最大收益 II

收益为 $G(c)=c[1-2F(c)]$。一阶条件\n\n$$\nG'(c)=1-2F(c)-2c f(c)=0\\Rightarrow \\boxed{1-2F(c)=2c f(c)}.\n$$\n\n指数分布 $F(c)=1-e^{-\\lambda c}$ 下方程为\n\n$$\n2e^{-\\lambda c}(1-\\lambda c)=1,\n$$\n\n数值解 $c\\approx 0.315/\\lambda$，最大期望约 $0.145/\\lambda$。

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英文解析

a. With the strategy c, the gain (actually, loss) will be equal to - c with probability $\mathrm{P(U< c) = F(c)}$ , and it will be equal to $+\mathrm{c}$ with probability $\mathrm{P(U > c) =}$ 1- F(c). Weighing these two cases by their respective probabilities, we get

$$\begin{array}{l}{G(c) = -c\cdot F(c) + c\cdot [1 - F(c)]}\\ {\quad = c\cdot [1 - 2F(c)]} \end{array}$$

Figure 3.6 shows the qualitative behavior of G.

Differentiating G with respect to c and setting this derivative equal to zero yields the optimum condition

$$\frac{dG(c)}{dc} = 1 - 2[cf(c) + F(c)] = 0$$

Figure 6.2: The relation between the expected gain G (c) and the number c that was chosen. The example assumes that U has an exponential distribution f $(\mathbf{t}) = \mathbf{A}\mathbf{e}\mathbf{x}\mathbf{p}$ $(- \lambda \mathbf{t})$ ，with mean $1 / \lambda = 50.$ The optimal choice of c is then 15.75，in which case the expected gain equals 7.24.

b.Note from the form of G (c) that if c is smaller than the median of U, then G (c) is positive. Therefore, the global maximum of G cannot be negative.

c.For an exponential rv with rate $\lambda$ ，the optimum condition yields

$$2e^{-\lambda c}(1 - \lambda c) = 1$$

The solution to this equation is $\mathrm{c} = 0.315 / \mathrm{A}$ . Compared to Problem 1.9, the less favorable rules of the current game clearly suggest a rather more cautious strategy; compare Figure 3.6 to Figure 3.5. Correspondingly, the expected gain with the optimal strategy decreases to just $0.145 / \lambda$ ，roughly $40\%$ of what the player could expect under the rules of the previous problem. On the other hand, although less attractive, the game should still be accepted by the player, because its expected gain is positive when the optimal strategy is chosen.