# A different discrete normal distribution

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In summary, the article discusses the discrete normal distribution (dNormal) and its relationship to the normal variate X. It explores the possibility of a closed form for the expectation and variance of X, and whether a dNormal variate of a different form is possible. The article concludes that there is no closed form for the expectation or variance of X due to the use of the cumulative normal distribution function in the probability mass function. Additionally, the expectation of X is equal to μ, but the variance may not be equal to σ^2. Further research is needed to determine the exact relationship between the two distributions.

TL;DR Summary
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##.

Last edited:
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

What about my seecond question?