \mathbf{C}\}$  in  $\mathcal{G}^{\mathcal{L}}(\mathbf{c})_{\underline{X}}$ .
Proof. Let  $M \in \mathcal{M}_{\langle \mathcal{G}^{\mathcal{L}} \rangle}$  and  $P^{M}(\mathbf{O}) = P(\mathbf{O})$ , i.e.,  $M$  induces the observational distribution  $P(\mathbf{O})$  be an arbitrary model. It suffices to show that there is a policy  $\pi \in \Pi$  uniquely computable from  $P(\mathbf{O})$  such that  $P^{M}(y|do(\pi)) = P^{M}(y)$ . To do so, first, partition  $\mathcal{D}_{\mathbf{C}}$  into two subsets  $\mathcal{D}_{\mathbf{C}}^{1}$  and  $\mathcal{D}_{\mathbf{C}}^{2}$  such that  $\mathcal{D}_{\mathbf{C}} = \mathcal{D}_{\mathbf{C}}^{1} \cup \mathcal{D}_{\mathbf{C}}^{2}$  and for every  $\mathbf{c} \in \mathcal{D}_{\mathbf{C}}^{1}$  the first condition of the lemma holds, whereas for every  $\mathbf{c} \in \mathcal{D}_{\mathbf{C}}^{2}$  the second condition does.
For each  $\mathbf{c} \in \mathcal{D}_{\mathbf{C}}^{1}$ , construct model  $M_{\pi_{\mathbf{c}}}$  by substituting the model of  $X$  given its parents by  $\pi_{\mathbf{c}}$  in  $M$ . Now, we get the following for each  $\mathbf{c} \in \mathbf{D}_{\mathbf{C}}^{1}$ ,
$$
\begin{array}{l} P ^ {M} (y | d o (\pi_ {\mathbf {c}}), \mathbf {c}) = P ^ {M _ {\pi_ {\mathbf {c}}}} (y | \mathbf {c}) = \sum_ {\mathbf {s} _ {\mathbf {c}} \in \mathcal {D} _ {\mathbf {S} _ {\mathbf {c}}}} P ^ {M _ {\pi_ {\mathbf {c}}}} (y | \mathbf {c}, \mathbf {s} _ {\mathbf {c}}) P ^ {M _ {\pi_ {\mathbf {c}}}} (\mathbf {s} _ {\mathbf {c}} | \mathbf {c}) \\ = \sum_ {\mathbf {s} _ {\mathbf {c}} \in \mathcal {D} _ {\mathbf {S} _ {\mathbf {c}}}} P ^ {M _ {\pi_ {\mathbf {c}}} (y | d o (x), \mathbf {c}, \mathbf {s} _ {\mathbf {c}}) P ^ {M _ {\pi_ {\mathbf {c}}} (\mathbf {s} _ {\mathbf {c}} | \mathbf {c})}} \\ = \sum_ {\mathbf {s} _ {\mathbf {c}} \in \mathcal {D} _ {\mathbf {S} _ {\mathbf {c}}}} P ^ {M} (y | d o (x), \mathbf {c}, \mathbf {s} _ {\mathbf {c}}) P ^ {M _ {\pi_ {\mathbf {c}}}} (\mathbf {s} _ {\mathbf {c}} | \mathbf {c}) \tag {19} \\ = \sum_ {\mathbf {s} _ {\mathbf {c}} \in \mathcal {D} _ {\mathbf {S} _ {\mathbf {c}}}} P ^ {M} (y | \mathbf {c}, \mathbf {s} _ {\mathbf {c}}) P ^ {M _ {\pi_ {\mathbf {c}}}} (\mathbf {s} _ {\mathbf {c}} | \mathbf {c}) \\ \stackrel {\mathrm {(c)}} {=} \sum_ {\mathbf {s _ {c}} \in \mathcal {D} _ {\mathbf {S _ {c}}}} P ^ {M} (y | \mathbf {c}, \mathbf {s _ {c}}) P ^ {M} (\mathbf {s _ {c}} | \mathbf {c}) = P ^ {M} (y | \mathbf {c}). \\ \end{array}
$$
Algorithm 3 Find Possible  $\pi$ -backdoor admissible set
1: function FINDSEP  $(\mathcal{G},\Pi ,X,Y,\mathbf{C})$   
2: if C is not specified then  
3:  $\mathbf{C}\gets \emptyset$   
4: Set  $\mathbf{Z} = \mathbf{A}\mathbf{n}(\{X,Y\} \cup \mathbf{C})\cap (\mathbf{Pa}^{\mathrm{II}})$   
5: if TESTSEP  $(\mathcal{G}_{\underline{X}},X,Y,\mathbf{Z})$  then  
6: return Z  
7: else  
8: return FAIL
where (a) and (b) hold since  $Y \perp X|\mathbf{S}_{\mathbf{c}}$ ,  $\mathbf{C}$  in  $\mathcal{G}^{\mathcal{L}}(\mathbf{c}) \cup \Pi$  and in  $\mathcal{G}^{\mathcal{L}}(\mathbf{c})$ , respectively (or  $Y \perp X|\mathbf{S}_{\mathbf{c}}$ ,  $\mathbf{c}$  in  $M^{\pi_{\mathbf{c}}}$  and  $M$ ). Moreover, (c) holds because  $P^{M}(\mathbf{S}_{\mathbf{c}}|do(\pi_{\mathbf{c}}), \mathbf{c}) = P^{M}(\mathbf{S}_{\mathbf{c}}|\mathbf{c})$ .
