where the denominator is called the Hurwitz zeta function, a fast-converging series. At this stage, the Bayesian statistician would compute the maximum a posterior estimation (MAP) given by the maximum of the distribution (which is at n=4n = 4n=4), or the mean nˉ=∑n≥4n1−k∑m≥4m−k=ζ(k−1,4)ζ(k,4)≃4.26\bar{n} = \frac{\sum_{n \geq 4} n^{1-k}}{\sum_{m \geq 4} m^{-k}} = \frac{\zeta(k-1, 4)}{\zeta(k, 4)} \simeq 4.26nˉ=∑m≥4m−k∑n≥4n1−k=ζ(k,4)ζ(k−1,4)≃4.26. A credible interval can be obtained now by just looking at the cumulative distribution function for the posterior distribution F(N)=∑s=4NP(n=s∣X)F(N) = \sum_{s=4}^N P(n = s | X)F(N)=∑s=4NP(n=s∣X) and finding the values [4,nR][4, n_R][4,nR] for which it covers 95% of the probability mass. For this problem we can just do it for a few values and see where it stops, leading to the interval [4,5]:
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fly launch --name your-app-name