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I want to discretize the Truncated Normal to describe the spread in the Grades of students in a class sitting the same exam.

Let an arbitrary number of students sit the same exam. The grades are usually said to be normally distrubutted (they couldn't possibly be distributed with the truncated normal and much less by a simply normal because the grade are bound from both ends and are discrete).

The minimum grade is a parametre a (even negative grades are allowed)

The maximum grade is a parametre b

The minimum difference possible between two grades is a parametre c

x (measured in X axis) is the grade

y (measured in Y axis) is the probability of a student achieving X grade or the frequency that students in the class achieve grade X.

μ is a transformation of the mean, it is the mode if μ lies between a and b and is also the mean if μ lies halfway between a and b

σ is a transformation of the SD, it is the SD when $a\to-\infty\cap b\to\infty \cap c=0$

When $c=0$ we have a Truncated Normal

When $a\to-\infty\cap b\to\infty \cap c=0$ we have a Normal.

I want my distribution (f(x,μ,σ,a,b,c)) to meet some criteria

1 Being supported on the terms of an arithmetic progression. Both the initial and final terms and the common difference are parametres

2 Being unimodal in all cases but 1. Being bimodal in case the mode is equidistant from 2 values in the support.

3 Having probabilites strictly decreasing as X distances itself from the mode. The decrease in probability should be equal for x equally distant from the mode. The change in the difference should first decrease and then increase after reaching a quasi-inflection point. The change in the difference should also be equal for x equally distant from the mean. I.e if the mode coincides with the midrange the distribution is symmetrical while if the mode is closer to one bound the distribution is symmetrical for all values closer than the nearest to the mode bound (short of like the Truncated Normal if you fold it at the mode all the values that are closer that the nearest bound coincide 1 to 1 and only what is further away does not coincide because there is nothing left for them to correspond to)

Hopefully the first 3 criteria are all compatible with each other.

4 If the above three criteria allow for it the distribution should belong to the Natural Exponential Family (better) or the Exponential Family (Plan B)

5 The distribution should be the Maximum entropy distribution that the above 4 criteria allow for (ignore criterion 4 if the first 3 make it impossible for the distribution to belong to the Exponential Family)

Such that the probability mass assigned to any point is proportional to the kernel of the normal distribution (leaving the mean and the standard error a parametre) calculated at that point.

E.g when a=0, b=25 and c=1

$f(0,\mu,\sigma, 0, 25, 1)\propto\mathrm{e}^{-\frac{{\mu}^2}{2{\sigma}^2}}$

$f(1,\mu,\sigma, 0, 25, 1)\propto\mathrm{e}^{-\frac{\left({\mu}-1\right)^2}{2{\sigma}^2}}$

$f(2,\mu,\sigma, 0, 25, 1)\propto\mathrm{e}^{-\frac{\left({\mu}-2\right)^2}{2{\sigma}^2}}$

$f(3,\mu,\sigma, 0, 25, 1)\propto\mathrm{e}^{-\frac{\left({\mu}-3\right)^2}{2{\sigma}^2}}$

.

.

.

$f(25,\mu,\sigma, 0, 25, 1)\propto\mathrm{e}^{-\frac{\left({\mu}-25\right)^2}{2{\sigma}^2}}$

of course

$\displaystyle \sum_{x=0}^{25} f(x,\mu,\sigma, 0, 25, 1) = 1$

or more generally

$\displaystyle \sum_{x=a}^b f(x,\mu,\sigma, a, b, c)=1$

What is the closed form solution (probably in terms of Theta functions)

