An Estimate of the Probability Density Function of the Sum of a Random NumberN of Independent Random Variables

The probability density function (PDF) of the sum of a random number of independent random variables is important for many applications in the scientific and technical area [1]. Such a problem is not at all straightforward and has a theoretical solution only in some cases [2–5]. Further, in some cases the theoretical solution is not engineering viable (see, e.g., [2– 6] and the references therein, especially in [4, 5] where the rates of convergence in various metrics are studied as well). The purpose of this paper is to find a viable and good estimate of the PDF of a random variable (RV) Ywhich is the sum of a random number N of independent and identically distributed (IID) real RVs X i (i = 1, 2, . . . , N). The new model has also the benefit of providing an intuitive support to physical interpretation. In simple terms, the sum PDF is represented as a sum of normal PDFs weighted according to theN PDF. Two particular results are achieved. They are the conditions under which the PDF of the sum approximates a normal PDF again and those under which it approximates the envelope of the PDF ofN, except at the extremes when they are high. The theory can be applied all the time where a random number of IID RVs has to be summed. For instance, the total energy stored in the volume of reverberation chambers (RCs) can be statistically achieved by the sum of a random number N of IID RVs representing the energy stored in as many appropriate subvolumesN. By following this rationale, the PDF of the total stored energy in an RC can be achieved as well as the PDF of the relevant quality factor Q. A similar rationale is used to achieve a generalized stochastic field model for RCs [7]. Another important applicative case is experienced in coherent narrow-band sensor when the speckle arises and an appropriate model cannot be achieved to best mitigate it or exploit it as a geophysical signal [8].


Introduction
The probability density function (PDF) of the sum of a random number of independent random variables is important for many applications in the scientific and technical area [1].Such a problem is not at all straightforward and has a theoretical solution only in some cases [2][3][4][5].Further, in some cases the theoretical solution is not engineering viable (see, e.g., [2][3][4][5][6] and the references therein, especially in [4,5] where the rates of convergence in various metrics are studied as well).
The purpose of this paper is to find a viable and good estimate of the PDF of a random variable (RV)  which is the sum of a random number  of independent and identically distributed (IID) real RVs   ( = 1, 2, . . ., ).The new model has also the benefit of providing an intuitive support to physical interpretation.In simple terms, the sum PDF is represented as a sum of normal PDFs weighted according to the  PDF.Two particular results are achieved.They are the conditions under which the PDF of the sum approximates a normal PDF again and those under which it approximates the envelope of the PDF of , except at the extremes when they are high.
The theory can be applied all the time where a random number of IID RVs has to be summed.For instance, the total energy stored in the volume of reverberation chambers (RCs) can be statistically achieved by the sum of a random number  of IID RVs representing the energy stored in as many appropriate subvolumes .By following this rationale, the PDF of the total stored energy in an RC can be achieved as well as the PDF of the relevant quality factor .A similar rationale is used to achieve a generalized stochastic field model for RCs [7].
Another important applicative case is experienced in coherent narrow-band sensor when the speckle arises and an appropriate model cannot be achieved to best mitigate it or exploit it as a geophysical signal [8].

The Model
The RV  is expressed as follows: where the real IID RVs   are assumed with finite variance. is an integer and positive RV independent of the   ; the values of  are denoted with   , ( = min, min+1, min+2, . . ., min+ ), where  min and  min+ =  max are the minimum and maximum value of , respectively.The value  min cannot be too small since the central limit theorem (CLT) has to be applied for each  value. min ≥ 30 can be certainly considered an adequate value [9].The  PDF () can be written as follows: where (  ) is the discrete probability relevant to   and (⋅) is the delta function.
The mean   and variance  2  of  are as follows [2]: where  0 and   are the mean and standard deviation of , respectively;   and   are the mean and standard deviation of   , respectively.The random sum  represents  mutually exclusive random sums, which are selected by the  values; they are The set of the values of , which is the certain event, is given by the sum of the mutually exclusive events, which are the subsets formed by the values of the  RVs    ; that is, it can be written as follows: where  is the set of the values of  and    ,   =  min , . . .,  min+ , are the subsets of the values of the corresponding    .Therefore, it results in The PDF of the random sum  is given by the sum of  PDFs, (   ), which are weighted by the relevant probability (  ).That is, it can be written as follows: Considering the CLT, the PDFs (   ) can be assumed normal [9,10].Hence [9], By considering ( 8) and ( 9), the estimate of () can be written as follows: It can be noted that if   = 0, then the  PDF is the sum of  normal PDFs which are centered at zero with increasing variances    2  and weighted by the relevant probability (  ).
