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pdf of sum of two uniform random variables

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Sums of uniform random values - johndcook.com >> Pdf of the sum of two independent Uniform R.V., but not identical I was hoping for perhaps a cleaner method than strictly plotting. endstream Then you arrive at ($\star$) below. Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. Would My Planets Blue Sun Kill Earth-Life? MATH 107 0 obj We shall discuss in Chapter 9 a very general theorem called the Central Limit Theorem that will explain this phenomenon. The best answers are voted up and rise to the top, Not the answer you're looking for? So how might you plot the pdf of a difference of two uniform variables? Google Scholar, Bolch G, Greiner S, de Meer H, Trivedi KS (2006) Queueing networks and markov chains: modeling and performance evaluation with computer science applications. /Filter /FlateDecode $\endgroup$ - Xi'an. i.e. Ann Stat 33(5):20222041. Then Z = z if and only if Y = z k. So the event Z = z is the union of the pairwise disjoint events. PDF of the sum of two random variables - YouTube \end{aligned}$$, \(\ln \left( (q_1e^{ 2\frac{t}{\sigma }}+q_2e^{ \frac{t}{\sigma }}+q_3)^n\right) \), $$\begin{aligned} \ln \left( (q_1e^{ 2\frac{t}{\sigma }}+q_2e^{ \frac{t}{\sigma }}+q_3)^n\right)= & {} \ln \left( q_1+q_2+q_3\right) {}^n+\frac{ t \left( 2 n q_1+n q_2\right) }{\sigma (q_1+q_2+q_3)}\\{} & {} \quad +\frac{t^2 \left( n q_1 q_2+n q_3 q_2+4 n q_1 q_3\right) }{2 \sigma ^2\left( q_1+q_2+q_3\right) {}^2}+O\left( \frac{1}{n^{1/2}}\right) \\= & {} \frac{ t \mu }{\sigma }+\frac{t^2}{2}+O\left( \frac{1}{n^{1/2}}\right) . 10 0 obj uniform random variables I Suppose that X and Y are i.i.d. So, if we let $Y_1 \sim U([1,2])$, then we find that, $$f_{X+Y_1}(z) = Commun Stat Theory Methods 47(12):29692978, Article The sign of $Y$ follows a Rademacher distribution: it equals $-1$ or $1$, each with probability $1/2$. A simple procedure for deriving the probability density function (pdf) for sums of uniformly distributed random variables is offered. }q_1^{x_1}q_2^{x_2}q_3^{n-x_1-x_2}, \end{aligned}$$, $$\begin{aligned}{} & {} P(2X_1+X_2=k)\\= & {} P(X_1=0,X_2=k,X_3=n-k)+P(X_1=1,X_2=k-2,X_3=n-k+1)\\{} & {} +\dots +P(X_1=\frac{k}{2},X_2=0,X_3=n-\frac{k}{2})\\= & {} \sum _{j=0}^{\frac{k}{2}}P(X_1=j,X_2=k-2j,X_3=n-k+j)\\= & {} \sum _{j=0}^{\frac{k}{2}}\frac{n!}{j! Consequently. Finding PDF of sum of 2 uniform random variables. >>/ProcSet [ /PDF /ImageC ] What positional accuracy (ie, arc seconds) is necessary to view Saturn, Uranus, beyond? Two MacBook Pro with same model number (A1286) but different year. pdf of a product of two independent Uniform random variables Intuition behind product distribution pdf, Probability distribution of the product of two dependent random variables. Accessibility StatementFor more information contact us atinfo@libretexts.org. \end{cases} of \((X_1,X_2,X_3)\) is given by. /Length 15 Sep 26, 2020 at 7:18. . /BBox [0 0 362.835 5.313] /Group << /S /Transparency /CS /DeviceGray >> A well-known method for evaluating a bridge hand is: an ace is assigned a value of 4, a king 3, a queen 2, and a jack 1. By Lemma 1, \(2n_1n_2{\widehat{F}}_Z(z)=C_2+2C_1\) is distributed with p.m.f. The distribution for S3 would then be the convolution of the distribution for \(S_2\) with the distribution for \(X_3\). This leads to the following definition. endstream Indian Statistical Institute, New Delhi, India, Indian Statistical Institute, Chennai, India, You can also search for this author in Google Scholar, Kordecki W (1997) Reliability bounds for multistage structures with independent components. New blog post from our CEO Prashanth: Community is the future of AI, Improving the copy in the close modal and post notices - 2023 edition. This page titled 7.2: Sums of Continuous Random Variables is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Charles M. Grinstead & J. Laurie Snell (American Mathematical Society) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. What does 'They're at four. /XObject << /Fm5 20 0 R >> Also it can be seen that \(\cup _{i=0}^{m-1}A_i\) and \(\cup _{i=0}^{m-1}B_i\) are disjoint. The best answers are voted up and rise to the top, Not the answer you're looking for? \end{aligned}$$, $$\begin{aligned} \phi _{2X_1+X_2}(t)&=E\left[ e^{ (2tX_1+tX_2)}\right] =(q_1e^{ 2t}+q_2e^{ t}+q_3)^n. Show that you can find two distributions a and b on the nonnegative integers such that the convolution of a and b is the equiprobable distribution on the set 0, 1, 2, . /ProcSet [ /PDF ] /Filter /FlateDecode stream endobj If you sum X and Y, the resulting PDF is the convolution of f X and f Y E.g., Convolving two uniform random variables give you a triangle PDF. /Type /XObject Pdf of the sum of two independent Uniform R.V., but not identical. /Shading << /Sh << /ShadingType 2 /ColorSpace /DeviceRGB /Domain [0 1] /Coords [0 0.0 0 8.00009] /Function << /FunctionType 2 /Domain [0 1] /C0 [0 0 0] /C1 [1 1 1] /N 1 >> /Extend [false false] >> >> >> /Filter /FlateDecode J Am Stat Assoc 89(426):517525, Haykin S, Van Veen B (2007) Signals and systems. /XObject << stream mean 0 and variance 1. << << /Filter /FlateDecode /S 100 /O 156 /Length 146 >> MathWorks is the leading developer of mathematical computing software for engineers and scientists. :). /BBox [0 0 337.016 8] >> I'm familiar with the theoretical mechanics to set up a solution. Pdf of the sum of two independent Uniform R.V., but not identical. /Length 15 The \(X_1\) and \(X_2\) have the common distribution function: \[ m = \bigg( \begin{array}{}1 & 2 & 3 & 4 & 5 & 6 \\ 1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 \end{array} \bigg) .\]. Well, theoretically, one would expect the solution to be a triangle distribution, with peak at 0, and extremes at -1 and 1. Let \(T_r\) be the number of failures before the rth success. Save as PDF Page ID . We see that, as in the case of Bernoulli trials, the distributions become bell-shaped. endobj Multiple Random Variables 5.5: Convolution Slides (Google Drive)Alex TsunVideo (YouTube) In section 4.4, we explained how to transform random variables ( nding the density function of g(X)). /BBox [0 0 16 16] I fi do it using x instead of y, will I get same answer? of standard normal random variable. << This is clearly a tedious job, and a program should be written to carry out this calculation. K. K. Sudheesh. (k-2j)!(n-k+j)!}q_1^jq_2^{k-2j}q_3^{n-k+j}. Sums of a Random Variables 47 4 Sums of Random Variables Many of the variables dealt with in physics can be expressed as a sum of other variables; often the components of the sum are statistically indepen-dent. /Filter /FlateDecode >> . /LastModified (D:20140818172507-05'00') What are the advantages of running a power tool on 240 V vs 120 V? The random variable $XY$ is the symmetrized version of $20$ times the exponential of the negative of a $\Gamma(2,1)$ variable. Something tells me, there is something weird here since it is discontinuous at 0. PDF of mixture of random variables that are not necessarily independent, Difference between gaussian and lognormal, Expectation of square root of sum of independent squared uniform random variables. $$f_Z(z) = Would My Planets Blue Sun Kill Earth-Life? Running this program for the example of rolling a die n times for n = 10, 20, 30 results in the distributions shown in Figure 7.1. /Parent 34 0 R \end{cases}$$. Stat Probab Lett 34(1):4351, Modarres M, Kaminskiy M, Krivtsov V (1999) Reliability engineering and risk analysis. Sum of two independent uniform random variables in different regions. PB59: The PDF of a Sum of Random Variables - YouTube endobj The error of approximation is shown to be negligible under some mild conditions. xP( ;) However, you do seem to have made some credible effort, and you did try to use functions that were in the correct field of study. Find the distribution of \(Y_n\). For this to be possible, the density of the product has to become arbitrarily large at $0$. I5I'hR-U&bV&L&xN'uoMaKe!