<p>I am not sure what exactly you are asking because of notational clutter, but I can take a stab. As I see it, there is no reason for the $g$'s and the $h$'s since they are never exchanged. So lets just use $g_1 \ldots g_k$ and $P_1 \ldots P_k$ as our lotteries (over a set $X = \{x,y\ldots\}$) and probabilities. Further, there is no reason make the second index be the odd one out.</p>

<p>Now, reduction of compound lotteries says that if we have $n$ lotteries $g_1 \ldots g_n$ and $\sum P_i = 1$ and $P_i \geq 0$ then we can identify $P_1g + \ldots + P_ng'$ with the lottery that assigns $P_1g(x) + \ldots + P_ng_n(x)$ to each $x \in X$. We only care about the total probability assigned to final consumption outcomes and not how we got there--for example, we do not care about the timing of the resolution of uncertainty.$^1$ </p>

<p>Let $P_1g_1 + \ldots + P_kg_k$ by a lottery of size $k$. From our reduction principle this is the lottery that assigns 
\begin{equation}
\tag{1}
P_1g_1(x) + \ldots + P_kg_k(x)
\end{equation}
to each $x \in X$.</p>

<p>Then consider the lottery $$g^* = \frac{P_2}{1-P_1}g_2 + \ldots + \frac{P_k}{1-P_2}g_k.$$ (Make sure you see why is this a well defined lottery, in other words why do the probabilities necessarily sum to 1?). We can, by the reduction principle treat this as the lottery that assigns
\begin{equation}
\tag{2}
g^*(x) = \frac{P_2}{1-P_1}g_2(x) + \ldots + \frac{P_k}{1-P_2}g_k(x)
\end{equation}
to each $x \in X$. </p>

<p>Now, what if we mix this new reduced lottery with $g_1$? Take a look at $g^{**} = P_1g_1 + (1-P_1)g^*$. Well, this is the lottery that assigns
$$g^{**}(x) =  P_1g_1(x) + (1-P_1)g^*(x)$$ to each $x \in X$. Look: this in the lottery of interest (if I read your question correctly)! From equation (2) we can re-write this as
$$g^{**}(x) =  P_1g_1(x) + (1-P_1)\big(\frac{P_2}{1-P_1}g_2(x) + \ldots + \frac{P_k}{1-P_2}g_k(x)\big).$$
Of course this is a garden variety real valued equation, so lets go ahead a cancel the $(1-P_1)$'s leaving us with exactly equation (1). So these two lotteries are indeed the same.</p>

<p>[1] If you are asking how to move from a definition of reduction that only uses 2 lotteries at a time to the (equivalent) one above, this is a different question. Hint: the proof is by induction and the base case is the definition. In spirit its whats going on below.</p>

CC BY-SA 3.0
2018-04-23T19:34:32.037Z
2018-04-23T19:34:32.037Z
201p
<p>Suppose that a binary relation satisfies only:</p>

<p>Independence axiom: $L≿L′⟺α\circ L+(1−α)\circ L′′≿α\circ L′+(1−α) \circ L′′$
 Reduction to simple lotteries: For all $g$, $g~g'$, $g'$ is the simple lotteries associated with g.</p>

<p>Can i state the following relation:</p>

<p>$(P1 \circ h1,P2\circ g2, P3\circ g3...PK\circ gk) \sim (P2\circ g2, ((1-P2)\circ ((P1/1-P2)\circ h1,(P3/1-P2)\circ g3,..,(PK/1-P2)\circ gk))$</p>

<p>where $h1$ to hk and $g1$ to $gk$ are lotteries, and $P1$ to $Pk$ are probabilities.  </p>

<p>I tried to show this only using reduction to simple lotteries, but i dont know if im doing another assumption when i write the second lotterie as 
$(P2\circ g2, P1\circ h1,P3 \circ g3,....,PK\circ gk))$</p>

CC BY-SA 3.0
2018-04-21T16:27:36.813Z
Please format mathematical expressions using [MathJax](https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference). Also, showing what you've tried so far will significantly increase the chances of your question being answered.
Herr K.
<preferences><uncertainty>
Indiference between two lotteries
mmramame
# Indiference between two lotteries

(Asked by: username_0 on 2018-04-21)

Suppose that a binary relation satisfies only:

Independence axiom: $L≿L′⟺α\circ L+(1−α)\circ L′′≿α\circ L′+(1−α) \circ L′′$
Reduction to simple lotteries: For all $g$, $g~g'$, $g'$ is the simple lotteries associated with g.

