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 $(P2\circ g2, P1\circ h1,P3 \circ g3,....,PK\circ gk))$

  • 2
    $\begingroup$ Please format mathematical expressions using MathJax. Also, showing what you've tried so far will significantly increase the chances of your question being answered. $\endgroup$ – Herr K. Apr 21 '18 at 16:27

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.