*R'ojYY:`9T+_:8h);-mWaQ9~:|%(Lw. . 26 0 obj 35 0 obj \frac{5}{4} - \frac{1}{4}z, &z \in (4,5)\\ Are there any constraint on these terms? Use MathJax to format equations. Learn more about Stack Overflow the company, and our products. PDF Lecture Notes 3 Multiple Random Variables - Stanford University They are completely specied by a joint pdf fX,Y such that for any event A (,)2, P{(X,Y . To learn more, see our tips on writing great answers. - 158.69.202.20. \frac{1}{4}z - \frac{1}{2}, &z \in (2,3) \tag{$\dagger$}\\ /Resources 19 0 R 23 0 obj \(\square \). Uniform Random Variable - an overview | ScienceDirect Topics To formulate the density for w = xl + x2 for f (Xi)~ a (0, Ci) ;C2 >Cl, where u (0, ci) indicates that random variable xi . If this is a homework question could you please add the self-study tag? The price of a stock on a given trading day changes according to the distribution. \nonumber \]. >> \end{aligned}$$, $$\begin{aligned} E\left[ e^{ t\left( \frac{2X_1+X_2-\mu }{\sigma }\right) }\right] =e^{\frac{-\mu t}{\sigma }}(q_1e^{ 2\frac{t}{\sigma }}+q_2e^{ \frac{t}{\sigma }}+q_3)^n=e^{\ln \left( (q_1e^{ 2\frac{t}{\sigma }}+q_2e^{ \frac{t}{\sigma }}+q_3)^n\right) -\frac{\mu t}{\sigma }}. Why refined oil is cheaper than cold press oil? << /Resources << Society of Actuaries, Schaumburg, Saavedra A, Cao R (2000) On the estimation of the marginal density of a moving average process. Assume that the player comes to bat four times in each game of the series. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. endobj >> /LastModified (D:20140818172507-05'00') Can J Stat 28(4):799815, Sadooghi-Alvandi SM, Nematollahi AR, Habibi R (2009) On the distribution of the sum of independent uniform random variables. Why is my arxiv paper not generating an arxiv watermark? Book: Introductory Probability (Grinstead and Snell), { "7.01:_Sums_of_Discrete_Random_Variables" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "7.02:_Sums_of_Continuous_Random_Variables" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "00:_Front_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "01:_Discrete_Probability_Distributions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "02:_Continuous_Probability_Densities" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "03:_Combinatorics" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "04:_Conditional_Probability" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "05:_Distributions_and_Densities" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "06:_Expected_Value_and_Variance" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "07:_Sums_of_Random_Variables" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "08:_Law_of_Large_Numbers" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "09:_Central_Limit_Theorem" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "10:_Generating_Functions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "11:_Markov_Chains" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "12:_Random_Walks" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "zz:_Back_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, [ "article:topic", "showtoc:no", "license:gnufdl", "Discrete Random Variables", "Convolutions", "authorname:grinsteadsnell", "licenseversion:13", "source@https://chance.dartmouth.edu/teaching_aids/books_articles/probability_book/book.html" ], https://stats.