<p>As a first remark: the Anscombe-Aumann axioms, in particular Independence, are defined over acts taking the state space to a linear space (generally simple lotteries over consumption objects). Even when we consider the restriction of the model to purely subjectively uncertain acts, we still need to employ the full model or we will lose information. </p>

<p>That being said: Lets let $S$ be a finite state space, and $X$ a finite set of alternatives. Let $\Delta(X)$ denote all the lotteries over $X$ and $f: S \to \Delta(X)$ is an act. For an event $E \subseteq S$, let $f_{-E}g$ be the act defined by
$$f_{-E}g \begin{cases} f(s) \text{ if } x \in E \\ g(s) \text{ if } x \notin E. \end{cases}$$</p>

<p>Now, we can say that our model satisfies the <strong>sure thing principle</strong> if $f_{-E}h \succsim g_{-E}h$ and $f_{-E^c}h \succsim g_{-E^c}h$ then $f \succsim g.$ This definition is valid for all acts, not just ones without objective risk, but clearly you can consider only the relevant projection.</p>

<p>Assume the antecedent of the STP. From $f_{-E}h \succsim g_{-E}h$ and independence we have that 
$$\frac12 f_{-E}h + \frac12 f_{-E^c}h \succsim \frac12 g_{-E}h + \frac12 f_{-E^c}h.$$ 
Notice we can rewrite this as
$$\frac12 f + \frac12 h \succsim \frac12 g_{-E}f + \frac12h$$
and, applying independence again, we get
\begin{equation}
\tag{1}
f \succsim g_{-E}f.
\end{equation}</p>

<p>In an analogous fashion, from $f_{-E^c}h \succsim g_{-E^c}h$ and independence we have that 
$$\frac12 f_{-E^c}h + \frac12 g_{-E}h \succsim \frac12 g_{-E^c}h + \frac12 g_{-E}h.$$ 
Again, we can rewrite as
$$\frac12 g_{-E}f + \frac12 h \succsim \frac12 g + \frac12h$$
and, applying independence again, we get
\begin{equation}
\tag{2}
g_{-E}f \succsim g.
\end{equation}</p>

<p>Combining (1) and (2) via transitivity yields the desired relations. Going back to the prefatory remark, notice that to apply independence, we need to mix acts, appealing to objective risk. Thus, even when $f$, $g$, and $h$ have no objective risk, we still need risky acts to serve as an intermediary in the proof. In a sense, this is the grand insight to the whole AA framework---using objective risk to get around the necessity of an infinite state space by using the linearity of expectations to force the STP.</p>

<p>Notice only independence and transitivity were used. This should indicated that even state-dependent EU (where monotonicity / state-independence fails) or Bewley EU (where completeness is relaxed)  will still satisfy the STP. </p>

<hr>

<p><strong>Edit</strong> in response to a comment: Lets call the above notion of the Sure Thing Principle <strong>STP1</strong> and say the preference satisfies <strong>STP2</strong> if $f_{-E}h \succsim g_{-E}h \iff f_{-E}h' \succsim g_{-E}h'$ for all $f,g,h,h'$. Then if  $\succsim$ is a preorder, it satisfies STP1 if and only if it satisfies STP2. </p>

<p>First assume STP2 holds and that $f_{-E}h \succsim g_{-E}h$ and $f_{-E^c}h \succsim g_{-E^c}h$. Then by STP2 we have 
$$f = f_{-E}f \succsim g_{-E}f \qquad \text{ and } \qquad g_{-E}f = f_{-E^c}g \succsim g.$$ 
Transitivity implies $f \succsim g$; STP1 holds.</p>

<p>Next, assume STP1 holds and $f_{-E}h \succsim g_{-E}h$. Define $\hat f = f_{-E}h'$ and $\hat g$ analogously. By definition 
$$\hat f_{-E}h = f_{-E}h \qquad \text{ and } \qquad \hat g_{-E}h = g_{-E}h,$$
so our assumption is identically that 
\begin{equation}
\tag{3}
\hat f_{-E}h \succsim \hat g_{-E}h.
\end{equation}
Further $\hat f_{-E^c}h = \hat g_{-E^c}h = h'_{-E}h$ so we have, by the reflexivity of preference, that
\begin{equation}
\tag{4}
\hat f_{-E^c}h \succsim \hat g_{-E^c}h.
\end{equation}
Now we can apply STP1 to (3) and (4) to obtain that $\hat f \succsim \hat g$, which, given their definition, exactly what we need to show for STP2 to hold.</p>

