xc```, fa`2Y&0*.ngN4{Wu^$-YyR?6S-Dz c` That is clearly what we see. }q_1^jq_2^{k-2j}q_3^{n-k+j}, &{} \text{ if } k\le n\\ \sum _{j=k-n}^{\frac{1}{4} \left( 2 k+(-1)^k-1\right) }\frac{n!}{j! >> Correspondence to It's too bad there isn't a sticky section, which contains questions that contain answers that go above and beyond what's required (like yours in the link). They are completely specied by a joint pdf fX,Y such that for any event A (,)2, P{(X,Y . Finally, the symmetrization replaces $z$ by $|z|$, allows its values to range now from $-20$ to $20$, and divides the pdf by $2$ to spread the total probability equally across the intervals $(-20,0)$ and $(0,20)$: $$\eqalign{ Why condition on either the r.v. Generate a UNIFORM random variate using rand, not randn. Here we have \(2q_1+q_2=2F_{Z_m}(z)\) and it follows as below; ##*************************************************************, for(i in 1:m){F=F+0.5*(xf(i*z/m)-xf((i-1)*z/m))*(yf((m-i-2)*z/m)+yf((m-i-1)*z/m))}, ##************************End**************************************. Which language's style guidelines should be used when writing code that is supposed to be called from another language? endobj $\endgroup$ - Xi'an. /BBox [0 0 8 8] Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. Sums of independent random variables. /FormType 1 /Filter /FlateDecode << We thank the referees for their constructive comments which helped us to improve the presentation of the manuscript in its current form. /Type /Page Requires the first input to be the name of a distribution. >>>> offers. the statistical profession on topics that are important for a broad group of stream Chapter 5. endobj Thus, since we know the distribution function of \(X_n\) is m, we can find the distribution function of \(S_n\) by induction. &=\frac{\log\{20/|v|\}}{40}\mathbb{I}_{-20\le v\le 20} % /SaveTransparency false To find \(P(2X_1+X_2=k)\), we consider four cases. Since $X\sim\mathcal{U}(0,2)$, $$f_X(x) = \frac{1}{2}\mathbb{I}_{(0,2)}(x)$$so in your convolution formula Learn more about Institutional subscriptions, Atkinson KE (2008) An introduction to numerical analysis. /Type /XObject /Resources << xP( MATH - 158.69.202.20. stream /Matrix [1 0 0 1 0 0] Other MathWorks country Suppose the \(X_i\) are uniformly distributed on the interval [0,1]. \end{aligned}$$, $$\begin{aligned} E\left( e^{(t_1X_1+t_2X_2+t_3X_3)}\right) =(q_1e^{t_1}+q_2e^{t_2}+q_3e^{t_3})^n. Sep 26, 2020 at 7:18. stream /BBox [0 0 362.835 5.313] . /Trans << /S /R >> 108 0 obj 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. /DefaultRGB 39 0 R Accelerating the pace of engineering and science. \end{aligned}$$, \(\sup _{z}|{\widehat{F}}_X(z)-F_X(z)|\rightarrow 0 \), \(\sup _{z}|{\widehat{F}}_Y(z)-F_Y(z)|\rightarrow 0 \), \(\sup _{z}|A_i(z)|\rightarrow 0\,\,\, a.s.\), \(\sup _{z}|B_i(z)|,\,\sup _{z}|C_i(z)|\), $$\begin{aligned} \sup _{z} |{\widehat{F}}_Z(z) - F_{Z_m}(z)|= & {} \sup _{z} \left| \frac{1}{2}\sum _{i=0}^{m-1}\left\{ A_i(z)+B_i(z)+C_i(z)+D_i(z)\right\} \right| \\\le & {} \frac{1}{2}\sum _{i=0}^{m-1} \sup _{z}|A_i(z)|+ \frac{1}{2}\sum _{i=0}^{m-1} \sup _{z}|B_i(z)|\\{} & {} +\frac{1}{2}\sum _{i=0}^{m-1} \sup _{z}|C_i(z)|+\frac{1}{2}\sum _{i=0}^{m-1} \sup _{z}|D_i(z)| \\\rightarrow & {} 0\,\,\, a.s. \end{aligned}$$, $$\begin{aligned} \sup _{z} |{\widehat{F}}_Z(z) - F_{Z}(z)|\le \sup _{z} |{\widehat{F}}_Z(z) - F_{Z_m}(z)|+\sup _{z} | F_{Z_m}(z)-F_Z(z) |. Thus, \[\begin{array}{} P(S_2 =2) & = & m(1)m(1) \\ & = & \frac{1}{6}\cdot\frac{1}{6} = \frac{1}{36} \\ P(S_2 =3) & = & m(1)m(2) + m(2)m(1) \\ & = & \frac{1}{6}\cdot\frac{1}{6} + \frac{1}{6}\cdot\frac{1}{6} = \frac{2}{36} \\ P(S_2 =4) & = & m(1)m(3) + m(2)m(2) + m(3)m(1) \\ & = & \frac{1}{6}\cdot\frac{1}{6} + \frac{1}{6}\cdot\frac{1}{6} + \frac{1}{6}\cdot\frac{1}{6} = \frac{3}{36}\end{array}\]. \frac{5}{4} - \frac{1}{4}z, &z \in (4,5)\\ Google Scholar, Panjer HH, Willmot GE (1992) Insurance risk models, vol 479. Let $X$ ~ $U(0,2)$ and $Y$ ~ $U(-10,10)$ be two independent random variables with the given distributions. /Filter /FlateDecode /Filter /FlateDecode