We stress that ( 9) and ( 10) are written under the condition of identically distributed RVs   ; it is precautionary with respect to the validity of the CLT [9][10][11][12][13].However, even though the extension of ( 9) is practicable when the RVs   are not identically distributed, by considering the further conditions on the RVs   for the possible convergences different from the normal distribution [9][10][11][12][13], the corresponding extension of (10) requires more stringent conditions as specified below (see the end of Section 4).Note that when the PDF of the sum of the RVs   is exactly expressible for any constant value   apart from the PDFs of   , which could also be different, then (10) can be exactly written and no validation procedure is necessary.The sums of independent exponentially distributed RVs with different scale parameters as well [14][15][16][17] (and not only them) are an example where the   are not identically distributed and ( 9) and (10) can also be exactly achieved.

Model Applicability
Equation ( 10) can be easy implemented by a simple algorithm, which can be solved by a calculator to achieve the () values; they are denoted here as analytical solution values.
The error on a single distribution (   ) depends on   and on the form of the density () [9,[18][19][20][21]; it is maximized for   =  min ; however, under the assumed hypothesis that the random variables   are IID, the value   = 30 is adequate for most applications [9].The total error depends on  min and  max , and it can be estimated by considering the error on each distribution (   ).The approximation in (10) improves as  min increases.However, the analytical determination of the total error is too arduous.Here, in order to find the conditions such that the overall approximation is adequate for applications, we compare the histogram achieved by (10) and that achieved by numerical simulation.This last one is achieved starting from the generation of the RVs having the same PDFs of   and , which are known, and then by summing the   according to the generated values of .The histogram originated from the numerical simulation is used as a null hypothesis for the comparison.The comparison is accomplished by Chi-Square Goodness-of-Fit test to the significance of 0.05.The analytical solution, which includes the implementation of (10), the numerical simulation, and Chi-Square Goodness-of-Fit test are accomplished by the software LabVIEW of National Instruments.Clearly, the width of the bins and the relevant centers along the -axes is the same both for the analytical histogram and for the numerically simulated one.
Even though the results of the tests shown here are achieved by using as a null hypothesis the PDF of the simulated data, it has been verified that the outcome of the test is the same when one chooses as a null hypothesis the achieved model (10).Actually, since ( 10) is an analytical model, it allows us to calculate probability and samples for each bin with no statistical fluctuations, so that it could be used as a null hypothesis for the comparison.This inversion of rationale is again consistent with the theory of the Chi-Square Goodness-of-Fit test.In fact, it corresponds to consider the achieved model as an exact model so that it can originate the theoretical frequencies, which are compared with the frequencies achieved by numerical simulations; these last ones have the typical statistical fluctuations.In order to calculate the number of degrees of freedom for Chi-Square test, it is specified that  0 ,  2  ,   , and  2  are theoretical parameters as the PDF of  is known as well as the mean and variance of   .When (10) is considered exact, its mean and variance are promptly achieved; they turn out to be equal to (3) and (4), respectively.
It is specified that the comparisons are made so that the bins whose samples are less than four are discarded.It is also specified that the sample number of , which is denoted with , is equal to 10000 and the bin number is 60.
In some cases, in order to avoid  values close to zero (too much low) or simply to test different values of  min and  max (namely, different  0 values) with the same shape of the  PDF, the mean of the  PDF is shifted by adding a fixed value to it.In these cases, the relevant PDF is shown with the same nomenclature except for the addition of the adjective "shifted" as we have a shift of the mean value only.
The comparison is made by using various pairs of PDFs for  and   with different means and variances.In particular, two meaningful PDFs are tested, which are the uniform distribution and the one chi squared with two freedom degrees; this last one is shifted along the -axis (shifted exponential distribution).However, both for  and   , PDFs as a triangular distribution, a chi distribution with two degrees of freedom (Rayleigh distribution), and a chi square distribution with six degrees of freedom are tested as well.To save space, only the results relative to PDFs as the uniform, (un)shifted exponential and (un)shifted Rayleigh are shown here.
All the tests made are accepted at the significance level of 0.05; some results are shown below, Figures 1-19.

