Graduate A different discrete normal distribution

[Ad VanderVen]

TL;DR

Is a particular variant of Roy's discrete normal distribution also possible?

In the article A Discrete Normal Distribution of Dilip Roy in the journal COMMUNICATION IN STATISTICS Theory and methods Vol. 32, no. 10, pp. 1871-1883, 2003 one can read:

A discrete normal ($dNormal$) variate, $dX$, can be viewed as the
discrete concentration of the normal variate $X$ following $N(\mu,\sigma)$ when the corresponding probability mass function of $dX$ can be written as
$$\displaystyle p \left(x \right) \, = \, \Phi((x+1-\mu)/\sigma) - \Phi((x-\mu)/\sigma) \hspace{3ex} x \, = \, \dots, \, -1, \, 0, \, +1, \, \dots$$
where $\Phi(x)$ represents the cumulative distribution function of the normal deviate $Z$.
My first question is, is there a closed form for the expectation and variance of $X$?
My second question is, is a discrete normal ($dNormal$) variate of the form
$$\displaystyle p \left(x \right) \, = \, \Phi((x+1/2-\mu)/\sigma) - \Phi((x-1/2-\mu)/\sigma) \hspace{3ex} x \, = \, \dots, \, -1, \, 0, \, +1, \, \dots$$
also possible and what is then the expectation and variance of $X$? I suspect that the expectation of $X$ is equal to $\mu$. But I don't know if the variance is equal to $\sigma^2$.

 

[BWV]

If E[X]=Σ p(X)*X, there is no closed end form for p(x) as it contains the cumulative normal dist function, so there should not be a closed end form for the expectation or variance

For the expectation =μ
set μ = 0 and σ = 1
then

$\displaystyle p \left(x \right) \, = \, \Phi(x+1) - \Phi(x) \hspace{3ex} x \, = \, \dots, \, -1, \, 0, \, +1, \, \dots$

then look at x=-1,0,1
on a continuous standard normal, the expectation of p(x) for X = [-1,1] = 0, same as for any interval symmetric around zero
but for above on the discrete normal,
E(x) = p(-1)*-1 + p(0)*0 +p(1)*1
E(x) = -.341 + 0 + .136 <> μ

 

[Ad VanderVen]

What about my seecond question?