<p>Take as example the set of lotteries over the outcomes <span class="math-container">$(1, 5, 10)$</span> consider the lotteries <span class="math-container">$L_1, L_2, L_3$</span> where,
<span class="math-container">$$
\Pr(L_1 = 1) = 0.5\\
\Pr(L_1 = 5) = 0\\
\Pr(L_1 = 10) = 0.5
$$</span></p>
<p><span class="math-container">$$
\Pr(L_2 = 1) = 0.1\\
\Pr(L_2 = 5) = 0.9\\
\Pr(L_2 = 10) = 0
$$</span></p>
<p><span class="math-container">$$
\Pr(L_3 = 1) = 0.2\\
\Pr(L_3 = 5) = 0.5\\
\Pr(L_3 = 10) = 0.3
$$</span></p>
<p>Lottery <span class="math-container">$L_i$</span> FOSD <span class="math-container">$L_j$</span> if <span class="math-container">$\Pr(L_i \le x) \le \Pr(L_j \le x)$</span> for all <span class="math-container">$x$</span>.</p>
<p>Note that</p>
<ul>
<li><p><span class="math-container">$\Pr(L_1 \le 5) = 0.5 &lt; \Pr(L_2 \le 5) = 1$</span> and <span class="math-container">$\Pr(L_2 \le 1) = 0.2 &lt; \Pr(L_1 \le 1) = 0.5$</span>, so there is no FOSD between <span class="math-container">$L_1$</span> and <span class="math-container">$L_2$</span>.</p>
</li>
<li><p><span class="math-container">$\Pr(L_2 \le 1) = 0.1 &lt; \Pr(L_3 \le 1) = 0.2$</span> and <span class="math-container">$\Pr(L_3 \le 5) = 0.7 &lt; \Pr(L_2 \le 5) = 1$</span>, so there is no FOSD between <span class="math-container">$L_2$</span> and <span class="math-container">$L_3$</span>.</p>
</li>
<li><p>Finally, <span class="math-container">$\Pr(L_3 \le 1)  = 0.2 &lt; \Pr(L_1 \le 1) = 0.5$</span> and <span class="math-container">$\Pr(L_1 \le 5) = 0.5 &lt; \Pr(L_3 \le 0.5) = 0.7$</span>, so there is no FOSD between <span class="math-container">$L_1$</span> and <span class="math-container">$L_3$</span>.</p>
</li>
</ul>
<p>I think you need lotteries on at least 3 outcomes to get such examples. If there are only two outcomes, the FOSD relation is complete.</p>

CC BY-SA 4.0
2023-12-05T10:04:30.413Z
Great answer I verify that it works (finding three lotteries not FOSD each other). Though just one humble cent: lottery usual means a probability distribution of a real interval, say $[0,1]$. What you have here is called simple lottery: the lottery with finite number of outcomes, which is the counterpart of simple random variable.
High GPA
CC BY-SA 4.0
2023-12-05T10:06:46.943Z
May I ask what is the specific preference relation you have in mind for the set of all lotteries?
High GPA
CC BY-SA 4.0
2023-12-04T07:08:24.543Z
2023-12-04T07:08:24.543Z
tdm
<p>Let <span class="math-container">$\succsim$</span> be a complete, non transitivity preference relation. I wonder if FOSD-transitivity implies transitivity.</p>
<p>The primitive is the space of lotteries <span class="math-container">$p_1,p_2,p_3,...$</span>. We say the preference satisfies FOSD-transitivity if <span class="math-container">$p_1\succsim p_2 \succsim...\succsim p_n$</span> implies <span class="math-container">$p_n$</span> does not FOSD <span class="math-container">$p_1$</span>.</p>
<p>I think the answer is no, as we can easily find three lotteries <span class="math-container">$p,q,r$</span> such that they don't FOSD each other. Then we let <span class="math-container">$p\succsim q\succsim r\succsim p$</span>.</p>
<p>However, when finding a specific numeric example, I failed.</p>

<microeconomics><theory><risk>
Can FOSD-transitivity replace the transitivity in utility representation theorem?
High GPA
# Can FOSD-transitivity replace the transitivity in utility representation theorem?

(Asked by: username_0 on 2023-12-02)

Let $\succsim$ be a complete, non transitivity preference relation. I wonder if FOSD-transitivity implies transitivity.

The primitive is the space of lotteries $p\_1,p\_2,p\_3,...$. We say the preference satisfies FOSD-transitivity if $p\_1\succsim p\_2 \succsim...\succsim p\_n$ implies $p\_n$ does not FOSD $p\_1$.

I think the answer is no, as we can easily find three lotteries $p,q,r$ such that they don't FOSD each other. Then we let $p\succsim q\succsim r\succsim p$.

However, when finding a specific numeric example, I failed.

**Tags:** microeconomics, theory, risk


---

## 1 answers

### Answer by username_1 on 2023-12-04. Score: 2

Take as example the set of lotteries over the outcomes $(1, 5, 10)$ consider the lotteries $L\_1, L\_2, L\_3$ where,
$$
\Pr(L\_1 = 1) = 0.5\\
\Pr(L\_1 = 5) = 0\\
\Pr(L\_1 = 10) = 0.5
$$

$$
\Pr(L\_2 = 1) = 0.1\\
\Pr(L\_2 = 5) = 0.9\\
\Pr(L\_2 = 10) = 0
$$

$$
\Pr(L\_3 = 1) = 0.2\\
\Pr(L\_3 = 5) = 0.5\\
\Pr(L\_3 = 10) = 0.3
$$

Lottery $L\_i$ FOSD $L\_j$ if $\Pr(L\_i \le x) \le \Pr(L\_j \le x)$ for all $x$.

Note that

* $\Pr(L\_1 \le 5) = 0.5 < \Pr(L\_2 \le 5) = 1$ and $\Pr(L\_2 \le 1) = 0.2 < \Pr(L\_1 \le 1) = 0.5$, so there is no FOSD between $L\_1$ and $L\_2$.
* $\Pr(L\_2 \le 1) = 0.1 < \Pr(L\_3 \le 1) = 0.2$ and $\Pr(L\_3 \le 5) = 0.7 < \Pr(L\_2 \le 5) = 1$, so there is no FOSD between $L\_2$ and $L\_3$.
* Finally, $\Pr(L\_3 \le 1) = 0.2 < \Pr(L\_1 \le 1) = 0.5$ and $\Pr(L\_1 \le 5) = 0.5 < \Pr(L\_3 \le 0.5) = 0.7$, so there is no FOSD between $L\_1$ and $L\_3$.

I think you need lotteries on at least 3 outcomes to get such examples. If there are only two outcomes, the FOSD relation is complete.

> **Comments:**
> **username_0**: Great answer I verify that it works (finding three lotteries not FOSD each other). Though just one humble cent: lottery usual means a probability distribution of a real interval, say $[0,1]$. What you have here is called simple lottery: the lottery with finite number of outcomes, which is the counterpart of simple random variable.  
> **username_0**: May I ask what is the specific preference relation you have in mind for the set of all lotteries?  

---