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fstats.libretexts.org%2FBookshelves%2FProbability_Theory%2FBook%253A_Introductory_Probability_(Grinstead_and_Snell)%2F07%253A_Sums_of_Random_Variables%2F7.01%253A_Sums_of_Discrete_Random_Variables, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), source@https://chance.dartmouth.edu/teaching_aids/books_articles/probability_book/book.html. given in the statement of the theorem. https://doi.org/10.1007/s00362-023-01413-4, DOI: https://doi.org/10.1007/s00362-023-01413-4. First, simple averages . Using the comment by @whuber, I believe I arrived at a more efficient method to reach the solution. If the \(X_i\) are all exponentially distributed, with mean \(1/\lambda\), then, \[f_{X_i}(x) = \lambda e^{-\lambda x}. \end{aligned}$$, $$\begin{aligned} E\left[ e^{ t\left( \frac{2X_1+X_2-\mu }{\sigma }\right) }\right] =\frac{t^2}{2}+O\left( \frac{1}{n^{1/2}}\right) . /BBox [0 0 8 87.073] \end{aligned}$$, \(A_i\cap A_j=B_i\cap B_j=\emptyset ,\,i\ne j=0,1m-1\), \(A_i\cap B_j=\emptyset ,\,i,j=0,1,..m-1,\), \(\{\cup _{i=0}^{m-1}A_i,\,\cup _{i=0}^{m-1}B_i,\,\left( \cup _{i=0}^{m-1}(A_i\cup B_i) \right) ^c\}\), $$\begin{aligned}{} & {} C_1=\text {Number of elements in }\cup _{i=0}^{m-1}B_i,\\{} & {} C_2=\text {Number of elements in } \cup _{i=0}^{m-1}A_i \end{aligned}$$, $$\begin{aligned} C_3=\text {Number of elements in } \left( \cup _{i=0}^{m-1}(A_i\cup B_i) \right) ^c=n_1n_2-C_1-C_2. \[ p_X = \bigg( \begin{array}{} -1 & 0 & 1 & 2 \\ 1/4 & 1/2 & 1/8 & 1/8 \end{array} \bigg) \]. Hence, endobj stream /Length 183 }\sum_{0\leq j \leq x}(-1)^j(\binom{n}{j}(x-j)^{n-1}, & \text{if } 0\leq x \leq n\\ 0, & \text{otherwise} \end{array} \nonumber \], The density \(f_{S_n}(x)\) for \(n = 2, 4, 6, 8, 10\) is shown in Figure 7.6. Anyone you share the following link with will be able to read this content: Sorry, a shareable link is not currently available for this article. Let us regard the total hand of 13 cards as 13 independent trials with this common distribution. Reload the page to see its updated state. Marcel Dekker Inc., New York, Moschopoulos PG (1985) The distribution of the sum of independent gamma random variables. 15 0 obj For certain special distributions it is possible to find an expression for the distribution that results from convoluting the distribution with itself n times. Sums of independent random variables. Modified 2 years, 6 months ago. /FormType 1 stream the statistical profession on topics that are important for a broad group of endstream Uniform Random Variable PDF - MATLAB Answers - MATLAB Central - MathWorks Making statements based on opinion; back them up with references or personal experience. Based upon his season play, you estimate that if he comes to bat four times in a game the number of hits he will get has a distribution, \[ p_X = \bigg( \begin{array}{} 0&1&2&3&4\\.4&.2&.2&.1&.1 \end{array} \bigg) \]. /Length 15 \begin{cases} . \\&\left. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Stat Neerl 69(2):102114, Article (a) X 1 (b) X 1 + X 2 (c) X 1 + .+ X 5 (d) X 1 + .+ X 100 11/12 The PDF p(x) is the derivative of the random variable's CDF, 19 0 obj Google Scholar, Belaghi RA, Asl MN, Bevrani H, Volterman W, Balakrishnan N (2018) On the distribution-free confidence intervals and universal bounds for quantiles based on joint records. If the null hypothesis is never really true, is there a point to using a statistical test without a priori power analysis? with peak at 0, and extremes at -1 and 1. It only takes a minute to sign up. Which ability is most related to insanity: Wisdom, Charisma, Constitution, or Intelligence? where the right-hand side is an n-fold convolution. xP( . >> endstream /Resources 15 0 R Other MathWorks country Then, \[f_{X_i}(x) = \Bigg{\{} \begin{array}{cc} 1, & \text{if } 0\leq x \leq 1\\ 0, & \text{otherwise} \end{array} \nonumber \], and \(f_{S_n}(x)\) is given by the formula \(^4\), \[f_{S_n}(x) = \Bigg\{ \begin{array}{cc} \frac{1}{(n-1)! 