Copy the n-largest files from a certain directory to the current one, Are these quarters notes or just eighth notes? /Subtype /Form 21 0 obj Thank you! In your derivation, you do not use the density of $X$. Suppose X and Y are two independent discrete random variables with distribution functions \(m_1(x)\) and \(m_2(x)\). Example 7.5), \[f_{X_i}(x) = \frac{1}{\sqrt{2pi}} e^{-x^2/2}, \nonumber \], \[f_{S_n}(x) = \frac{1}{\sqrt{2\pi n}}e^{-x^2/2n} \nonumber \]. $$f_Z(z) = What are you doing wrong? Which ability is most related to insanity: Wisdom, Charisma, Constitution, or Intelligence? /ColorSpace << The results of the simulation study are reported in Table 6.In Table 6, we report MSE \(\times 10^3\) as the MSE of the estimators is . The distribution for S3 would then be the convolution of the distribution for \(S_2\) with the distribution for \(X_3\). $$h(v) = \int_{y=-\infty}^{y=+\infty}\frac{1}{y}f_Y(y) f_X\left (\frac{v}{y} \right ) dy$$. endobj /Resources 19 0 R The probability of having an opening bid is then, Since we have the distribution of C, it is easy to compute this probability. Using the program NFoldConvolution find the distribution for your total winnings after ten (independent) plays. x=0w]=CL?!Q9=\ ifF6kiSw D$8haFrPUOy}KJul\!-WT3u-ikjCWX~8F+knT`jOs+DuO /FormType 1 The convolution of k geometric distributions with common parameter p is a negative binomial distribution with parameters p and k. This can be seen by considering the experiment which consists of tossing a coin until the kth head appears. Probability function for difference between two i.i.d. /BBox [0 0 353.016 98.673] HTiTSY~I(6E@E!$I,m8ahElDADVY*$}pA6YDEMI m3?L{U$VY(DL6F ?_]hTaf @JP D%@ZX=\0A?3J~HET,)p\*Z&mbkYZbUDk9r'F;*F6\%sc}. << Using the comment by @whuber, I believe I arrived at a more efficient method to reach the solution. 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. /Resources 25 0 R Let \(Y_3\) be the maximum value obtained. /Resources 21 0 R Much can be accomplished by focusing on the forms of the component distributions: $X$ is twice a $U(0,1)$ random variable. Accessibility StatementFor more information contact us atinfo@libretexts.org. stream xZKs6W|ud&?TYz>Hi8i2d)B H| H##/c@aDADra&{G=RA,XXoP!%. The probability that 1 person arrives is p and that no person arrives is \(q = 1 p\). (c) Given the distribution pX , what is his long-term batting average? Now let \(R^2 = X^2 + Y^2\), Sum of Two Independent Normal Random Variables, source@https://chance.dartmouth.edu/teaching_aids/books_articles/probability_book/book.html. /Type /XObject IEEE Trans Commun 43(12):28692873, Article /Subtype /Form Is that correct? It is possible to calculate this density for general values of n in certain simple cases. Um, pretty much everything? 0. 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)! (Sum of Two Independent Uniform Random Variables) . Also it can be seen that \(\cup _{i=0}^{m-1}A_i\) and \(\cup _{i=0}^{m-1}B_i\) are disjoint. 17 0 obj << $$h(v)= \frac{1}{20} \int_{-10}^{10} \frac{1}{|y|}\cdot \frac{1}{2}\mathbb{I}_{(0,2)}(v/y)\text{d}y$$(I also corrected the Jacobian by adding the absolute value). .. /Matrix [1 0 0 1 0 0] (This last step converts a non-negative variate into a symmetric distribution around $0$, both of whose tails look like the original distribution.). For instance, this characterization gives us a way to generate realizations of $XY$ directly, as in this R expression: Thsis analysis also reveals why the pdf blows up at $0$. What is the distribution of $V=XY$? For this reason we must negate the result after the substitution, giving, $$f(t)dt = -\left(-\log(z) e^{-(-\log(z))} (-dz/z)\right) = -\log(z) dz,\ 0 \lt z \lt 1.$$, The scale factor of $20$ converts this to, $$-\log(z/20) d(z/20) = -\frac{1}{20}\log(z/20)dz,\ 0 \lt z \lt 20.$$. . endobj Consider if the problem was $X \sim U([1,5])$ and $Y \sim U([1,2] \cup [4,5] \cup [7,8] \cup [10, 11])$. uniform random variables I Suppose that X and Y are i.i.d. $$h(v)=\frac{1}{40}\int_{y=-10}^{y=10} \frac{1}{y}dy$$. rev2023.5.1.43405. f_{XY}(z)dz &= -\frac{1}{2}\frac{1}{20} \log(|z|/20),\ -20 \lt z\lt 20;\\ . Choose a web site to get translated content where available and see local events and Two MacBook Pro with same model number (A1286) but different year. 15 0 obj << . 