Are A and B independent?

I'm assuming weights are constant scalars - correct?

I think variance of a linear combination of random variables should be just:

V(Y) = (c^2)V(A) + (k^2)V(B) + 2(c)(k)COV(A,B)

If independent third term will drop out, as covariance = 0.

I may be missing something critical about the question though -

Let’s consider the distribution of normal variable y with parameters m and s (mean and standard deviation, respectively) truncated at point y=c (equation 1).

Then, the moment-generating function is defined as in equation 2.

By taking the inverse Mills ratio (equation 3), where alpha = - m/s, the first two moments of y (i.e., E(y) and var(y)) are given in equation 4:

As corollary, for y>c the left-truncated normal distribution of y with parameters m_b and s_b, has the moment-generating function reported in equation 5:

while equation 6 shows, for y not greater than c, the right-truncated normal of y with parameters m_a and s_a:

The mixture Y of left and right truncated normal distributions yields the moment-generating function of equation 7, where the weights are normalized to unity

Mirko, hi. I suppose that the intersection of A and B is null. My results when working additive models like yours is that it is not needed to assume a normal distribution for A nor B, and if medias are different then the total media must be between A and B. The distribution data of A and B may be given as quantils tables. So if data is made of 1)medias A,B, 2) population fractions c,k that you may call "population weights"and 3) Endogenous quantils series for both sectors, you have all you need to build your total results with media Y, population 1 and distribution of total -this last one usually result completely different from the components A, B, eventhough they conform it. My thesis is that the addition of two normal distributions does not produce a new normal distribution, even if arbitrary cuts of tails are made to try getting normality. When I add two Pareto's distributions, result is not another Pareto´s distribution. It is easier to explain the point working real data than with words. Send the data I need to work them. emilio

Pietro - very nice...

One comment, I think inverse Mills ratio should be lower case phi in numerator - ie

pdf(alpha)/(1-cdf(alpha))

also - shouldn't alpha = (c - m)/s ?

Milan, your notes are right. Thank you !

First of all, thanks for replying!

@Milan, they are independent. I tried that but the results were not satisfying.

@Pietro and Emilio, I don't know if I didn't understand your answer or if my question wasn't clear enough.

Let me try to express my needs more clearly...

I am trying to replicate what was described here: http://www.agenarisk.com/resources/technology_articles/FentonRankedNodes_TKDE-0095-0206_R2.pdf.

I am trying to perform calculations on a Bayesian network. I am working with ordinal nodes and converting them to Truncated Normal distributions (TNormal). This process is simple: represent the ordinal values as intervals of the same length between 0 and 1 (for example, for 3 states: State 1, between 0 and 1/3; State 2, 1/2 and 2/3 and State 3, 2/3 and 1). Afterwards, this is converted to a TNormal distribution with 0 and 1 as the boundaries.

I am using a tool that calculates the mixture of two or more Truncated Normal distributions (TNormal). I want to figure out which are the steps to perform this procedure and compare the results with the tool's (in the article, they explain how to calculate the mean correctly but they don't explain the variance...).

Say, C = Ka*A + Kb*B, where A, B and C are TNormal distributions truncated between 0 and 1, and Ka and Kb are "weights" that indicate the correlation between a variable and C. Consider that we use the notation TNormal(mean, variance, lower boundary, upper boundary).

Say, that A = TNorm(0.83,0,0,1) and B = TNorm(0.17,0,0,1). I attached a file with each variable's PDF and some more information.

Using the tool, we can set a "nominal" variance for C. In this case, I used 0.0005.

For Ka=Kb=1, using the tool, we have C = TNorm(.5,0.005,0,1). For Ka=2 and Kb=1, we have C = TNorm(.6,.023,0,1).

To calculate the mean is straighforward: MeanC = (Ka*MeanA) + Kb*Meanb)/(Ka+Kb). The problem is to calculate the variance. By performing some tests, I figured it depends on:

- C's "nominal" variance

- A and B variance

- A and B mean

- Ka and Kb

I hope I made my question clear now and that you could point me to the correct direction. Thank you!

Mirko, I tried to understand your referenced article and could not understand it. Article speaks of normal distributions based in qualitative answers that are quantified with some subjective weights and scales. You did not mention that at your initial question. My opinion is that the article is promoting another one of the many commercial packages available in market. How is it possible that authors speak of a biparametric normal model and can not give both parameters needed to define it? Also, in the same way that media of normal distribution is based on errors measures, then any standandard deviation must contain a SD-measurement error too, and if they come from censored data, authors add more problems to the paper. I see no outlet to your question if it is based in that paper, so I understand your reasons to make it. emilio

Hi Emilio, the main point is that it mixes Truncated Normal distributions (not Normal... http://en.wikipedia.org/wiki/Truncated_normal_distribution). I used R (http://www.r-project.org/) to generate TNormal distributions (http://cran.r-project.org/web/packages/truncnorm/truncnorm.pdf) given the mean, variance and lower and upper boundaries.

I am confused on how to mix the distributions. I am not sure how the variances, means and weights influence on the resulting variance. I can do that calculation for the mean, but not for the variance.

