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<h2 class="hd hd-2 unit-title">The Principle of Countable Additivity</h2>
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<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"> Some philosophers and mathematicians think that the (finite) Additivity principle we considered in Section 6.2.1 needs to be strengthened to the following:</span></p>
<p style="padding-left: 30px;"><span style="font-family: 'book antiqua', palatino;"><strong>Countable Additivity</strong></span><br /><span style="font-family: 'book antiqua', palatino;">Let <span class="math inline">\(A_1,A_2,A_3,\dots\)</span> be a countable list of propositions, and suppose that <span class="math inline">\(A_i\)</span> and <span class="math inline">\(A_j\)</span> are incompatible whenever <span class="math inline">\(i \neq j\)</span>. Then: <span class="math display">\[p\left(A_1 \mbox{ or } A_2 \mbox{ or } A_3 \mbox{ or } \ldots\right) = p(A_1) + p(A_2) + p(A_3) + \ldots\]</span></span></p>
<p><span style="font-family: 'book antiqua', palatino;">There are two main reasons to hope for the adoption of Countable Additivity. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">The first is that it is needed to prove a number of important results in probability theory, including certain versions of the Law of Large Numbers. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">The second is that probability functions that fail to be countably additive can have mathematically awkward properties. We’ll consider one of these properties below. We’ll also talk about the fact that adopting Countable Additivity is not without costs. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">My own view—and that of many others—is that <em>on balance</em> the advantages outweigh the costs. But it is important to be clear that the issue is not entirely straightforward.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">(Incidentally, why stop with <em>countable</em> Additivity? Why not go all the way, and require probability distributions to be additive with respect to <em>any</em> infinite set of mutually exclusive propositions, regardless of cardinality? The answer is that the resulting mathematics is not nice. Consider, for example, a dart on a random trajectory to the unit interval, [mathjaxinline] [0, 1]. [/mathjaxinline] A principle of uncountable additivity would entail that, for some number [mathjaxinline]r \in [0,1][/mathjaxinline], the probability that the dart lands on [mathjaxinline]r[/mathjaxinline] is [mathjaxinline]x[/mathjaxinline] for some [mathjaxinline]x > 0[/mathjaxinline], which would be an awkward result, since [mathjaxinline]r[/mathjaxinline] is smaller than a subinterval of [mathjaxinline][0,1][/mathjaxinline] of size [mathjaxinline]x[/mathjaxinline]. We’ll return to this sort of issue in Lecture 7.)</span></p>
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<h2 class="hd hd-2 unit-title">Life with Countable Additivity</h2>
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<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"> In this section I'll try to give you a sense of why the issue is not entirely straightforward. I’ll start by mentioning an awkward consequence of <em>accepting</em> Countable Additivity. Then I'll point to an awkward consequence of <em>not accepting</em> Countable Additivity.</span></p>
<h4><span style="font-family: 'book antiqua', palatino;">An awkward consequence of accepting Countable Additivity</span></h4>
<p><span style="font-family: 'book antiqua', palatino;">Imagine that God has selected a positive integer, and that you have no idea which. For <span class="math inline">\(n\)</span> a positive integer, what credence should you assign to the proposition, <span class="math inline">\(G_n\)</span>, that God selected <span class="math inline">\(n\)</span>? </span></p>
<p><span style="font-family: 'book antiqua', palatino;">Countable Additivity entails that your credences should remain undefined, unless you’re prepared to give different answers for different choices of <span class="math inline">\(n\)</span>. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">To see this, suppose otherwise. Suppose that for some real number <span class="math inline">\(r\)</span> between 0 and 1, you assign credence <span class="math inline">\(r\)</span> to each proposition <span class="math inline">\(G_n\)</span>.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">What real number could <span class="math inline">\(r\)</span> be? It must either be equal to zero or greater than zero. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">First, suppose that <span class="math inline">\(r\)</span> is equal to zero. Then Countable Additivity entails: <span class="math display">\[c_S(G_1 \text{ or } G_2 \text{ or } G_3 \text{ or } \ldots) = c_s(G_1) + c_s(G_2) + c_s(G_3) + \ldots = \underbrace{0 + 0 + 0 + 0 \ldots}_{\text{once for each integer}} = 0\]</span> But <span class="math inline">\(G_1 \text{ or } G_2 \text{ or } G_3 \text{ or } \ldots\)</span> is just the proposition that God selects some positive integer. So we would end up with the unacceptable conclusion that you’re certain that God won’t select a positive integer after all.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Now suppose that <span class="math inline">\(r\)</span> is greater than zero. Then Countable Additivity entails: <span class="math display">\[c_S(G_1 \text{ or } G_2 \text{ or } G_3 \text{ or } \ldots) = c_s(G_1) + c_s(G_2) + c_s(G_3) + \ldots = \underbrace{r + r + r + r \ldots}_{\text{once for each integer}} = \infty\]</span> But since probabilities are always real numbers between 0 and 1, this contradicts the assumption that <span class="math inline">\(c_S\)</span> is a probability function.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">The moral is that when Countable Additivity is in place there is no way of assigning probabilities to the <span class="math inline">\(G_n\)</span>, unless one is prepared to assign different probabilities to different <span class="math inline">\(G_n\)</span>. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">More generally: Countable Additivity entails that there is no way of distributing probability <em>uniformly</em> across a countably infinite set of (mutually exclusive and jointly exhaustive) propositions.</span></p>
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<span style="font-family: 'book antiqua', palatino;">As before, God has selected a number. But this time your credences are as follows:</span>
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<span style="font-family: 'book antiqua', palatino;">Your credence that God selected the number <span class="math inline">\(1\)</span> is <span class="math inline">\(1/2\)</span>, your credence that God selected the number <span class="math inline">\(2\)</span> is <span class="math inline">\(1/4\)</span>, your credence that God selected the number <span class="math inline">\(3\)</span> is <span class="math inline">\(1/8\)</span>, and so forth. (In general, your credence that God selected positive natural number <span class="math inline">\(n\)</span> is <span class="math inline">\(1/2^{n}\)</span>.)</span>
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<span style="font-family: 'book antiqua', palatino;">Assuming your credence function satisfies Countable Additivity, what is your credence that God selected a natural number?</span>
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<h2 class="hd hd-2 unit-title">Infinitesimals to the rescue?</h2>
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<p><span style="font-family: 'book antiqua', palatino;">One might be tempted to think that there is an easy way out of this difficulty. Perhaps we could reject the assumption that a probability must be a real number between 0 and 1, and instead allow for <em>infinitesimal</em> probabilities: probabilities that are greater than zero but smaller than every positive real number.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">On the face of it, bringing in infinitesimal values could make all the difference. For suppose we had an infinitesimal value <span class="math inline">\(\iota\)</span> with the following property: <span class="math display">\[\underbrace{\iota + \iota + \iota + \iota + \iota + \ldots}_{\text{once for each natural number}} = 1\]</span> Now assign probability <span class="math inline">\(\iota\)</span> to each proposition, <span class="math inline">\(G_n\)</span>, that God has selected number <span class="math inline">\(n\)</span>. Doesn’t that mean that we will have assigned the same probability to each <span class="math inline">\(G_n\)</span> without having to give up on Countable Additivity?</span></p>
<p><span style="font-family: 'book antiqua', palatino;">It is not clear that we have. The problem is not to do with talk of infinitesimals, which were shown to allow for rigorous theorizing by mathematician Abraham Robinson. The problem is to do with what we hope infinitesimals might do for us in the present context. For it is hard to see how they could help us theorize about a uniform probability distributions on a countably infinite set unless infinite sums of infinitesimals are defined using limits, in the usual way: <br />\[\underbrace{\iota + \iota + \iota + \iota + \iota + \ldots}_{\text{once for each natural number}} = \lim_{n \to \infty} n \cdot \iota\] But this leads to the conclusion that \(\iota + \iota + \iota + \iota + \iota + \ldots\) cannot equal 1, as required in the present context. For on any standard treatment of infinitesimals we have \(n \cdot \iota < \frac{1}{m}\) for any positive integers \(n\) and \(m\). And from this it follows that \(\lim_{n \to \infty} n \cdot \iota\) cannot converge to a positive real number.