35 0 obj /Subtype /Form stream Then, the pdf of $Z$ is the following convolution general solution sum of two uniform random variables aY+bX=Z? Something tells me, there is something weird here since it is discontinuous at 0. . 20 0 obj Let \(\{\cup _{i=0}^{m-1}A_i,\,\cup _{i=0}^{m-1}B_i,\,\left( \cup _{i=0}^{m-1}(A_i\cup B_i) \right) ^c\}\) be a partition of \((0,\infty )\times (0,\infty )\). \nonumber \]. So, if we let $\lambda$ be the Lebesgue measure and notice that $[1,2]$ and $[4,5]$ disjoint, then the pdfs are, $$f_X(x) = >> << So then why are you using randn, which produces a GAUSSIAN (normal) random variable? xP( Then the distribution function of \(S_1\) is m. We can write. What does 'They're at four. /XObject << Should there be a negative somewhere? >> Summing two random variables I Say we have independent random variables X and Y and we know their density functions f . 103 0 obj It's not them. /ProcSet [ /PDF ] Provided by the Springer Nature SharedIt content-sharing initiative, Over 10 million scientific documents at your fingertips, Not logged in Thank you for the link! /Group << /S /Transparency /CS /DeviceGray >> Find the distribution of \(Y_n\). Find the treasures in MATLAB Central and discover how the community can help you! It doesn't look like uniform. /ProcSet [ /PDF ] The random variable $XY$ is the symmetrized version of $20$ times the exponential of the negative of a $\Gamma(2,1)$ variable. Indeed, it is well known that the negative log of a $U(0,1)$ variable has an Exponential distribution (because this is about the simplest way to generate random exponential variates), whence the negative log of the product of two of them has the distribution of the sum of two Exponentials. >> stream Their distribution functions are then defined on these integers. Wiley, Hoboken, Beaulieu NC, Abu-Dayya AA, McLane PJ (1995) Estimating the distribution of a sum of independent lognormal random variables. \sum _{i=0}^{m-1}\left[ \left( \#X_v's \text { between } \frac{iz}{m} \text { and } \frac{(i+1) z}{m}\right) \times \left( \#Y_w's\le \frac{(m-i-1) z}{m}\right) \right] \right\} \\&=\frac{1}{2n_1n_2}(C_2+2C_1)\,(say), \end{aligned}$$, $$\begin{aligned} C_1=\sum _{i=0}^{m-1}\left[ \left( \#X_v's \text { between } \frac{iz}{m} \text { and } \frac{(i+1) z}{m}\right) \times \left( \#Y_w's\le \frac{(m-i-1) z}{m}\right) \right] \end{aligned}$$, $$\begin{aligned} C_2=\sum _{i=0}^{m-1}\left[ \left( \#X_v's \text { between } \frac{iz}{m} \text { and } \frac{(i+1) z}{m}\right) \times \left( \#Y_w's\text { between } \frac{(m-i-1) z}{m} \text { and } \frac{(m-i) z}{m}\right) \right] . Use MathJax to format equations. In this paper, we obtain an approximation for the distribution function of sum of two independent random variables using quantile based representation. endobj << Why did DOS-based Windows require HIMEM.SYS to boot? Learn more about Stack Overflow the company, and our products. >> (2023)Cite this article. endobj endstream xr6_!EJ&U3ohDo7 I=RD }*n$zy=9O"e"Jay^Hn#fB#Vg!8|44%2"X1$gy"SI0WJ%Jd LOaI&| >-=c=OCgc Is there such a thing as aspiration harmony? $|Y|$ is ten times a $U(0,1)$ random variable. MathJax reference. I Sum Z of n independent copies of X? By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. /PieceInfo << /Resources 17 0 R We might be content to stop here. << \frac{1}{2}z - \frac{3}{2}, &z \in (3,4)\\ A sum of more terms would gradually start to look more like a normal distribution, the law of large numbers tells us that. >> What are the advantages of running a power tool on 240 V vs 120 V? Extensive Monte Carlo simulation studies are carried out to evaluate the bias and mean squared error of the estimator and also to assess the approximation error. )f{Wd;$&\KqqirDUq*np 2 *%3h#(A9'p6P@01 v#R ut Zl0r( %HXOR",xq=s2-KO3]Q]Xn"}P|#'lI >o&in|kSQXWwm`-5qcyDB3k(#)3%uICELh YhZ#DL*nR7xwP O|. << To learn more, see our tips on writing great answers. /Subtype /Form By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Would My Planets Blue Sun Kill Earth-Life? /Length 15 plished, the resultant function will be the pdf, denoted by g(w), for the sum of random variables stated in conventional form. /PTEX.PageNumber 1 This is a preview of subscription content, access via your institution. You want to find the pdf of the difference between two uniform random variables. 