Discussion on the Behaviour of the 𝑌 PDF and Two of Its Useful Approximations
In this section, we consider the behavior of the  PDF by some parameters defined in terms of mean and variance of  and   .In other words, we write the general conditions in order that the  PDF approximates the normal PDF and the  one; they include implicitly the conditions why the  PDF progressively changes from a normal PDF to the  one.The conditions, why the weighted normal PDFs approximating the  PDF are separated, are also examined.
To start from (4), by also taking a cue from ( 10) and (24) in [2], and by considering that where  1 and  2 are adequate threshold values depending on the PDFs of the RVs  and   , then we can again assume () approximately normal.Note that  2  / 0 =  0 V  , where V  is the variation coefficient of .Normally, the thresholds  1 and  2 are both less than one; however, if   = 0 then  1 can turn out to be considerably greater than one.By setting  3 =  2  / 2  , it is noted that  2 =  1  3 ; namely, as ( 11)-( 12) show, the thresholds  1 and  2 are not independent in sense that if one decreases, the other can increase and vice versa; clearly, it can occur within adequate limits depending on the PDFs of  and   .It can be noted that ( 11)-( 12) are similar to (10) and (24) in [2]; actually, they are as a whole less stringent than (10) and (24) in [2].However, ( 11)-( 12) are very useful for the practical applications, where it is sufficient to know an approximate solution and not just that exact or asymptotically exact.
Similarly, by considering that where  3 and  4 are adequate threshold values depending on the PDFs of  and   , then () approximates the envelope of the  PDF, as it will be cleared below.The thresholds  3 and  4 depend on the  3 ; as this last one increases,  3 decreases and  4 increases.
For both inferences ( 11)-( 12) and ( 13)-( 14), it is necessary that the next weighted normal PDFs forming the  PDF markedly interfere.In order to find the conditions under Therefore, it has to be     = √     ≅>   (the double symbol ≅> means "the same order of magnitude and greater than"); that is Indeed, if ( 16) is satisfied for  min it is satisfied for any  belonging to its distribution.For the inference ( 11)- (12), in (16) the symbol meaning "greater than" is prevalent, whereas, for the inference ( 13)-( 14), in (16) the symbol meaning "the same order of magnitude" is prevalent.Note that condition ( 16) is included in condition (12).Note also that for a given pair of PDFs of  and   , by fixing a certain value of  1 , as  3 increases,  2 increases as well; so, gradually  4 is exceeded and condition (16)  separation increases with the increasing of  3 with respect to  min , so that (10) always approximates the  PDF, but this last one does not approximate the envelope of the  PDF again.We stress that the limitation  4 is due to condition (16).
Clearly, ( 15) and ( 16) can be written for the cases where the difference between two next  values is greater than 1; that is, it can be written as follows: where  is an integer given by the difference between  +1 and   .Therefore, it has to be     = √     ≅>   ; that is, Shortly, the conditions in order that the  PDF approximates a normal PDF again or the envelope of the  PDF are ( 11)-( 12) and ( 13), (14), and ( 16), respectively.Below, both the inferences will be verified by (10) and numerical simulations; the values of the thresholds for which () approximates a normal PDF and those for which () approximates the envelope of the  PDF will also be shown for some examples.Y values at the bin center  We recall that if the  PDF is normal then () is approximately normal apart from the conditions given here [2].

Experimental Results
In order to verify the model achieved for the  PDF, some simulation results are shown; various pairs of PDFs are used for  and   with different means and variances.In particular, two PDFs with very different kurtosis index are tested, so as to make the tests very meaningful.Such PDFs are the uniform and chi squared with two freedom degrees shifted along the -axis (shifted exponential distribution).The uniform distribution is platykurtic whereas the exponential one is leptokurtic.
A PDF formed by three delta of Dirac only ( = 3;  = 1, 2, 3) is first used for ; see Figures 1-6.To each delta is coupled the relative probability.The   PDF is chi square distribution with two degrees of freedom so that  3 is a constant equal to 1. Since the forms of histograms and relevant PDFs are the same, the marker points representing the height of the histogram bins are interpolated by a simple line in all the figures shown here.
When the normal approximation is tested, it is specified that the null hypothesis is the theoretical normal with mean and variance given by ( 3) and ( 4), respectively.Similarly,  when the envelope of the  PDF is tested, it is specified that the null hypothesis is the  PDF, which is considered continuous, with mean and variance given by ( 3) and ( 4) again.However, the captions of all the figures are exhaustive and the null hypothesis is specified.
