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


\begin{equation}
\alpha + p\_6 = 1, \quad \quad \quad 0.66 < p\_6 < 1,
\end{equation}
\begin{equation}
\alpha p\_6^2 = 1 / 2 \pi, \quad \quad \quad 0.34 < p\_6 < 0.66.
\end{equation}


How is such a thing possible in the first place as these relations are not even dependent on $c\_1$ and $c\_2$?



To prove a relation involving a probability distribution
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
\begin{equation}
\alpha p\_6^2 = 1 / 2 \pi, \quad \quad \quad 0.34 < p\_6 < 0.66,
\end{equation}
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 $2\pi\alpha p\_6^2=0.90785\ldots\ne1$.