<br /> <br />What if we were to sidestep standard treatments of infinitesimals altogether? Suppose we introduce a new quantity <span class="math inline">\(\iota\)</span> with the double stipulation that <span class="math inline">\((a)\)</span> <span class="math inline">\(\iota\)</span> is greater than zero but smaller than any positive real number, and <span class="math inline">\((b)\)</span> the result of adding <span class="math inline">\(\iota\)</span> to itself countably many times is equal to 1. Even if the details of such a theory could be successfully spelled out, it is not clear that it would put us in a position to assign the same probability to each <span class="math inline">\(G_n\)</span> without giving up on Countable Additivity. For suppose we assume both that <span class="math inline">\(p(G_n) = \iota\)</span>, for each <span class="math inline">\(n\)</span>, and that Countable Additivity is in place. Then we are left with the unacceptable conclusion that God’s selecting an even number and God’s selecting an odd number are both events of probability 1. Here is a proof of one half of that assertion: <span class="math display">\[\begin{aligned} p(G_1 \text{ or } G_3 \text{ or } G_5 \text{ or }\ldots) &= p(G_1) + p(G_3) + p(G_5) \ldots & \text{[Countable Additivity]}\\ &= \underbrace{\iota + \iota + \iota + \iota + \iota + \ldots}_{\text{once for each natural number}} & \text{[$p(G_n) = \iota$]}\\ &= 1 & \text{[stipulation $(b)$]}\end{aligned}\]</span> I have not shown that it is impossible to use infinitesimals to construct a probability theory that allows for both Countable Additivity and a uniform probability distribution over a countably infinite set of mutually exclusive propositions. But I hope to have made clear why the prospects of doing so are not as rosy as they might appear.<br /></span></p>
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<h2 class="hd hd-2 unit-title">An awkward consequence of rejecting Countable Additivity</h2>
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<p><span style="font-family: 'book antiqua', palatino;">Wouldn’t it be better to give up on Countable Additivity altogether? </span></p>
<p><span style="font-family: 'book antiqua', palatino;">By giving up Countable Additivity, one would be free to assign credence 0 to each of a countable set of mutually exclusive propositions. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">For example, one could assign credence 0 to each of our propositions <span class="math inline">\(G_n\)</span>. (Keep in mind that saying that assigning an event credence 0 is not the same as taking the event to be impossible. It is just to say that your degree of belief is so low that no positive real number is small enough to measure it.)</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Unfortunately, giving up Countable Additivity can lead to a theory with undesirable mathematical properties. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">Here is an example, which I borrow from Kenny Easwaran. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">Let <span class="math inline">\(X\)</span> and <span class="math inline">\(Y\)</span> be sets of positive integers. Let <span class="math inline">\(p(X)\)</span> be the probability that God selects a number in <span class="math inline">\(X\)</span>, and let <span class="math inline">\(p(X|Y)\)</span> be the probability that God selects a number in <span class="math inline">\(X\)</span> given that She selects a number in <span class="math inline">\(Y\)</span>. One might think that the following is an attractive way of characterizing <span class="math inline">\(p(X)\)</span> and <span class="math inline">\(p(X|Y)\)</span>:</span></p>
<p><span style="font-family: 'book antiqua', palatino;"><span class="math display">\[p(X|Y) =_{df} \lim \limits_{n \to \infty}\frac{|X\cap Y \cap \{1,2,\ldots,n\}|}{|Y\cap\{1,2,\ldots,n\}|}\]</span> <span class="math display">\[p(X) =_{df} p(X|\mathbb{Z}^+)\]</span> (Notice that <span class="math inline">\(p(X)\)</span> is finitely additive but not countably additive, since <span class="math inline">\(p(\mathbb{Z}^+)=1\)</span> but <span class="math inline">\(p(\{k\}) = 0\)</span> for each <span class="math inline">\(k\in\mathbb{Z}^+\)</span>. Notice also that <span class="math inline">\(p\)</span> is not well-defined for arbitrary sets of integers, since <span class="math inline">\(\lim \limits_{n \to \infty}\dfrac{|X\cap\{1,2,\ldots,n\}|}{|\{1,2,\ldots,n\}|}\)</span> is not generally well-defined; for instance, it is not well-defined when <span class="math inline">\(X\)</span> consists of the integers <span class="math inline">\(k\)</span> such that <span class="math inline">\(2^m \leq k < 2^{m+1}\)</span>, for some even <span class="math inline">\(m\)</span>.)</span></p>