In Figures 1-6, the behavior of the  PDF with the increasing of the ratio  1 for the same probability (  ) can be noted.In particular, when (11) and ( 12) are satisfied the  PDF approximates the normal one; see Figure 1; then as the ratios  1 and  2 increase, the  PDF converges to the envelope of the PDF of except at its extremes when they are high; see Figures 2 and 3; finally, the  PDF becomes a sequence of separated approximate normal PDFs; see Figures 5 and 6.In Figure 4, the normal PDFs forming the  PDF are only partially separated.Figure 1 shows that if the thresholds  1 and  2 are limited to 0.3 and the PDFs of  and   are as specified in the caption of the same figures, then the  PDF can be considered approximately normal again.
In Figures 5 and 6, it can be noted that each weighted normal PDF forming the PDF of the sum lowers and widens with the increasing of the relative   , as expected.In this connection, it is worth to recall that the variation coefficient of each weighted normal component tends to zero as the relative   tends to infinite.In Figure 6, note that the probabilities (  ) are different and that in Figures 5 and  6, the weighted normal components are separated as the condition ( 18) is not satisfied.Clearly, it is why the  PDF is not formed by consecutive deltas of Dirac but the next deltas of Dirac are far.In particular,  3 −  2 =  2 −  1 =  = 150 in Figures 5 and 6.
It is specified that different PDFs for   do not change the essence of the results, except the thresholds  1 and  2 for which () can be considered approximately normal again and those for which () approximates the envelope of the  PDF, with mean and variance given by ( 3) and (4).As an example, in Figures 7(a), 7(b), and 7(c), the normal approximation is tested where the PDFs of  and   are both uniform distributions; their parameters are shown in the captions of the relative figures.In Figures 8-10, the PDFs of  and   are both uniform distributions again; the results show that () approximates gradually the envelope of the  PDF.In particular, in Figure 9, where  1 = 10, it is shown that  3 ≅ 30 and  4 ≅ 80.In Figure 11, the case where ( 16) is not satisfied so that the single normal PDFs are separated is shown; in this case, in order to resolve well the PDF graphs, the bin number is 1000 and  is 100000.
Figures 12-15 show the results where the PDF of  is a shifted chi square distribution with two degrees of freedom and the one of   is a uniform distribution.It can be noted again that () converges to the envelope of the  PDF as  1 and  2 increase.In Figure 15, the thresholds  3 and  4 for  1 = 71 are shown.
Similarly, Figures 16-18 show the results where the PDFs of  and   are a shifted chi distribution with two degrees of freedom and a chi square distribution with two degrees of freedom, respectively.In Figure 19, the PDF of   is a uniform distribution.In Figure 19, the thresholds  3 and  4 for  1 = 54.79 are also shown.
It is understood that ( 13), (14), and ( 16) or (18) are satisfied; it is important to note that with the increasing of  2 the  PDF converges to the envelope of the PDF of  except at the extremes when they are high.In fact, the extremes of () are always tapered and the tapering at the lower extremity is steeper than the one at the upper extremity, as the theoretical model (10) shows.It is shown in Figures 9-10 where the envelope of  PDF is uniform.Similarly, when the envelope of the  PDF has an extreme high as that of a shifted exponential PDF, where the first extreme is high and the second tends to zero, then for values sufficiently high of  2 , the  PDF assumes the same form of the envelope of the PDF of , except for a starting steep rise to which is associated a small amount of probability only.However, for high values of  2 , the probabilities associated with the tapers are very low in all cases apart from the  PDF, so that the Chi-Square test is as a rule accepted at the significance level of 0.05, even if the null hypothesis is with no tapers; it is shown in Figures 10, 15, and 19.
