VNM Expected Utility Proof

Cheatsheet Content

### Introduction to VNM Expected Utility This cheatsheet demonstrates the derivation of the expected utility formula for a simple gamble using the Von Neumann-Morgenstern (VNM) axioms. We will follow a step-by-step process, mirroring a typical slide presentation. **Given:** - Three outcomes: $a_1$ (best), $a_2$ (middle), $a_3$ (worst). - Normalized VNM utilities: $u(a_1)=1$, $u(a_2)=0.4$, $u(a_3)=0$. - A simple gamble: $g_s=(0.2\circ a_1,\;0.5\circ a_2,\;0.3\circ a_3)$. - This is of the form $g_s=(p_1\circ a_1,\dots,p_n\circ a_n)$, where $p_1=0.2$, $p_2=0.5$, $p_3=0.3$. ### Step 1: Define $q^i$ By definition, for each outcome $a_i$, we can find a simple gamble $q^i$ such that $a_i \sim q^i$, where $q^i$ is a lottery over the best and worst outcomes, $a_1$ and $a_n$ (in our case, $a_3$): $a_i \sim \big(u(a_i)\circ a_1,\;(1-u(a_i))\circ a_n\big)\equiv q^i$ Applying this to our outcomes: - For $a_1$: $a_1 \sim (1\circ a_1,\;0\circ a_3)\equiv q^1$ - For $a_2$: $a_2 \sim (0.4\circ a_1,\;0.6\circ a_3)\equiv q^2$ - For $a_3$: $a_3 \sim (0\circ a_1,\;1\circ a_3)\equiv q^3$ ### Step 2: Substitute $q^i$ into $g_s$ The VNM axioms (specifically, the Substitutability axiom) allow us to replace outcomes in a gamble with indifferent gambles. Given $g_s=(p_1\circ a_1,\dots,p_n\circ a_n)$, we can substitute each $a_i$ with its equivalent $q^i$: $g_s \sim (p_1\circ q^1,\dots,p_n\circ q^n)\equiv g'$ Using our numbers: $g_s=(0.2\circ a_1,\;0.5\circ a_2,\;0.3\circ a_3)$ becomes $g'=(0.2\circ q^1,\;0.5\circ q^2,\;0.3\circ q^3)$ Now, substitute the actual $q^i$ expressions: $$g'= \Big( 0.2\circ(1\circ a_1,0\circ a_3),\; 0.5\circ(0.4\circ a_1,0.6\circ a_3),\; 0.3\circ(0\circ a_1,1\circ a_3) \Big)$$ This $g'$ is a compound gamble. ### Step 3: Find the Simple Gamble Induced by $g'$ Using the VNM axiom of Reduction of Compound Gambles, we can simplify $g'$ into an equivalent simple gamble $g'_s$ over $a_1$ and $a_n$. The probability of ending up with $a_1$ in $g'$ is the sum of probabilities of reaching $a_1$ through each branch: $\sum_{i=1}^n p_i \cdot (\text{probability of } a_1 \text{ in } q^i) = \sum_{i=1}^n p_i u(a_i)$ And the probability of ending up with $a_n$ (which is $a_3$ in our case) is: $\sum_{i=1}^n p_i \cdot (\text{probability of } a_n \text{ in } q^i) = \sum_{i=1}^n p_i (1-u(a_i)) = 1 - \sum_{i=1}^n p_i u(a_i)$ Let's calculate $\sum_{i=1}^3 p_i u(a_i)$: $\sum_{i=1}^3 p_i u(a_i) = 0.2(1) + 0.5(0.4) + 0.3(0)$ $= 0.2 + 0.2 + 0 = 0.4$ So, the induced simple gamble is: $g'_s = (0.4\circ a_1,\; (1-0.4)\circ a_3) = (0.4\circ a_1,\; 0.6\circ a_3)$ ### Step 4: Use Reduction and Transitivity From the previous steps, we have: $g_s \sim g'$ (by Substitutability) $g' \sim g'_s$ (by Reduction of Compound Gambles) By the axiom of Transitivity, if $g_s \sim g'$ and $g' \sim g'_s$, then $g_s \sim g'_s$. Therefore: $g_s \sim (0.4\circ a_1,\;0.6\circ a_3)$ ### Step 5: Compare with the Definition of $u(g_s)$ The VNM theorem states that for any gamble $g_s$, there exists a unique utility value $u(g_s)$ such that $g_s$ is indifferent to a simple gamble over $a_1$ and $a_n$ with probabilities $u(g_s)$ and $1-u(g_s)$ respectively: $g_s \sim \big(u(g_s)\circ a_1,\;(1-u(g_s))\circ a_n\big)$ We found in Step 4 that: $g_s \sim (0.4\circ a_1,\;0.6\circ a_3)$ By comparing these two expressions, we conclude that: $u(g_s) = 0.4$ ### Step 6: Check the Expected Utility Formula The VNM expected utility formula states: $u(g_s) = \sum_{i=1}^n p_i u(a_i)$ Let's calculate this using our initial values: $u(g_s) = 0.2u(a_1) + 0.5u(a_2) + 0.3u(a_3)$ $u(g_s) = 0.2(1) + 0.5(0.4) + 0.3(0)$ $u(g_s) = 0.2 + 0.2 + 0$ $u(g_s) = 0.4$ This matches the result from Step 5. Thus, the boxed formula is confirmed: $$\boxed{u(g_s)=\sum_{i=1}^n p_i u(a_i)}$$ ### Summary of Proof Flow The proof can be summarized as follows: 1. **Define $q^i$**: $a_i \sim q^i = \big(u(a_i)\circ a_1,\;(1-u(a_i))\circ a_n\big)$ 2. **Substitute**: $g_s=(p_1\circ a_1,\dots,p_n\circ a_n)\sim (p_1\circ q^1,\dots,p_n\circ q^n)=g'$ 3. **Reduce to Simple Gamble**: $g'_s = \left( \left(\sum_{i=1}^n p_i u(a_i)\right)\circ a_1,\; \left(1-\sum_{i=1}^n p_i u(a_i)\right)\circ a_n \right)$ 4. **Transitivity**: $g_s\sim g'\sim g'_s$ 5. **Compare with Definition**: We know $g_s\sim \big(u(g_s)\circ a_1,\;(1-u(g_s))\circ a_n\big)$ 6. **Conclusion**: By comparison, $\boxed{u(g_s)=\sum_{i=1}^n p_i u(a_i)}$