A better statement of jwmerril&#x27;s razor: points drawn from a probability distribution have a higher likelihood than points coming from far away.<p>I don&#x27;t really know why you <i>don&#x27;t</i> think that the asymptotic forms are evidence in favor of this - a prototypical probability distribution on the real line is a bump somewhere with a decaying tail. And that &quot;somewhere&quot; is far closer to the origin than points out in some arbitrarily distant tail.<p>Now obviously if you want to make stronger claims about a specific origin, you&#x27;ll need to specify a particular probability distribution, and justify why that&#x27;s the right one. I agree that a non-asymptotic Occams razor is an additional assumption.<p>But you also get pretty far with the asymptotic theory. Consider a theory of &quot;green&quot; as compared to a theory of &quot;bleen&quot; (namely that green turns to blue after some time T). You have a prior with some probability that only green exists (say 50%), and also a 50% chance that green turns to blue after some time T. But now you have a continuous distribution over T.<p>Now suppose you want to make a prediction - e.g., H = &quot;the grass will be green, not blue, at t=50&quot;. When you compute a posterior, you reject all values of T &lt; 0 (supposing the present time is 0). Also, all values of T &gt; 50 actually yield the same prediction as &quot;only green exists&quot;. So the only way you can get a prediction of blue at time 50 is if 0 &lt; T &lt; 50. Of course, the more time you spend gathering data, the further into the tail you move and the less likely it is that your posterior will predict blue. I.e., Bayesian stats even with very few assumptions gets sensible results eventually.<p>I do in fact hold the view that Bayesian probability is a consistent theory of the scientific method, and also of how humans should update their beliefs when new evidence is gathered.<p>(Minor nit: your continuous asymptotic form isn&#x27;t slightly wrong for this purpose, f(x) need not approach zero. Counterexample: f(x) = 1 for x \in [1, 1+2^{-1}], [2, 2+2^{-2}], etc, f(x) = 0 elsewhere. That integrates out to 1&#x2F;2 + 1&#x2F;4 + ... = 1, but lim_{x -&gt; \infty} f(x) doesn&#x27;t exist.)<p>[I&#x27;m also a bit surprised you are being so heavily downvoted. I don&#x27;t think you are right, but you are hardly so crazily wrong that you should be greyed out.]