Hence, the  PDF changes from a normal PDF to the envelope of the  PDF, by assuming various forms between the two, and finally, it gradually becomes a sequence of separated approximate normal PDFs according to the ratios  1 ,  2 , and  3 .Note that the envelope of the normal PDFs forming the  PDF has the same form of the envelope of the  PDF also when they are totally separated as they are weighted according to the probabilities (  ).    3) and ( 4), respectively.The means and standard deviations of  and   are   = 10000;   = 400;   = 0;   = 57.73.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 15.96; 2 = 0;  3 = 0. (b)  histogram: cross markers (black color), numerical simulation; circle markers (red color), analytical solution.The PDFs of  and   are both uniform distributions. min and  max are 9307 and 10693, respectively.The null hypothesis is the normal PDF with mean and variance given by ( 3) and (4), respectively.The means and standard deviations of  and   are   = 10000;   = 400;   = 8;   = 57.73.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 15.9; 2 = 0.3;  3 = 0.019.(c)  histogram: cross markers (black color), numerical simulation; circle markers (red color), analytical solution.The PDFs of  and   are both uniform distributions. min and  max are 9948 and 10052, respectively.The null hypothesis is the normal PDF with mean and variance given by ( 3) and (4), respectively.The means and standard deviations of  and   are   = 10000;   = 30;   = 100;   = 57.73.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 0.09;  2 = 0.27;  3 = 3.     18:  histogram: cross markers (black color), numerical simulation; circle markers (red color), analytical solution.The PDF of  is a chi distribution with two degrees of freedom, which is shifted by one unit, and the one of   is a uniform distribution. min and  max are 1 and 43, respectively.The null hypothesis is the PDF of the simulated data with mean and variance given by ( 3) and (4), respectively.The means and standard deviations of  and   are   = 13.53;  = 6.55;   = 100;   = 57.73.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 3.17;  2 = 9.51;  3 = 3.
In order to give results ready for the applications, the next progress of this work is to map the areas of a Cartesian plane ( 1 ,  2 ) to show where the  PDF is approximately normal for the pairs of PDFs of  and   most common and useful for applications.Similarly, the areas of the Cartesian plane ( 1 ,  2 ) where the  PDF converges to the envelope of the  PDF should be mapped.In this last one case,  1 can be taken as a parameter so that the areas can be mapped as a function of  3 and  4 .
In principle, by considering the PDF of  as a continuous function, the analytical model for the PDF of  could also be written as an integral; that is, it could be written as follows: Clearly, the continuity of  affects the quality of the approximation on ().However, the analytical solution of the integral in (19) is enough difficult also in the case in which () is uniform; therefore, the attempt to achieve an analytical solution of ( 19) is not very feasible.
The above developments could be extended to the case where the RVs   are not identically distributed.In this case, by also considering the above on the further conditions  for the convergence to the normal distribution [9][10][11][12][13], the estimates of (   ) and () can be written as follows:

Y
values at the bin center sample number per bin) Y values at the bin center sample number per bin)

Figure 5 :
Figure 5:  histogram: cross markers (black color), numerical simulation; circle markers (red color), analytical solution.The PDF of  is formed by three delta of Dirac, with  1 = 350,  2 = 500, and  3 = 650; ( 1 ) = ( 2 ) = ( 3 ) = 0.3(3); the PDF of   is a chi square distribution with two degrees of freedom.The null hypothesis is the PDF of the simulated data with mean and variance given by (3) and (4), respectively.The means and standard deviations of  and   are   = 500;   = 122.5;  = 20000;   = 20000.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 30.0; 2 = 30.0; 3 = 1.
sample number per bin) Y values at the bin center bin (histogram)Y values at the bin center

Figure 7 :
Figure7: (a)  histogram: cross markers (black color), numerical simulation; circle markers (red color), analytical solution.The PDFs of  and   are both uniform distributions. min and  max are 9307 and 10693, respectively.The null hypothesis is the normal PDF with mean and variance given by (3) and (4), respectively.The means and standard deviations of  and   are   = 10000;   = 400;   = 0;   = 57.73.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 15.96; 2 = 0;  3 = 0. (b)  histogram: cross markers (black color), numerical simulation; circle markers (red color), analytical solution.The PDFs of  and   are both uniform distributions. min and  max are 9307 and 10693, respectively.The null hypothesis is the normal PDF with mean and variance given by (3) and (4), respectively.The means and standard deviations of  and   are   = 10000;   = 400;   = 8;   = 57.73.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 15.9; 2 = 0.3;  3 = 0.019.(c)  histogram: cross markers (black color), numerical simulation; circle markers (red color), analytical solution.The PDFs of  and   are both uniform distributions. min and  max are 9948 and 10052, respectively.The null hypothesis is the normal PDF with mean and variance given by (3) and (4), respectively.The means and standard deviations of  and   are   = 10000;   = 30;   = 100;   = 57.73.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 0.09;  2 = 0.27;  3 = 3.

