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<h1 id="lecture-12">Lecture 12</h1>
<h2 id="检测evolution-stable例子">检测evolution stable例子</h2>
<p>收益矩阵：</p>
<table>
<thead>
<tr>
<th style="text-align:center"></th>
<th style="text-align:center">a</th>
<th style="text-align:center">b</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align:center">a</td>
<td style="text-align:center">1, 1</td>
<td style="text-align:center">1, 1</td>
</tr>
<tr>
<td style="text-align:center">b</td>
<td style="text-align:center">1, 1</td>
<td style="text-align:center">0, 0</td>
</tr>
</tbody>
</table>
<p>(a, a)是其中的一个纳什均衡点，但是并不是strict NE，因此a并不是evolution stable策略。</p>
<p>u(a, a) = u(b, a) = 1</p>
<p>下一步，我们验证u(a, b) = 1和u(b, b) = 0的大小关系，发现u(a, b)>u(b, b)因此a是evolution stable的。</p>
<h2 id="evolution-of-social-convention-drving-on-the-left-or-right">Evolution of social convention: drving on the left or right</h2>
<table>
<thead>
<tr>
<th></th>
<th>L</th>
<th>R</th>
</tr>
</thead>
<tbody>
<tr>
<td>L</td>
<td>2, 2</td>
<td>0, 0</td>
</tr>
<tr>
<td>R</td>
<td>0, 0</td>
<td>1, 1</td>
</tr>
</tbody>
</table>
<p> (L, L) (R, R)都是strict 纳什均衡的,因此都是evolution stable。</p>
<p>lesson:</p>
<p>从上面开车的结果来看，社会上可以存在着多种的evolution stable。并不一定需要效率最高。</p>
<h2 id="另一个例子">另一个例子:</h2>
<table>
<thead>
<tr>
<th></th>
<th>a</th>
<th>b</th>
</tr>
</thead>
<tbody>
<tr>
<td>a</td>
<td>0, 0</td>
<td>2, 1</td>
</tr>
<tr>
<td>b</td>
<td>1, 2</td>
<td>0, 0</td>
</tr>
</tbody>
</table>
<p>该博弈赋予的含义：两个车相向而行，a表示暴力往前开，b表示掉头开。</p>
<p>(a, b)和(b, a)都是纳什均衡点，但是他们是不对称的（我们之前的所有有关evolution stable的研究都是对称博弈）。</p>
<p><strong>没有一个纯的策略，使得该策略在整个族群中稳定的保存。因此我们必须开始考虑混合策略了(mixed strategy)。</strong></p>
<p>在lecture 10有关约会的例子中，NE = [(2/3, 1/3), (1/3, 2/3)]是一种混合策略的纳什均衡。</p>
<p>在这个例子中，也就是一个种群中，有2/3的生物是好斗的，而1/3的生物是温顺的。</p>
<p>实际上，在自然界中确实存在多样的选择，也就是生物的多态性，这就是为什么我们接下来讨论mixed strategy 是合理的。</p>
<p>有趣的例子：SLF：sneaky little fucker</p>
<h2 id="definition-of-mixed-strategy-in-evolution">definition of mixed strategy in evolution</h2>
<p>在混合纳什均衡中，<strong>他们的值不是严格最大的</strong>。因为不论怎么偏移两种选择之间的概率，他们的混合策略期望都是不变的。</p>
<p>正是有了这一条，我们在lecture 11最后的关于evolution stable的第二个定义中，我们需要验证<script type="math/tex; ">u(\hat p, p^{'}) > u(p^{'}, p^{'})</script>。（因为第一个验证条件必定是满足的）</p>
<h2 id="例子：-鹰鸽之争">例子： 鹰鸽之争</h2>
<table>
<thead>
<tr>
<th style="text-align:center"></th>
<th style="text-align:center">H</th>
<th style="text-align:center">D</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align:center">H</td>
<td style="text-align:center"><script type="math/tex; ">\frac{v-c}{2}, \frac{v-c}{2}</script></td>
<td style="text-align:center">v, 0</td>
</tr>
<tr>
<td style="text-align:center">D</td>
<td style="text-align:center">0, v</td>
<td style="text-align:center"><script type="math/tex; ">\frac{v}{2}, \frac{v}{2}</script></td>
</tr>
</tbody>
</table>
<p>prize = v > 0</p>
<p>costs of fight = c > 0</p>
<p>is D evolution stable strategy(ESS)?</p>
<p>首先我们要搞明白(D, D)是否是纳什均衡的的？并不是的。</p>
<hr>
<p>那么(H, H)是ESS吗？</p>