Ok Mirko, just publish here a table with less than N=50 of your data made of 1) frequences, and 2) The C values of your dataset. Then I would give you my opinion of your data as it comes. I do not need to use St.Deviation, only medias are enough. The point is that if you censor just one point from a normal distribution, the whole distributive structure changes and the model used too, and it is no more a normal one. Let´s try it and see what it results. emilio

Hi! I kinda figured out how they did... I posted a question regarding the procedure's validity in StackExchange (attached here). It would be great if you could take a loot and check if it makes sense. In a way, it is similar to this (https://www.youtube.com/watch?v=IkXxHwQZU2g) approach to mix Normal distributions. Cheers!

http://stats.stackexchange.com/questions/95750/is-this-method-for-mixing-doubly-truncated-normal-distribution-acceptable

Mirko, I will create two samples for two sectors. Each sector must have its media, its own fraction of total population, and its own clearly defined inner distribution. I will use different Pareto´s distributions for the two sectors because it is fast and easy to handle and to create datasets. Later you may try the same with your own data. OK, emilio

I forgot a key question: are A and B variables of the same nature and units that represent two sectors that conform the whole population? Or is this a bivariate case where you create an artificial new third variable Y=Ka.A + Kb.B that you want to explore? OK, thanks, emilio

Hi Emilio, it is the same problem as this (https://www.youtube.com/watch?v=IkXxHwQZU2g). The difference is that, in the video, the problem is to mix Normal distributions and I want to mix double truncated normal distributions. A and B are constants that represent the influence of each distribution on the mixture. For example for, C = 2A + B... Ka = 2, Kb = 1... which means that A is two times "more important" to the mixture than B... take a look at the video. It might make things clearer. Thanks

Mirko. I understand what you seek. OK, put here 50 values of A,B,C after your double truncation plus each one's U media -or expected value-. I will work the dataset as it comes and send you resulting graphs. You are right about that medias influence results, as also your weighting values; I just doubt that censoring is needed at all because those points portray valuable information and research time, but when analyzing we accept data such as it comes, we assume it is representative and do not ask about gathering and presentation details. Justifying truncation and weights belongs to your criteria about methods of research; in some moment you will be asked by others so it is honest to declare it at the start. emilio

OK. I usually used a huge sample (100000)to perform the calculations to have a better representation. Anyways, I did what you asked and sampled 50 elements from TNormal dist. for A and B (between 0 and 1).

MediaA = .5; VarianceA = 1

MediaB = .3; VarianceB = .1

A = { 0.966312563 0.819610073 0.227720154 0.642912969 0.177716373 0.217717879 0.291993825 0.479140207 0.100085891 0.249922618 0.510924171 0.873465574 0.711158088 0.443859265 0.524944058 0.051389732 0.693885463 0.683726507 0.513533667 0.461656685 0.793654035 0.358754477 0.591100198 0.035848238 0.671586962 0.287581170 0.192476030 0.565166095 0.627128467 0.226876266 0.901421547 0.140655615 0.004367106 0.868436975 0.674726161 0.273244401 0.722086970 0.536286633 0.078905157 0.591184001 0.810062857 0.739842533 0.737932087 0.787460518 0.525685090 0.536114297 0.290546324 0.247636412 0.539123941 0.470332965 }

B = {0.49151419 0.76556361 0.33128501 0.39610377 0.57485818 0.04039092 0.46055906 0.16366037 0.23005112 0.40944434 0.28146683 0.59215095 0.36649373 0.21558912 0.72268438 0.44988211 0.10893268 0.03755334 0.35435792 0.30470190 0.60170998 0.48689857 0.37145738 0.04290611 0.23154944 0.62178638 0.06988486 0.50682217 0.46720474 0.65583994 0.43106729 0.02421803 0.89362141 0.28074149 0.30501197 0.41772809 0.39699799 0.17999318 0.61175803 0.41763369 0.93058439 0.08100566 0.43044886 0.21029881 0.12426542 0.17798407 0.76309226 0.43641942 0.22792990 0.63706361 }

After mixing A and B (C=2A+B) with varianceC = .5:

C = 0.25580073 0.54385067 0.22441061 0.89873021 0.32035550 0.35456228 0.06270970 0.94912314 0.89584568 0.24891357 0.18814636 0.80427062 0.25181618 0.70341948 0.19068862 0.17992578 0.46651196 0.86840285 0.45474363 0.43931125 0.24048701 0.16938053 0.15213627 0.78736091 0.40467748 0.55885389 0.54147885 0.01679001 0.86562806 0.61536589 0.99023492 0.25737764 0.38217505 0.12490029 0.13857468 0.34393520 0.24273922 0.38923261 0.98270894 0.17490201 0.40085231 0.41605699 0.45134483 0.45639801 0.71122002 0.21251358 0.23083002 0.71463643 0.38813014 0.40613309 }

Thank you!

Mirco, hi. I worked your two datasets of C: one is made with formula C=2A+B and the other one with your C data. They have two different medias: Uc formula=1,36534 and Uc data=0,44713.

From the analysis it results that their adimentional structures -or W(p)- are quite different. The W(p) function was obtained from their Lorenz curves with W(p) = ln(Li)/ln(P). And even if the two datasets had same media, given that Wformula is diferent from Wdata, it is enough to advance that the two CDF would be different too.

I attach here a one page Excel file with your 50 points tables and graphs. If you want to replicate it for 1000 you may do it with the same procedure. I think that results may change somewhat for U and graphs, but not too much from the graphs shown. Standard deviations are not important for this method, so I did not estimate them. If you make this new text, please tell us about your results.

OK, Myrko, your interpretation is the next step. Regards, emilio