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    $\begingroup$ There is little justification in defining a pmf this way when compared with other functions from $\{0,...,y\}$ to $(0,1)$. The Normal density function is a density for a continuous variate and its values at a finite number of values such as the above integers cannot be interpreted as probabilities. In addition, most of the nice properties of a Normal distribution do not transfer to this new distribution. $\endgroup$ – Xi'an Dec 9 '20 at 19:44
  • $\begingroup$ @Xi'an The justification is simple I hate approximations. My data are bound both from below and above and much more significantly are discrete. I don't want a distribution which is easy to work with I want an acurate distribution (or the closest possible to that) even if I end up with something hard to work with. What other distributions f(μ,σ,x,z,y) are there (which are both discrete, and bounded from both sides). What properties do those distributions have that my constructed distributtion would lack? $\endgroup$ – George Ntoulos Dec 9 '20 at 20:19
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    $\begingroup$ In which sense an integer-Normal distribution is less of an approximation? more accurate? And what properties are you looking for in your model? $\endgroup$ – Xi'an Dec 9 '20 at 21:01
  • $\begingroup$ @Xi'an The properties I am looking for in my distribution would be having a discrete support, being bound both from above and below, being unimodal (or at worst bi-modal in case the mean is exactly in the mid-range between two points in the support), having probabilities strictly decreasing as X distances itself from the mode, being symmetrical if the mean coincides with the midrange (in this case with y=25 if the mean is 12.5). Hopefully the difference in probability would first decrease as we got more distant from the mean and then start increasing back when reaching a quasiinflextion point. $\endgroup$ – George Ntoulos Dec 9 '20 at 22:07
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    $\begingroup$ This question is confusing. You repeatedly state you "hate approximations," but ironically you have yet to characterize the quantities you wish to compute! All you have said explicitly is "I need a discrete distribution." Could you characterize what you need in clear, mathematical terms so that people can know what might constitute an "approximation" and how to evaluate it? What do you mean by the "general problem" of which this might be one instance? $\endgroup$ – whuber Dec 9 '20 at 22:21
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The most natural way would be as follows:

Let $X$ have support $0,1,...T$, where $T \in \{2,3,...,\}$ (i.e T can be taken as the limit to $\infty$, since the infinite sum can be bounded by the integral over the kernel of a gaussian pdf, but it will be intractable to work with in practice).

$$f_X(X=x;\mu, \sigma) = \frac{exp(-\frac{(\mu -x)^2}{2\sigma^2})}{\sum_{y=0}^Texp(-\frac{(\mu -y)^2}{2\sigma^2})}$$

This is a special case of the softmax function ($\frac{exp(z_i)}{\sum_{i=1}^Nexp(z_i)})$, which is commonly used to map the reals to the interval $(0,1)$ with the property that the elements sum to 1. There are many choice models which take this form, such as the multinomial logit, with $T < \infty$.

I agree with @Xi'an, however, that defining a pmf this way is unlikely to be justified. It is hard to imagine a principled construction of what this would represent and as Xi'an stated, the properties of a gaussian that make it nice to work with are not inherited by this new model.

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  • $\begingroup$ The justification is simple I hate approximations. My data are bound both from below and above and much more significantly are discrete. I don't want a distribution which is easy to work with I want an acurate distribution (or the closest possible to that) even if I end up with something hard to work with. What other distributions f(μ,σ,x,z,y) are there (which are both discrete, and bounded from both sides). What properties do those distributions have that my constructed distributtion would lack? $\endgroup$ – George Ntoulos Dec 9 '20 at 20:22
  • $\begingroup$ reading the other comments, are you modelling the responses to each question i for data set j? Or the number of correct responses in each data set j? $\endgroup$ – Tyrel Stokes Dec 9 '20 at 23:28
  • $\begingroup$ Most probably the second. There are i students (and i exams) that each received the same 25 questions (different each year). The results are positively dependent (students who know/believe that they will do well tend to sit the exams more often, so on years with better teaching and/or a smarter cohort the sample is larger and the mean is larger too because of self-choice students themselves choose to sit the exam or not). I consider each year a different population (different students unless some failed and choose to retake it and different 25 question). $\endgroup$ – George Ntoulos Dec 10 '20 at 7:17
  • $\begingroup$ I want to infer 2 things a) how good was the performance of the specific student (my metric is [Their score / (The mean of all other students + The standard error of all other students)] and I take the weighted harmonic mean of the "Standardized" score between all their subjects. b I want to make inference on the Teaching that year. the mean might have dropped either because the test makers gave difficult questions or because the teachers taught poorly that year (or the cohort were stupid). $\endgroup$ – George Ntoulos Dec 10 '20 at 7:24
  • $\begingroup$ You could use a binomial model, where p_i is the probability of success for each class. If you wanted to try and separate the effects of teaching and the effect of the cohort it will be tricky. The two main approaches would be to find covariates with sufficient information to control for the strength of the class (maybe mark from previous year or gpa generally) and model p_i as an inverse logit which is a function of a parameter for the class and then the covariate information. The other typical route would be to model the selection process itself and use a weighted estimator. $\endgroup$ – Tyrel Stokes Dec 11 '20 at 23:42

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