JaynesProbabilityTheory/chapter13.html at master · MaksimIM/JaynesProbabilityTheory · GitHub

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<h1 id="decision-theory-historical-background.">Decision theory, historical background.</h1>
<p><span class="math inline">\(\leftarrow\)</span> <a href="./index.html">Back to Chapters</a></p>
<h3 id="exercise-13.1">Exercise 13.1</h3>
<p>We interpret this problem as follows: Starting from current fortune <span class="math inline">\(M\)</span> one has to bet unknown amount <span class="math inline">\(N\)</span> which with probability <span class="math inline">\(p&gt;0.5\)</span> will yield success (and new fortune of <span class="math inline">\(M+N\)</span>) and <span class="math inline">\(1-p\)</span> will yield failure (and new fortune of <span class="math inline">\(M-N\)</span>). The question is to find <span class="math inline">\(y\)</span> which maximizes expected log of the new fortune.</p>
<p>Thus the expected log fortune is</p>
<p><span class="math display">\[p  \log (M+N)+ (1-p)\log(M-N)\]</span></p>
<p>Maximizing this in <span class="math inline">\(N\)</span> yields</p>
<p><span class="math display">\[\frac{p}{M+N}=\frac{1-p}{M-N}.\]</span></p>
<p>Denoting <span class="math inline">\(r=\frac{N}{M}\)</span> we get <span class="math inline">\(p(1-r)=(1-p)(1+r)\)</span>, or <span class="math inline">\(p-rp=1-p+r-rp\)</span>, and <span class="math inline">\(r=2p-1\)</span>.</p>
<p>The expected log fortune is then</p>
<p><span class="math display">\[\log M+ p \log 2p +(1-p) \log 2(1-p)=\log M + H_{0.5}-H_p,\]</span></p>
<p>where <span class="math inline">\(H_p\)</span> is the entropy of the entropy of the Bernoulli distribution with parameter <span class="math inline">\(p\)</span>.</p>
<p>(The fact that this is higher than <span class="math inline">\(\log M\)</span> which is what one would obtain at <span class="math inline">\(N=0\)</span> (and of course higher than <span class="math inline">\(-\infty\)</span> that one would obtain at <span class="math inline">\(N=1\)</span>) also shows that this is a global maximum (as opposed to just local stationary point), which we could’ve verified by second derivative test, but were too lazy to do.)</p>
<p>After <span class="math inline">\(n\)</span> iterations of this the expected log fortune is <span class="math inline">\(\log M +n(H_{0.5}-H_p)\)</span>, and the exponent of that is <span class="math inline">\(M\exp\{n \alpha\}\)</span> with <span class="math inline">\(\alpha=H_{0.5}-H_p\)</span>, as per the exercise.</p>
<p>Demonstration that this is actually the best (with respect to expected log fortune) strategy not only for single step, but over any number <span class="math inline">\(n\)</span> of steps is via dynamic programming aka Bellman optimality principle, and can be found in Bellman-Kalaba, “On the Role of Dynamic Programming in Statistical Communication Theory”, Sections VII-IX.</p>
<h3 id="comments-on-13.6">Comments on 13.6</h3>
<p>Some of this seems strange.</p>
<ol type="1">
<li><p>If we reformulate St. Petersburg paradox to pay out in utility rather than money, some of the same issues remain. Of course, it may be that these are not the issues Jaynes cares about, but, then, maybe they are the ones that Feller and others care about – so much so that the observation of non-linear utility of money does nothing to resolve their concerns with this paradox.</p></li>
<li><p>“Admissible” does not mean “good”, it just means others are bad. So this line of criticism seems moot. (And besides, if our prior information is a certainty that <span class="math inline">\(\theta=5\)</span> than the strategy based on the estimate <span class="math inline">\(\theta^*=5\)</span> seems perfect.)</p></li>
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