Comments by "EebstertheGreat" (@EebstertheGreat) on "Euler’s Pi Prime Product and Riemann’s Zeta Function" video.

  1. z is a "constant" here in the sense that it is the same for every term. Both sides of the equation are just being multiplied by 1/3^z, whatever z happens to be. That term will still distribute over the series. In the same way, since e^x = 1 + x + x²/2 + x³/3! + · · · + xⁿ/n! + · · · , that means e^x/x = 1/x + 1 + x/2 + x²/3! + · · · + xⁿ⁻¹/n! + · · · .
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  2. There are indeed no uniform probability distributions on N that satisfy the usual axioms of probability distributions. However, there are some that exist by weakening one or more axioms, usually countable additivity. Countable additivity says that the probability of any countable set of disjoint events is equal to the sum of the probabilities of each event on its own. So for instance, in a Poisson distribution on a variable X where P(X=n) = 1/(e·n!), we can say that the probability that X is even is P(X even) = P(X=0) + P(X=2) + P(X=4) + · · · = 1/e (1 + 1/2 + 1/4! + · · · ) = cosh(1). This is sensible and does lead to the conclusion you posted above: that such a distribution over all natural numbers cannot be uniform. After all, if it were uniform, P(n) would be 0 for each n, yet P(n≥0) must be 1, which is definitely not the sum of 0 + 0 + 0 + · · ·. By weakening the requirement of countable additivity to finite additivity, we can define a generalized probability distribution (sometimes called a "probability charge") that does have most of the properties we want. Specifically, it will still be true that P(X=n) = 0 for all n, but it will no longer be required that the total probability be the sum of 0 + 0 + 0 + · · ·. Instead, the total probability can simply be 1, and the probabilities of other countably infinite sets of outcomes can similarly be larger than expected by conventional rules. Because we still require finite additivity, that means that finite sums work normally. For instance, P(X is even) + P(X is odd) = P(X is even or odd) = 1, and P(X=1 or X=2) = P(X=1) + P(X=2) = 0 + 0 = 0. We can define a "uniform" generalized distribution in a number of ways, but usually we require that the probabilities of subsets are equal to their asymptotic densities. So for instance, we would want P(X is even) = P(X is odd) = 1/2, P(X is divisible by 5) = 1/5, P(X is prime) = 0, etc. This distribution seems pretty reasonable at first, but picking at the edges reveals some very strange behaviors. For instance, in a paper by Kadane, Schervish, and Seidenfeld titled "Is Ignorance Bliss?" the authors ask us to imagine a gambler who has bet on one of two mutually exclusive and equally likely outcomes, A and B, one of which must be true. If A is true, he will lose $1, and if B is true, he will win $1. Suppose further that a random number will be selected according to one distribution if A is true and according to a different distribution if B is true. Specifically, the gambler knows that if A is true, then the number n will be selected with probability P(n|A) = (1/2)ⁿ, but if B is true, then the number n will be selected according to the uniform distribution described above, meaning P(n|B) = 0, whatever the n. Thus at the moment, before n is selected, both A and B are equally likely, and the gambler has an expected value of $0 for his bet. But after n is selected, no matter what it is, Bayes' Theorem shows that P(B|n) = 0. Therefore, the gambler should actually pay money not to know what number was selected, since if he does find out the number, he is guaranteed to lose.
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