Y
values at the bin center Sampled PDF (sample number per bin)

Y
values at the bin center Sampled PDF (sample number per bin) Figure 1:  histogram: cross markers (black color), numerical simulation; circle markers (red color), analytical solution.The PDF of  is formed by three delta of Dirac, with  1 = 485,  2 = 500, and  3 = 515; ( 1 ) = ( 2 ) = ( 3 ) = 0.3(3); the PDF of   is a chi square distribution with two degrees of freedom.The null hypothesis is the normal PDF with mean and variance given by (3) and (4), respectively.The means and standard deviations of  and   are   = 500;   = 12.3;   = 20000;   = 20000.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 0.3;  2 = 0.3;  3 = 1. Figure 2:  histogram: cross markers (black color), numerical simulation; circle markers (red color), analytical solution.The PDF of  is formed by three delta of Dirac, with  1 = 460,  2 = 500, and  3 = 540; ( 1 ) = ( 2 ) = ( 3 ) = 0.3(3); the PDF of   is a chi square distribution with two degrees of freedom.The null hypothesis is the PDF of the simulated data with mean and variance given by (3) and (4), respectively.The means and standard deviations of  and   are   = 500;   = 32.7;  = 20000;   = 20000.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 2.1;  2 = 2.1;  3 = 1. Figure 3:  histogram: cross markers (black color), numerical simulation; circle markers (red color), analytical solution.The PDF of  is formed by three delta of Dirac, with  1 = 450,  2 = 500, and  3 = 550; ( 1 ) = ( 2 ) = ( 3 ) = 0.3(3); the PDF of   is a chi square distribution with two degrees of freedom.The null hypothesis is the PDF of the simulated data with mean and variance given by (3) and (4), respectively.The means and standard deviations of  and   are   = 500;   = 40.8;= 20000;   = 20000.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 3.3;  2 = 3.3;  3 = 1.which the next weighted normal PDFs markedly interfere, the next unweighted normal PDFs can be considered.Mean and standard deviation of each normal component can be assumed equal to     =     and  2   =    2  , respectively; it can be written as follows: is no longer satisfied.Under these conditions, the weighted normal PDFs approximating the  PDF are partially separated; this Figure 4:  histogram: cross markers (black color), numerical simulation; circle markers (red color), analytical solution.The PDF of  is formed by three delta of Dirac, with  1 = 440,  2 = 500, and  3 = 560; ( 1 ) = ( 2 ) = ( 3 ) = 0.3(3); the PDF of   is a chi square distribution with two degrees of freedom.The null hypothesis is the PDF of the simulated data with mean and variance given by (3) and (4), respectively.The means and standard deviations of  and   are   = 500;   = 49;   = 20000;   = 20000.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 4.8;  2 = 4.8;  3 = 1. Figure 6:  histogram: cross markers (black color), numerical simulation; circle markers (red color), analytical solution.The PDF of  is formed by three delta of Dirac, with  1 = 350,  2 = 500, and  3 = 650; ( 1 ) = ( 3 ) = 0.2, ( 2 ) = 0.6; the PDF of   is a chi square distribution with two degrees of freedom.The null hypothesis is the PDF of the simulated data with mean and variance given by (3) and (4), respectively.The means and standard deviations of  and   are   = 500;   = 94.9;  = 20000;   = 20000.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 18.0;  2 = 18.0;  3 = 1.
Figure8:  histogram: cross markers (black color), numerical simulation; circle markers (red color), analytical solution.The PDFs of  and   are both uniform distributions.min and  max are 948 and 1052, respectively.The null hypothesis is the PDF of the simulated data with mean and variance given by (3) and (4), respectively.The means and standard deviations of  and   are   = 1000;   = 30;   = 10;   = 57.7.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 0.9;  2 = 2.7;  3 = 3.