Next, for each  $\mathbf{c} \in \mathcal{D}_{\mathbf{C}}^{2}$ , let  $\pi_{\mathbf{c}}(x|\mathbf{Pa}^{\Pi}) = P^{M}(x|\mathbf{Z}_{\mathbf{c}},\mathbf{c})$ , then we get
$$
\begin{array}{l} P ^ {M} (y | d o (\pi_ {\mathbf {c}}), \mathbf {c}) = \sum_ {x \in \mathcal {D} _ {X}, \mathbf {z} _ {\mathbf {c}} \in \mathcal {D} _ {\mathbf {Z} _ {\mathbf {c}}}} P ^ {M} (y | d o (x), \mathbf {z} _ {\mathbf {c}}, \mathbf {c}) P ^ {M} (x | \mathbf {z} _ {\mathbf {c}}, \mathbf {c}) P ^ {M} (\mathbf {z} _ {\mathbf {c}} | \mathbf {c}) \\ = \sum_ {x \in \mathcal {D} _ {X}, \mathbf {z} _ {\mathbf {c}} \in \mathcal {D} _ {\mathbf {Z} _ {\mathbf {c}}}} P ^ {M} (y | x, \mathbf {z} _ {\mathbf {c}}, \mathbf {c}) P ^ {M} (x | \mathbf {z} _ {\mathbf {c}}, \mathbf {c}) P ^ {M} (\mathbf {z} _ {\mathbf {c}} | \mathbf {c}) = P ^ {M} (y | \mathbf {c}), \\ \end{array}
$$
where the second equation holds since  $X\perp Y|\mathbf{Z}_{\mathbf{c}}$ $\mathbf{C}$  in  $\mathcal{G}^{\mathcal{L}}(\mathbf{c})_{\underline{X}}$
Now, define a policy  $\pi^{*}(x|\mathbf{Pa}^{\Pi})\coloneqq \sum_{\mathbf{c}\in \mathcal{D}_{\mathbf{C}}}\mathbb{1}_{\{(\mathbf{Pa})_{\mathbf{C}}^{\Pi} = \mathbf{c}\}}\pi_{\mathbf{c}}$
We get the following where the second line follows from Equation (19) and the fact that  $\mathbf{C} \cap \mathbf{D}\mathbf{e}(X) = \emptyset$ .
$$
\begin{array}{l} P ^ {M} (y | d o (\pi^ {*})) = \sum_ {\mathbf {c} \in \mathcal {D} _ {\mathbf {C}}} P ^ {M} (y | d o (\pi^ {*}), \mathbf {c}) P ^ {M} (\mathbf {c} | d o (\pi^ {*})) \\ = \sum_ {\mathbf {c} \in \mathcal {D} _ {\mathbf {C}}} P ^ {M} (y | d o (\pi_ {\mathbf {c}}), \mathbf {c}) P ^ {M} (\mathbf {c}) \\ = \sum_ {\mathbf {c} \in \mathcal {D} _ {\mathbf {C}}} P ^ {M} (y | \mathbf {c}) P ^ {M} (\mathbf {c}) = P ^ {M} (y). \\ \end{array}
$$
The above equation implies that there is a policy  $\pi^{*}\in \Pi$  such that  $P^{M}(y|do(\pi^{*})) = P^{M}(y)$  where  $M$  is an arbitrary model compatible with  $\mathcal{G}^{\mathcal{L}}$ . Therefore,  $P(y)$  is imitable w.r.t.  $\langle \mathcal{G}^{\mathcal{L}},\Pi \rangle$  and  $\pi^{*}$  is an imitating policy for that.