<p><span style="font-family: 'book antiqua', palatino;">As it turns out, <span class="math inline">\(p(X)\)</span> has the following awkward property: there is a set <span class="math inline">\(S\)</span> and a partition <span class="math inline">\(E_i\)</span> of <span class="math inline">\(\mathbb{Z}^+\)</span> such that <span class="math inline">\(p(S) = 0\)</span> even though <span class="math inline">\(p(S|E_i) \geq 1/2\)</span> for each <span class="math inline">\(E_i\)</span>. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">For example, let <span class="math inline">\(S\)</span> be the set of perfect squares, and for each <span class="math inline">\(i\)</span> which is not a power of any other positive integer, let <span class="math inline">\(E_i\)</span> be the set of powers of <span class="math inline">\(i\)</span>. In other words: <span class="math display">\[\begin{array}{ccc} S &= &\{1, 4, 9, 16, 25, \dots\} \\ E_1 &= &\{1\} \\ E_2 &= &\{2, 4, 8, 16, 32, \dots\}\\ E_3 &= &\{3, 9, 27, 81, 243, \dots\}\\ \text{[No $E_4$, since $4 = 2^2$]} \\ E_5 &= &\{5, 25, 125, 625, 3125, \dots\}\\ &\vdots \end{array}\]</span> It is easy to verify that <span class="math inline">\(p(S|E_1) = 1\)</span> and <span class="math inline">\(p(S|E_n) = 1/2\)</span> for each <span class="math inline">\(n > 1\)</span>, even though <span class="math inline">\(p(S) = 0\)</span>.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">To see why this is awkward, imagine that Susan’s credences are given by <span class="math inline">\(p\)</span>. Then Susan can be put in the following situation: there is a sequence of bets such that Susan thinks she ought to take each of the bets, but such that she believes to degree 1 that she will lose money if she takes them all. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">Here is one way of spelling out the details. For each <span class="math inline">\(E_i\)</span> Susan is offered the following bet:</span></p>
<p style="padding-left: 30px;"><span style="font-family: 'book antiqua', palatino;"><strong><span class="math inline" style="font-family: 'book antiqua', palatino;">\(\mathbf{B_{E_i}:}\)</span></strong></span><br /><span style="font-family: 'book antiqua', palatino;">Suppose God selects a number in <span class="math inline">\(E_i\)</span>. Then you’ll receive $2 if the selected number is in <span class="math inline">\(S\)</span>, and you’ll be forced to pay $1 if the selected number is not in <span class="math inline">\(S\)</span>. (If the selected number is not in <span class="math inline">\(E_i\)</span>, then the bet is called off, and no money exchanges hands.)</span></p>
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<p><span style="font-family: 'book antiqua', palatino;">We know that <span class="math inline">\(p(S|E_i) \geq 1/2\)</span> for each <span class="math inline">\(E_i\)</span>. So the expected value of <span class="math inline">\(B_{E_i}\)</span> on the assumption that the selected number is in <span class="math inline">\(E_i\)</span> will always be positive. (It is at least <span class="math inline">\(\$2 \cdot 1/2 -\$1 \cdot 1/2 = \$0.5\)</span>.) So Susan should be willing to accept <span class="math inline">\(B_{E_i}\)</span> for each <span class="math inline">\(E_i\)</span>. Notice, however, that Susan believes to degree 1 that if she does accept all such bets, she’ll loose money. For Susan believes to degree 1 that God will select a number <span class="math inline">\(x\)</span> outside <span class="math inline">\(S\)</span> (<span class="math inline">\(p(S)=0\)</span>). And if <span class="math inline">\(x\)</span> is outside <span class="math inline">\(S\)</span> then Susan looses $1 on bet <span class="math inline">\(B_{E^x}\)</span> (where <span class="math inline">\(E^x\)</span> is the <span class="math inline">\(E_i\)</span> such that <span class="math inline">\(x \in E_i\)</span>), with no money exchanging hands on any other bet.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">As it turns out, problems of this general form are inescapable: they will occur whenever a probability function on a countable set of possibilities fails to be countably additive. (See Section 6.6 for further details.) This might lead one to think that rejecting Countable Additivity is less attractive than one might have thought: it is true that doing so allows one to distribute probability uniformly across a countable set of mutually exclusive propositions, but the sorts of distributions one gets are not especially attractive.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">My own view is that, on balance, it is best to accept Countable Additivity. But it is important to be clear that the issues here are far from straightforward.</span></p>
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<h3 class="hd hd-2">Video Review: Rejecting Countable Additivity</h3>
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