Figure10:  histogram: cross markers (black color), numerical simulation; circle markers (red color), analytical solution.The PDFs of  and   are both uniform distributions.min and  max are 2292 and 57703, respectively.The null hypothesis is a uniform PDF with mean and variance given by (3) and (4), respectively.The means and standard deviations of  and   are   = 30000;   = 16000;   = 100;   = 57.73.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 8533;  2 = 25600;  3 = 3.Figure11:  histogram: cross markers (black color), numerical simulation; circle markers (red color), analytical solution.The PDFs of  and   are both uniform distributions.min and  max are 97 and 103, respectively.The null hypothesis is the PDF of the simulated data with mean and variance given by (3) and (4), respectively.The means and standard deviations of  and   are   = 100;   = 2;   = 100;   = 0.577.In this case, the sample number  is 100000 and the number of bins is 1000.The chi squared test is accepted at the significance level of 0.05. 1 = 0.04;  2 = 1200;  3 = 30000.Figure12:  histogram: cross markers (black color), numerical simulation; circle markers (red color), analytical solution.The PDF of  is a chi square distribution with two degrees of freedom, which is shifted by 10000 unit, and the one of   is a uniform distribution.minandmaxare10000and10164,respectively.The null hypothesis is the normal PDF with mean and variance given by (3) and (4), respectively.The means and standard deviations of  and   are   = 10018;   = 18;   = 100;   = 57.73.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 0.03;  2 = 0.09;  3 = 3. unit, and the one of   is a uniform distribution.minandmaxare1000and1645,respectively.The null hypothesis is the PDF of the simulated data with mean and variance given by (3) and (4), respectively.The means and standard deviations of  and   are   = 1072;   = 72;   = 100;   = 57.73.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 4.83;  2 = 14.50;  3 = 3.Figure14:  histogram: cross markers (black color), numerical simulation; circle markers (red color), analytical solution.The PDF of  is a chi square distribution with two degrees of freedom, which is shifted by 50 unit, and the one of   is a uniform distribution.minandmaxare50and845,respectively.The null hypothesis is the PDF of the simulated data with mean and variance given by (3) and (4), respectively.The means and standard deviations of  and   are   = 122;   = 72;   = 100;   = 57.73.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 42.49; 2 = 127.47; 3 = 3.Figure16:  histogram: cross markers (black color), numerical simulation; circle markers (red color), analytical solution.The PDF of  is a chi distribution with two degrees of freedom, which is shifted by 400 unit, and the one of   is a uniform distribution.minandmaxare 400 and 441, respectively.The null hypothesis is the normal PDF with mean and variance given by (3) and (4), respectively.The means and standard deviations of  and   are   = 412.53;  = 6.55;   = 100;   = 57.73.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 0.1;  2 = 0.3;  3 = 3.
are both uniform distributions.min and  max are 38 and 142, respectively.The null hypothesis is the PDF of the simulated data with mean and variance given by (3) and (4), respectively.The means and standard deviations of  and   are   = 90;   = 30;   = 100;   = 57.73.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 10;  2 = 30;  3 = 3;  1 = 10 ⇒  3 ≅ 30 and  4 ≅ 80. min and  max are 1 and 711, respectively.The null hypothesis is chi square with two degrees of freedom with mean and variance given by (3) and (4), respectively.The means and standard deviations of  and   are   = 73;   = 72;   = 100;   = 57.73.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 71.0; 2 = 213.0; 3 = 3;  1 = 71 ⇒  3 ≅ 450 and  4 ≅ 32000.shiftedby 10 unit, and the one of   is a uniform distribution.minandmaxare 10 and 56, respectively.The null hypothesis is the PDF of the simulated data with mean and variance given by (3) and (4), respectively.The means and standard deviations of  and   are   = 22.53;   = 6.55;   = 100;   = 57.73.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 1.9;  2 = 5.71;  3 = 3.
Figure19:  histogram: cross markers (black color), numerical simulation; circle markers (red color), analytical solution.The PDF of  is a chi distribution with two degrees of freedom and the one of   is a chi square distribution with two degrees of freedom.min and  max are 2 and 776, respectively.The null hypothesis is chi with two degrees of freedom with mean and variance given by (3) and (4), respectively.The means and standard deviations of  and   are   = 200.5;  = 104.8;  = 20000;   = 20000.The sample number  is 10000 and the number of bins is 60.The chi squared test is accepted at the significance level of 0.05. 1 = 54.79; 2 = 54.79; 3 = 1;  1 = 54.79 ⇒  3 ≅ 54.79 and  4 ≅ 20000.
(20)1 +   2 + ⋅ ⋅ ⋅ +    and  2 mean and variance of the sum of   ( = 1, 2, ...,   ).Practically,(20)imply that  min be very high so that     and  2    have stable values; otherwise, the situation is much more difficult as it is necessary to know the means and variances     and  2 ∑ =min  (  )  (   ) = are