B FindSep Algorithm
FindSep was used as a subroutine in Alg. 1. This function is summarized here as Algorithm 3, which takes a DAG  $\mathcal{G}$ , variables  $X$  and  $Y$ , and a subset of variables  $\mathbf{C}$  as inputs. It finds a separating set containing  $\mathbf{C}$ , if exists, in  $\mathcal{G}_{\underline{X}}$  between  $X$  and  $Y$ . To do so, it constructs  $\mathbf{Z}$  in line 4. According to lemma 3.4 of [40] where  $I = \mathbf{C}$  and  $R = \mathbf{P}\mathbf{a}^{\Pi}$ , if such separating set exists,  $\mathbf{Z}$  is a separating set.
C Example for Algorithm 2
We present a simple counterexample to illustrate that Algorithm 2 is not complete, in the sense that it may happen that an instance is imitable w.r.t.  $\langle \mathcal{G}^{\mathcal{L}},\Pi ,P(\mathbf{O})\rangle$  , but Algorithm 2 fails to find an imitating policy. To that end, take the instrumental variable graph depicted in Figure 6. It is evident that the set of equations
$$
\begin{array}{l} P (y | d o (\pi)) = P (y | X = 0, C = 0) \pi (X = 0) + P (y | X = 1, C = 0) \pi (X = 1) \\ = P (y \mid X = 0) P (X = 0) + P (y \mid X = 1) P (X = 1) = P (y) \\ \end{array}
$$
for both values of  $y = 0$  and  $y = 1$  is a set of linear equations with two equations in two parameters, namely  $\pi(X = 0)$  and  $\pi(X = 1)$ , which always has a solution. This indeed indicates that this
Figure 6: An example where Algorithm 2 fails to identify an imitating policy, despite the fact that there exists one such policy.
instance is immutable, i.e., there always exists at least one imitation policy. However, Algorithm 2 would not be able to find any such policy.
D Experimental setup
For the experiments of Section 5.2, we worked with an SCM in which
$$
C \sim B e (0. 0 5),
$$
$$
P (Y = 0 | C = 1) = 0. 2,
$$
$$
P (Y = 1 | C = 1) = 0. 5,
$$
$$
P (Y = 2 | C = 1) = 0. 3,
$$
$$
T \sim B e (0. 4),
$$
$$
U _ {1} \sim B e (0. 8),
$$
$$
X \mid T = 0, U _ {1} = 0 \sim B e (0. 7),
$$
$$
X \mid T = 0, U _ {1} = 1 \sim B e (0. 7),
$$
$$
X \mid T = 1, U _ {1} = 0 \sim B e (0), \text {a n d},
$$
$$
X \mid T = 1, U _ {1} = 1 \sim B e (1).
$$
Moreover, we had
$$
S | C = 0, X = 0, U _ {1} = 0 \sim B e (1),
$$
$$
S | C = 0, X = 0, U _ {1} = 1 \sim B e (0),
$$
$$
S \mid C = 0, X = 1, U _ {1} = 0 \sim B e (0),
$$
$$
S | C = 0, X = 1, U _ {1} = 1 \sim B e (1).
$$
And finally,
$$
P (Y = 0 | C = 0, S = 0) = 0. 8,
$$
$$
P (Y = 1 | C = 0, S = 0) = 0. 1,
$$
$$
P (Y = 2 | C = 0, S = 0) = 0. 1,
$$
$$
P (Y = 0 \mid C = 0, S = 1) = 0. 0 5,
$$
$$
P (Y = 1 | C = 0, S = 1) = 0. 2, \text {a n d},
$$
$$
P (Y = 2 \mid C = 0, S = 1) = 0. 7 5.
$$