11 0 obj Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. A sum of more terms would gradually start to look more like a normal distribution, the law of large numbers tells us that. /Trans << /S /R >> It becomes a bit cumbersome to draw now. Asking for help, clarification, or responding to other answers. >> stream What is Wario dropping at the end of Super Mario Land 2 and why? /Type /XObject endobj Did the drapes in old theatres actually say "ASBESTOS" on them? The convolution of two binomial distributions, one with parameters m and p and the other with parameters n and p, is a binomial distribution with parameters \((m + n)\) and \(p\). $$\begin{aligned}{} & {} {\widehat{F}}_Z(z) - F_{Z_m}(z)\\= & {} \left\{ \frac{1}{2}\sum _{i=0}^{m-1}\left( {\widehat{F}}_X\left( \frac{(i+1) z}{m}\right) -{\widehat{F}}_X\left( \frac{i z}{m}\right) \right) \left( {\widehat{F}}_Y\left( \frac{z (m-i-1)}{m}\right) +{\widehat{F}}_Y\left( \frac{z (m-i)}{m}\right) \right) \right\} \\{} & {} -\left\{ \frac{1}{2}\sum _{i=0}^{m-1}\left( F_X\left( \frac{(i+1) z}{m}\right) -F_X\left( \frac{i z}{m}\right) \right) \left( F_Y\left( \frac{z (m-i-1)}{m}\right) +F_Y\left( \frac{z (m-i)}{m}\right) \right) \right\} \\= & {} \frac{1}{2}\sum _{i=0}^{m-1}\left\{ \left( {\widehat{F}}_X\left( \frac{(i+1) z}{m}\right) -{\widehat{F}}_X\left( \frac{i z}{m}\right) \right) \left( {\widehat{F}}_Y\left( \frac{z (m-i-1)}{m}\right) +{\widehat{F}}_Y\left( \frac{z (m-i)}{m}\right) \right) \right\} \\{} & {} -\frac{1}{2}\sum _{i=0}^{m-1}\left\{ \left( F_X\left( \frac{(i+1) z}{m}\right) -F_X\left( \frac{i z}{m}\right) \right) \left( F_Y\left( \frac{z (m-i-1)}{m}\right) +F_Y\left( \frac{z (m-i)}{m}\right) \right) \right\} \end{aligned}$$, $$\begin{aligned}{} & {} {\widehat{F}}_Z(z) - F_{Z_m}(z)\nonumber \\= & {} \frac{1}{2}\sum _{i=0}^{m-1}\Big \{{\widehat{F}}_X\left( \frac{(i+1) z}{m}\right) {\widehat{F}}_Y\left( \frac{z (m-i-1)}{m}\right) -{\widehat{F}}_X\left( \frac{i z}{m}\right) {\widehat{F}}_Y\left( \frac{z (m-i-1)}{m}\right) \nonumber \\\ \quad \quad \quad{} & {} +{\widehat{F}}_X\left( \frac{(i+1) z}{m}\right) {\widehat{F}}_Y\left( \frac{z (m-i)}{m}\right) -{\widehat{F}}_X\left( \frac{i z}{m}\right) {\widehat{F}}_Y\left( \frac{z (m-i)}{m}\right) \nonumber \\ \quad \quad \quad{} & {} - F_X\left( \frac{(i+1) z}{m}\right) F_Y\left( \frac{z (m-i-1)}{m}\right) + F_X\left( \frac{i z}{m}\right) F_Y\left( \frac{z (m-i-1)}{m}\right) \nonumber \\ \quad \quad \quad{} & {} - F_X\left( \frac{(i+1) z}{m}\right) F_Y\left( \frac{z (m-i)}{m}\right) + F_X\left( \frac{i z}{m}\right) F_Y\left( \frac{z (m-i)}{m}\right) \Big \}\nonumber \\= & {} \frac{1}{2}\sum _{i=0}^{m-1}\Big \{\Big [{\widehat{F}}_X\left( \frac{(i+1) z}{m}\right) {\widehat{F}}_Y\left( \frac{z (m-i-1)}{m}\right) - F_X\left( \frac{(i+1) z}{m}\right) F_Y\left( \frac{z (m-i-1)}{m}\right) \Big ]\nonumber \\ \quad \quad \quad{} & {} +\Big [ F_X\left( \frac{i z}{m}\right) F_Y\left( \frac{z (m-i-1)}{m}\right) -{\widehat{F}}_X\left( \frac{i z}{m}\right) {\widehat{F}}_Y\left( \frac{z (m-i-1)}{m}\right) \Big ]\nonumber \\ \quad \quad \quad{} & {} +\Big [{\widehat{F}}_X\left( \frac{(i+1) z}{m}\right) {\widehat{F}}_Y\left( \frac{z (m-i)}{m}\right) - F_X\left( \frac{(i+1) z}{m}\right) F_Y\left( \frac{z (m-i)}{m}\right) \Big ]\nonumber \\ \quad \quad \quad{} & {} +\Big [ F_X\left( \frac{i z}{m}\right) F_Y\left( \frac{z (m-i)}{m}\right) -{\widehat{F}}_X\left( \frac{i z}{m}\right) {\widehat{F}}_Y\left( \frac{z (m-i)}{m}\right) \Big ]\Big \}\nonumber \\= & {} \frac{1}{2}\sum _{i=0}^{m-1}\left\{ A_i(z)+B_i(z)+C_i(z)+D_i(z)\right\} \quad (say).

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