<p>如果<script type="math/tex; ">\frac{v-c}{2}\ge 0</script>，那么(H, H)就是纳什均衡的。</p>
<p>如果<script type="math/tex; ">\frac{v-c}{2}\gt 0</script>，那么(H, H)就是严格纳什均衡的。那么此刻肯定是ESS。</p>
<p>下面讨论v = c的情况：</p>
<p>重要：</p>
<blockquote>
<p> 回顾一下u(D, H)的含义：这个值表示这个策略组合的期望，并且主角应该是使用了H的，入侵者是D。</p>
<p>玩家1使用了D，玩家2使用了H，计算玩家1的期望。</p>
<p><script type="math/tex; ">u(H, \hat p) </script>表示玩家2使用了组合策略（H, D），并且概率是<script type="math/tex; ">(p, 1-p)</script>，因此计算方法为<script type="math/tex; ">u(H, \hat p) = payoffs(H, H)*p+payoffs(H, D)*(1-p)</script>.</p>
<p>后面有关于鹰和鸽子的例题就是很好的例子。</p>
</blockquote>
<p>也就是u(H, H) = u(D, H) = 0 （Dove进来没有用）</p>
<p>并且u(H, D) = v > u(D, D) = <script type="math/tex; ">\frac{v}{2}</script> （有利的方向朝着Hawk不断进行发展）</p>
<p>综上，H是ESS。</p>
<hr>
<p>如果<script type="math/tex; ">c>v</script>呢？</p>
<p>我们知道H不是ESS</p>
<p>D不是ESS</p>
<p>那么混合 情况下呢？</p>
<table>
<thead>
<tr>
<th style="text-align:center"></th>
<th style="text-align:center">H</th>
<th style="text-align:center">D</th>
<th></th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align:center">H</td>
<td style="text-align:center"><script type="math/tex; ">\frac{v-c}{2}, \frac{v-c}{2}</script></td>
<td style="text-align:center">v, 0</td>
<td><script type="math/tex; ">\hat p</script></td>
</tr>
<tr>
<td style="text-align:center">D</td>
<td style="text-align:center">0, v</td>
<td style="text-align:center"><script type="math/tex; ">\frac{v}{2}, \frac{v}{2}</script></td>
<td><script type="math/tex; ">1-\hat p</script></td>
</tr>
<tr>
<td style="text-align:center"></td>
<td style="text-align:center"><script type="math/tex; ">\hat p</script></td>
<td style="text-align:center"><script type="math/tex; ">1-\hat p</script></td>
</tr>
</tbody>
</table>
<p>首先找到混合策略的纳什均衡点：</p>
<p>纯使用H的期望:<script type="math/tex; ">u(H, \hat p) = \frac{v-c}{2}*\hat p+v*(1-\hat p)</script></p>
<p>纯使用D的期望：<script type="math/tex; ">u(D, \hat p) = 0*\hat p+\frac{v}{2}*(1-\hat p)</script></p>
<p>令两者相等：<script type="math/tex; ">\hat p = \frac{v}{c}</script></p>
<p>因此NE = <script type="math/tex; ">(\frac{v}{c}, 1-\frac{v}{c})</script></p>
<p>混合的策略都不是strict NE。</p>
<p>因此我们要检查<script type="math/tex; ">(\hat p, p^{'})\gt (p^{'}, p^{'}) </script>，对任意的<script type="math/tex; ">p^{'}</script>都成立</p>
<p>带入上面的值，因此鸽派的混合收益为<script type="math/tex; ">\frac{v}{2}(1-\frac{v}{c}) < \frac{v}{2}</script></p>
<p>lesson:</p>
<p>如果<script type="math/tex; ">c>v</script>那么老鹰的比例为<script type="math/tex; ">\frac{v}{c}</script>:</p>
<p>a) 当v增加的时候，斗争的利益放大，在ESS中老鹰的比例更大。</p>
<p>同理，当c增加的时候，鸽子的比例更大。</p>
<p>b) <script type="math/tex; ">payoffs = \frac{v}{2}(1-\frac{v}{c})</script></p>
<p>当c增加的时候，payoffs是增加的！斗争看来是不好的。</p>
<p>c) 验证：在实际的实验中，我们能够统计出<script type="math/tex; ">\frac{v}{c}</script>的值。</p>
<p><strong>上面的例子主要就是为了讲mixed strategy在evolution stable strategy中的应用。</strong></p>
<h2 id="剪刀石头布在自然界中的应用">剪刀石头布在自然界中的应用</h2>
<p>通过验证剪刀石头布在NE的情况下，并不存在ESS状态。</p>
<p><strong>会进行一个此消彼长的循环状态。</strong></p>
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