A A different discrete normal distribution

AI Thread Summary
The discussion centers on the properties of a discrete normal distribution (dNormal) as defined in Dilip Roy's article. The first question addresses whether there is a closed form for the expectation and variance of the discrete normal variate, with the conclusion that no closed form exists due to the involvement of the cumulative normal distribution function. The second question explores an alternative form of the discrete normal variate and its implications for expectation and variance, with a suspicion that the expectation equals the mean (μ). However, calculations show that the expectation does not align with μ when evaluated for specific values. The conversation highlights the complexities in deriving statistical measures for discrete normal distributions.
Ad VanderVen
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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##.
 
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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?
 
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