A double lottery and non-normalized probabilities

Suppose a positive integer N is generated by a fair lottery.

Then, a random integer K is chosen between 1 and N (inclusive).

What information does this give you about N?

Obviously you now know that N ≥ K. Anything else?

Consider some specific pair of numbers n ≥ k, and suppose we’ve found out that K = k. What’s the probability that N = n? Of course P(N=n|K=k) = 0/0. But what if we do this as a limiting procedure. Suppose first that N is randomly chosen between 1 and M where M ≥ n, and let P_M be the probabilities for this case. Then

• P_M(N=n|K=k) = (1/M)(1/n)/[(1/M)Σ_j=k^Mj⁻¹] = (1/n)/Σ_j=k^Mj⁻¹.

Take the limit as M goes to infinity. Since Σ_n=k^∞j⁻¹ = ∞, the limit is zero, so we don’t have a meaningful distribution for N.

On the other hand, what if we independently choose two random integers K₁ and K₂ between 1 and N? Suppose n ≥ k_i for i = 1, 2. Let k^* = max (k₁,k₂). Then:

• PM(N=n|K1=k1,K2=k2) = (1/M)(1/n2)/[(1/M)Σj=k*Mj−2] = (1/n2)/Σj=k*Mj−2.

Take the limit as M → ∞ and call that P(N=n|K₁=k₁,K₂=k₂). The limit behaves like ck^*/n², for a constant c > 0, and generates a well-defined probability for N = n.

With zero samples, we don’t have a well-defined probability for N. With one sample, we still don’t. But with two samples (or more), now we do. This is a rummy thing: how is it that sampling turns probabilistic nonsense into sense?

This is making me more friendly to using non-normalized probabilities. After all, the fair lottery for N is easily modeled by the constant probability p₀(n) = 1. With one sample N = k, we have p₁(n) = 1/n for n ≥ k and p₁(n) = 0 for n < k. With two samples k₁, k₂, we have p₂(n) = 1/n² for n ≥ max (k₁,k₂) and p₂(n) otherwise. All this makes perfect sense. And there is a lovely mathematical feature of non-normalized probabilities: conditionalization is conjunction. The conditional probability of an event A on event B is just the probability of A ∩ B.

Non-normalized probabilities aren’t going to solve all problems with infinite fair lotteries. For instance, I toss a fair coin and generate a number N with the following rule. On heads, I choose N with my fair lottery on the positive integers. On tails, I choose N such that the probability of N = n is 2⁻ⁿ (e.g., I toss an independent fair coin and let N be the number of the first toss that gives heads). What’s my non-normalized probability p(x,n), where x is heads or tails and n is a positive integer? We surely want ∑_np(H,n) = ∑_np(T,n): the total probability of the heads options equals the total probability of the tails options. But clearly p(T,n) has to exponentially decrease so ∑_np(T,n) is finite and non-zero. On the other hand, p(H,n) is constant, so ∑_np(H,n) is zero or infinity. So they can’t be equal.

But I wonder if one could say something like this: Non-normalized probabilities make sense in certain cases, and in those cases it’s reasonable to use them?