Using the notation of Exercise 4, assume that \(\sum_{i=1}^{n} 1 / r_{i}=1\), but try to find a solution where one of the \(\alpha_{k}\) 's is zero. In particular, suppose the segments are ordered in such a way that \(p_{n} r_{n}

(a) Find a solution with \(\alpha_{n}=0\) and all other \(\alpha_{i}\) 's positive.

(b) Evaluate this solution for the wheel of Example 18.5.

Data from Exercises 4

Consider a wheel with \(n\) sectors. If the wheel pointer lands on sector \(i\), the payoff obtained is \(r_{i}\) for every unit bet on that sector. The chance of landing on sector \(i\) is \(p_{i}, i=1,2, \ldots, n\). Let \(\alpha_{i}\) be the fraction of one's capital bet on sector \(i\). We require \(\sum_{i=1}^{n} \alpha_{i} \leq 1\) and \(\alpha_{i} \geq 0\) for \(i=1,2, \ldots, n\).

Example

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Example 18.5 (The investment wheel) Let us compute the full optimal strategy for the investment wheel allowing for the possibility of investing on all sectors. For a strategy (a1, a2, 3) we find the results as follows: 1. If 1 occurs, R=1+2--3. 2. If 2 occurs, R = 1-1+-3. 3. If 3 occurs, R = 1-1-+53. To maximize the expected logarithm of this return structure, we maximize m=ln(1+21-2-3)+ln(1-1 +2-3)+ln(1-1-+53). If we assume that the solution has a; > 0 for each i = 1,2,3, we can find the solution by setting the derivatives with respect to each a; equal to zero. This gives the equations 2 1 1 3(1-1 +2-3) 6(1 1 2+53) 0 2(1+2-2-3) 1 1 1 + =0 2(1+21-a2-3) 3(1-1+-3) 6(1-1-a2+503) 1 1 5 + 0. 2(1+2a1-a2-3) 3(1-1+a2-03) 6(1-1 2+503)