Can i state the following relation:

$(P1 \circ h1,P2\circ g2, P3\circ g3...PK\circ gk) \sim (P2\circ g2, ((1-P2)\circ ((P1/1-P2)\circ h1,(P3/1-P2)\circ g3,..,(PK/1-P2)\circ gk))$

where $h1$ to hk and $g1$ to $gk$ are lotteries, and $P1$ to $Pk$ are probabilities.

I tried to show this only using reduction to simple lotteries, but i dont know if im doing another assumption when i write the second lotterie as
$(P2\circ g2, P1\circ h1,P3 \circ g3,....,PK\circ gk))$

**Tags:** preferences, uncertainty
> **Comments:**
> **username_1**: Please format mathematical expressions using [MathJax](https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference). Also, showing what you've tried so far will significantly increase the chances of your question being answered.  

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## 1 answers

### Answer by username_2 on 2018-04-23. Score: 0

I am not sure what exactly you are asking because of notational clutter, but I can take a stab. As I see it, there is no reason for the $g$'s and the $h$'s since they are never exchanged. So lets just use $g\_1 \ldots g\_k$ and $P\_1 \ldots P\_k$ as our lotteries (over a set $X = \{x,y\ldots\}$) and probabilities. Further, there is no reason make the second index be the odd one out.

Now, reduction of compound lotteries says that if we have $n$ lotteries $g\_1 \ldots g\_n$ and $\sum P\_i = 1$ and $P\_i \geq 0$ then we can identify $P\_1g + \ldots + P\_ng'$ with the lottery that assigns $P\_1g(x) + \ldots + P\_ng\_n(x)$ to each $x \in X$. We only care about the total probability assigned to final consumption outcomes and not how we got there--for example, we do not care about the timing of the resolution of uncertainty.$^1$

Let $P\_1g\_1 + \ldots + P\_kg\_k$ by a lottery of size $k$. From our reduction principle this is the lottery that assigns
\begin{equation}
\tag{1}
P\_1g\_1(x) + \ldots + P\_kg\_k(x)
\end{equation}
to each $x \in X$.

Then consider the lottery $$g^\* = \frac{P\_2}{1-P\_1}g\_2 + \ldots + \frac{P\_k}{1-P\_2}g\_k.$$ (Make sure you see why is this a well defined lottery, in other words why do the probabilities necessarily sum to 1?). We can, by the reduction principle treat this as the lottery that assigns
\begin{equation}
\tag{2}
g^\*(x) = \frac{P\_2}{1-P\_1}g\_2(x) + \ldots + \frac{P\_k}{1-P\_2}g\_k(x)
\end{equation}
to each $x \in X$.

Now, what if we mix this new reduced lottery with $g\_1$? Take a look at $g^{\*\*} = P\_1g\_1 + (1-P\_1)g^\*$. Well, this is the lottery that assigns
$$g^{\*\*}(x) = P\_1g\_1(x) + (1-P\_1)g^\*(x)$$ to each $x \in X$. Look: this in the lottery of interest (if I read your question correctly)! From equation (2) we can re-write this as
$$g^{\*\*}(x) = P\_1g\_1(x) + (1-P\_1)\big(\frac{P\_2}{1-P\_1}g\_2(x) + \ldots + \frac{P\_k}{1-P\_2}g\_k(x)\big).$$
Of course this is a garden variety real valued equation, so lets go ahead a cancel the $(1-P\_1)$'s leaving us with exactly equation (1). So these two lotteries are indeed the same.

[1] If you are asking how to move from a definition of reduction that only uses 2 lotteries at a time to the (equivalent) one above, this is a different question. Hint: the proof is by induction and the base case is the definition. In spirit its whats going on below.


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