# Clarification - DeGroot Proof on Transitivity Property of Subjective Probability

In developing axiomatic foundation for subjective probability DeGroot (Optimal Statistical Decision, 2004, p71) gives two axioms/assumptions:

SP1: For any two events A and B, exactly one of the following three relations must hold:$$A \preceq B, A \succeq B, A \sim B$$. By definition, $$A \preceq B$$ is when P(A) $$\leq$$ P(B)

SP2: If $$A_{1}, A_{2}, B_{1}, B_{2}$$ are four events such that $$A_{1} \bigcap A_{2}= \emptyset$$ and $$B_{1} \bigcap B_{2}=\emptyset$$ and $$A_{i} \preceq B_{i}$$ for i=1,2, then $$A_{1}\bigcup A_{2} \preceq B_{1} \bigcup B_{2}$$. In addition, if either $$A_{1}\prec A_{2}$$ or $$B_{1}\prec B_{2}$$ then $$A_{1}\bigcup A_{2} \prec B_{1} \bigcup B_{2}$$

Based on these two assumptions, he gives a lemma as a direct consequences of the above two assumptions: Lemma 1:Suppose that A, B, D are events such that A $$\bigcap$$ D = B $$\bigcap$$ D = $$\emptyset$$ then A $$\preceq$$B if and only if $$A\bigcup D \preceq B \bigcup D$$.

DeGroot continues to give a theorem of transitivity property: Theorem I: If A,B,D are event such that A $$\preceq$$ B and B$$\preceq$$D then A$$\preceq$$D

I need some help understanding the proof of this theorem which he gives as follow:

Consder the seven disjoint events shown in the figure bellow, whose union is $$A\bigcup B \bigcup D$$. Since A $$\preceq$$B it follows from Lemma 1 that
(1) $$( A\bigcap B ^{c} \bigcap D^{c} ) \bigcup ( A\bigcap B ^{c} \bigcap D ) \preceq ( A^{c}\bigcap B \bigcap D^{c} ) \bigcup ( A^{c}\bigcap B \bigcap D )$$.
Similarly, since B$$\preceq$$D it follows from lemma 1 that
(2) $$( A\bigcap B \bigcap D^{c} ) \bigcup ( A^{c}\bigcap B \bigcap D^{c}) \preceq ( A\bigcap B ^{c}\bigcap D) \bigcup ( A^{c}\bigcap B^{c} \bigcap D )$$.

The above lemma doesn't seem to justify this. What i understand from the lemma is that because A $$\preceq$$B then $$( A \bigcup D ) \preceq (( B \bigcup D )$$. How did exactly he go from A $$\preceq$$B into $$( A\bigcap B ^{c} \bigcap D^{c} ) \bigcup ( A\bigcap B ^{c} \bigcap D ) \preceq ( A^{c}\bigcap B \bigcap D^{c} ) \bigcup ( A^{c}\bigcap B \bigcap D )$$ ?

Next, he argues that since the left sides of the relations (1) and (2) are disjoint and the right sides are also disjoint, it follows from assumption SP2 that $$( A\bigcap B ^{c} \bigcap D^{c} ) \bigcup ( A\bigcap B ^{c} \bigcap D )\bigcup (A\bigcap B \bigcap D^{c} ) \bigcup ( A^{c}\bigcap B \bigcap D^{c}) \preceq$$

$$( A^{c}\bigcap B \bigcap D^{c} ) \bigcup ( A^{c}\bigcap B \bigcap D ) \bigcup ( A\bigcap B ^{c}\bigcap D) \bigcup ( A^{c}\bigcap B^{c} \bigcap D )$$.

If the common event $$( A\bigcap B^{c} \bigcap D ) \bigcup ( A^{c}\bigcap B \bigcap D^{c}$$ ) is eliminated from both sides of this relation, it follows from lemma 1 that:

$$( A\bigcap B ^{c} \bigcap D^{c} ) \bigcup ( A\bigcap B \bigcap D^{c} ) \preceq ( A^{c}\bigcap B \bigcap D ) \bigcup ( A^{c}\bigcap B^{c} \bigcap D )$$, he argues that it can now be seen from the above figure and lemma 1 that A $$\preceq$$D. What I dont understand is how we can justify the elimination of the common event as how he does here.

Any help would be very much appreciated, cheers:)

How did exactly he go from $A\preceq B$ into $( A\bigcap B ^{c} \bigcap D^{c} ) \bigcup ( A\bigcap B ^{c} \bigcap D ) \preceq ( A^{c}\bigcap B \bigcap D^{c} ) \bigcup ( A^{c}\bigcap B \bigcap D )$ ?

\begin{eqnarray*} A\preceq B &\iff & AB^c \cup AB \preceq A^cB \cup AB \qquad\text{since $A=AB^c \cup AB$ and $B=A^cB \cup AB$} \\ &\iff& AB^c \preceq A^cB \qquad\qquad\qquad\qquad\text{by Lemma 1} \\ &\iff& AB^cD^c \cup AB^cD \preceq A^cBD^c \cup A^cBD \qquad\text{since $AB^c=AB^cD^c \cup AB^cD$, etc.} \end{eqnarray*}

What I dont understand is how we can justify the elimination of the common event as how he does here.

Lemma 1 allows for elimination of a common event, $D$. Apply Lemma 1 with:

\begin{eqnarray*} A &=& AB^cD^c \cup ABD^c \\ B &=& A^cBD \cup A^cB^cD \\ D &=& AB^cD \cup A^cBD^c. \end{eqnarray*}

Then the lemma, in saying $A\cup D \preceq B\cup D \implies A \preceq B$, eliminates our common event.

The final step to reach $A\preceq D$ is the same as the "expansion" of $A\preceq B$ shown above but in reverse.