CC BY-SA 3.0
2018-05-01T15:03:25.630Z
(+1) A question: it has been shown that the STP requires that acts do not affect the probabilities over the events, otherwise it may not hold. Is this covered / guaranteed by the AA framework?
Alecos Papadopoulos
CC BY-SA 3.0
2018-05-01T16:55:21.650Z
@201p great answer, thanks a lot. One question: the standard definition of the STP is that $f_{E}h \succeq g_{E}h \Leftrightarrow f_{E}h' \succeq g_{E}h'$. Is your definition equivalent to this one?
Oliv
CC BY-SA 3.0
2018-05-01T16:57:19.370Z
@AlecosPapadopoulos isn't it the axiom P4 (instead of P2) which requires probabilities to be act-independent? Otherwise, do you have a reference for your claim?
Oliv
CC BY-SA 3.0
2018-05-01T17:02:11.973Z
@Oliv Sure, check ftp.cs.ucla.edu/pub/stat_ser/r466.pdf and the literature therein.
Alecos Papadopoulos
CC BY-SA 3.0
2018-05-01T17:19:40.747Z
@AlecosPapadopoulos thanks a lot, that's very useful.
Oliv
CC BY-SA 3.0
2018-05-01T19:19:45.600Z
@AlecosPapadopoulous by the way, the answer to your question is yes because the AA theorem delivers a subjective expected utility representation with act-independent probabilities, as the Savage theorem. How this is implied by the AA axioms is a good question though (it is not clear to me).
Oliv
CC BY-SA 4.0
2018-05-03T12:35:28.010Z
@201p Thanks a lot for editing your proof in response to my question. That's a beautiful and useful answer.
Oliv
CC BY-SA 4.0
2024-10-06T08:16:59.410Z
Is it possible to prove that AA indepdence implies P4?
High GPA
CC BY-SA 3.0
2018-05-01T14:50:50.527Z
2018-05-01T19:26:23.993Z
2018-05-01T19:26:23.993Z
201p
<p>Consider an Anscombe-Aumann setting, and assume that a preference relation satisfies all original Anscombe-Aumann's axioms (rationality, continuity, independence, and monotonicity).</p>

<p>If we restrict attention to pure horse races (that is, acts without any objective uncertainty), the Anscombe-Aumann model boils down to a Subjective Expected Utility representation a la Savage. Therefore, over pure horse races, the decision-maker satisfies all of Savage's axioms, notably the Sure-Thing Principle (P2 in Savage's terminology).</p>

<p>I fail to see the direct connection between Anscombe-Aumann's axioms and the Sure-Thing Principle. Does anyone see how the Sure-Thing Principle is implied by Anscombe-Aumann's axioms? In particular, does it result from Independence only, or are Independence AND Monotonicity required?</p>

<decision-theory>
Which of Anscombe-Aumann's axioms imply the Sure-Thing principle?
Oliv
# Which of Anscombe-Aumann's axioms imply the Sure-Thing principle?

(Asked by: username_0 on 2018-05-01)

Consider an Anscombe-Aumann setting, and assume that a preference relation satisfies all original Anscombe-Aumann's axioms (rationality, continuity, independence, and monotonicity).

If we restrict attention to pure horse races (that is, acts without any objective uncertainty), the Anscombe-Aumann model boils down to a Subjective Expected Utility representation a la Savage. Therefore, over pure horse races, the decision-maker satisfies all of Savage's axioms, notably the Sure-Thing Principle (P2 in Savage's terminology).

I fail to see the direct connection between Anscombe-Aumann's axioms and the Sure-Thing Principle. Does anyone see how the Sure-Thing Principle is implied by Anscombe-Aumann's axioms? In particular, does it result from Independence only, or are Independence AND Monotonicity required?

**Tags:** decision-theory

---

## 1 answers

### Answer by username_1 on 2018-05-01. Score: 10 | Accepted answer

As a first remark: the Anscombe-Aumann axioms, in particular Independence, are defined over acts taking the state space to a linear space (generally simple lotteries over consumption objects). Even when we consider the restriction of the model to purely subjectively uncertain acts, we still need to employ the full model or we will lose information.

That being said: Lets let $S$ be a finite state space, and $X$ a finite set of alternatives. Let $\Delta(X)$ denote all the lotteries over $X$ and $f: S \to \Delta(X)$ is an act. For an event $E \subseteq S$, let $f\_{-E}g$ be the act defined by
$$f\_{-E}g \begin{cases} f(s) \text{ if } x \in E \\ g(s) \text{ if } x \notin E. \end{cases}$$

Now, we can say that our model satisfies the **sure thing principle** if $f\_{-E}h \succsim g\_{-E}h$ and $f\_{-E^c}h \succsim g\_{-E^c}h$ then $f \succsim g.$ This definition is valid for all acts, not just ones without objective risk, but clearly you can consider only the relevant projection.

