# To prove a relation involving a probability distribution

I'm reading a book and have encountered a relation which seems to me to be impossible to prove, I would like to be sure if this is the case. The author gives a probability function as $$p_n = \frac{e^{-c_1 n - c_2/n}}{Z},$$ where $$c_1$$ and $$c_2$$ are constants and Z is a normalization factor and $$n \geq 3$$. Then by considering $$\langle n \rangle = 6$$ and defining $$\alpha$$ (the second moment) as $$\alpha = \sum_{n = 3}^{\infty} p_n (n - 6)^2$$, the author claims one can show that

$$$$\alpha + p_6 = 1, \quad \quad \quad 0.66 < p_6 < 1,$$$$ $$$$\alpha p_6^2 = 1 / 2 \pi, \quad \quad \quad 0.34 < p_6 < 0.66.$$$$

How is such a thing possible in the first place as these relations are not even dependent on $$c_1$$ and $$c_2$$?

• Do you want to tell us the name of the secret book? Jul 12, 2021 at 21:58

The equality $$\alpha + p_6 = 1$$ can be rewritten as $$\sum_{n=3}^\infty g(n)p_n=1,$$ where $$g(n):=(n-6)^2+1(n=6)\ge1(n\in\{5,6,7\})+4\times1(n\notin\{5,6,7\}).$$ Therefore and because $$p_n>0$$ for all $$n\ge3$$, we have $$\sum_{n=3}^\infty g(n)p_n>\sum_{n=3}^\infty p_n=1,$$ so that the equality $$\alpha + p_6 = 1$$ is always false.
As for the equality $$$$\alpha p_6^2 = 1 / 2 \pi, \quad \quad \quad 0.34 < p_6 < 0.66,$$$$ numerics suggest that it is also false in general. In particular, using Mathematica's numerical summation NSum[] command, for $$c_1=12$$ and $$c_2=500$$ we get $$p_6=0.50995\ldots\in(0.34,0.66)$$ but $$2\pi\alpha p_6^2=0.90785\ldots\ne1$$.