Assume the antecedent of the STP. From $f\_{-E}h \succsim g\_{-E}h$ and independence we have that
$$\frac12 f\_{-E}h + \frac12 f\_{-E^c}h \succsim \frac12 g\_{-E}h + \frac12 f\_{-E^c}h.$$
Notice we can rewrite this as
$$\frac12 f + \frac12 h \succsim \frac12 g\_{-E}f + \frac12h$$
and, applying independence again, we get
\begin{equation}
\tag{1}
f \succsim g\_{-E}f.
\end{equation}

In an analogous fashion, from $f\_{-E^c}h \succsim g\_{-E^c}h$ and independence we have that
$$\frac12 f\_{-E^c}h + \frac12 g\_{-E}h \succsim \frac12 g\_{-E^c}h + \frac12 g\_{-E}h.$$
Again, we can rewrite as
$$\frac12 g\_{-E}f + \frac12 h \succsim \frac12 g + \frac12h$$
and, applying independence again, we get
\begin{equation}
\tag{2}
g\_{-E}f \succsim g.
\end{equation}

Combining (1) and (2) via transitivity yields the desired relations. Going back to the prefatory remark, notice that to apply independence, we need to mix acts, appealing to objective risk. Thus, even when $f$, $g$, and $h$ have no objective risk, we still need risky acts to serve as an intermediary in the proof. In a sense, this is the grand insight to the whole AA framework---using objective risk to get around the necessity of an infinite state space by using the linearity of expectations to force the STP.

Notice only independence and transitivity were used. This should indicated that even state-dependent EU (where monotonicity / state-independence fails) or Bewley EU (where completeness is relaxed) will still satisfy the STP.

---

**Edit** in response to a comment: Lets call the above notion of the Sure Thing Principle **STP1** and say the preference satisfies **STP2** if $f\_{-E}h \succsim g\_{-E}h \iff f\_{-E}h' \succsim g\_{-E}h'$ for all $f,g,h,h'$. Then if $\succsim$ is a preorder, it satisfies STP1 if and only if it satisfies STP2.

First assume STP2 holds and that $f\_{-E}h \succsim g\_{-E}h$ and $f\_{-E^c}h \succsim g\_{-E^c}h$. Then by STP2 we have
$$f = f\_{-E}f \succsim g\_{-E}f \qquad \text{ and } \qquad g\_{-E}f = f\_{-E^c}g \succsim g.$$
Transitivity implies $f \succsim g$; STP1 holds.

Next, assume STP1 holds and $f\_{-E}h \succsim g\_{-E}h$. Define $\hat f = f\_{-E}h'$ and $\hat g$ analogously. By definition
$$\hat f\_{-E}h = f\_{-E}h \qquad \text{ and } \qquad \hat g\_{-E}h = g\_{-E}h,$$
so our assumption is identically that
\begin{equation}
\tag{3}
\hat f\_{-E}h \succsim \hat g\_{-E}h.
\end{equation}
Further $\hat f\_{-E^c}h = \hat g\_{-E^c}h = h'\_{-E}h$ so we have, by the reflexivity of preference, that
\begin{equation}
\tag{4}
\hat f\_{-E^c}h \succsim \hat g\_{-E^c}h.
\end{equation}
Now we can apply STP1 to (3) and (4) to obtain that $\hat f \succsim \hat g$, which, given their definition, exactly what we need to show for STP2 to hold.

> **Comments:**
> **username_2**: (+1) A question: it has been shown that the STP requires that acts do not affect the probabilities over the events, otherwise it may not hold. Is this covered / guaranteed by the AA framework?  
> **username_0**: @201p great answer, thanks a lot. One question: the standard definition of the STP is that $f_{E}h \succeq g_{E}h \Leftrightarrow f_{E}h' \succeq g_{E}h'$. Is your definition equivalent to this one?  
> **username_0**: @AlecosPapadopoulos isn't it the axiom P4 (instead of P2) which requires probabilities to be act-independent? Otherwise, do you have a reference for your claim?  
> **username_2**: @Oliv Sure, check ftp.cs.ucla.edu/pub/stat_ser/r466.pdf and the literature therein.  
> **username_0**: @AlecosPapadopoulos thanks a lot, that's very useful.  
> **username_0**: @AlecosPapadopoulous by the way, the answer to your question is yes because the AA theorem delivers a subjective expected utility representation with act-independent probabilities, as the Savage theorem. How this is implied by the AA axioms is a good question though (it is not clear to me).  
> **username_0**: @201p Thanks a lot for editing your proof in response to my question. That's a beautiful and useful answer.  
> **username_3**: Is it possible to prove that AA indepdence implies P4?  

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