107 0 obj /Length 29 We also know that $f_Y(y) = \frac{1}{20}$, $$h(v)= \frac{1}{20} \int_{y=-10}^{y=10} \frac{1}{y}\cdot \frac{1}{2}dy$$ i.e. What differentiates living as mere roommates from living in a marriage-like relationship? Products often are simplified by taking logarithms. /ProcSet [ /PDF ] /ProcSet [ /PDF ] In this section, we'll talk about how to nd the distribution of the sum of two independent random variables, X+ Y, using a technique called . 0, &\text{otherwise} Therefore $XY$ (a) is symmetric about $0$ and (b) its absolute value is $2\times 10=20$ times the product of two independent $U(0,1)$ random variables. We shall find it convenient to assume here that these distribution functions are defined for all integers, by defining them to be 0 where they are not otherwise defined. It only takes a minute to sign up. into sections: Statistical Practice, General, Teacher's Corner, Statistical /Subtype /Form endobj The journal is organized stream \[ \begin{array}{} (a) & What is the distribution for \(T_r\) \\ (b) & What is the distribution \(C_r\) \\ (c) Find the mean and variance for the number of customers arriving in the first r minutes \end{array}\], (a) A die is rolled three times with outcomes \(X_1, X_2\) and \(X_3\). /Filter /FlateDecode endobj /FormType 1 /Matrix [1 0 0 1 0 0] As I understand the LLN, it makes statements about the convergence of the sample mean, but not about the distribution of the sample mean. >> stream I had to plot the PDF of X = U1 U2, where U1 and U2 are uniform random variables . Building on two centuries' experience, Taylor & Francis has grown rapidlyover the last two decades to become a leading international academic publisher.The Group publishes over 800 journals and over 1,800 new books each year, coveringa wide variety of subject areas and incorporating the journal imprints of Routledge,Carfax, Spon Press, Psychology Press, Martin Dunitz, and Taylor & Francis.Taylor & Francis is fully committed to the publication and dissemination of scholarly information of the highest quality, and today this remains the primary goal. &= \frac{1}{40} \mathbb{I}_{-20\le v\le 0} \log\{20/|v|\}+\frac{1}{40} \mathbb{I}_{0\le v\le 20} \log\{20/|v|\}\\ Statistical Papers To do this, it is enough to determine the probability that Z takes on the value z, where z is an arbitrary integer.Suppose that X = k, where k is some integer. 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 \(p\). Please help. Could a subterranean river or aquifer generate enough continuous momentum to power a waterwheel for the purpose of producing electricity? /Shading << /Sh << /ShadingType 3 /ColorSpace /DeviceRGB /Domain [0 1] /Coords [8.00009 8.00009 0.0 8.00009 8.00009 8.00009] /Function << /FunctionType 2 /Domain [0 1] /C0 [1 1 1] /C1 [0 0 0] /N 1 >> /Extend [true false] >> >> }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! }$$. /Type /XObject 20 0 obj (Be sure to consider the case where one or more sides turn up with probability zero. ', referring to the nuclear power plant in Ignalina, mean? What is Wario dropping at the end of Super Mario Land 2 and why? Thus, we have found the distribution function of the random variable Z. Suppose X and Y are two independent random variables, each with the standard normal density (see Example 5.8). maybe something with log? The point count of the hand is then the sum of the values of the cards in the hand. of \(\frac{2X_1+X_2-\mu }{\sigma }\) converges to \(e^{\frac{t^2}{2}},\) which is the m.g.f. To me, the latter integral seems like the better choice to use. Therefore X Y (a) is symmetric about 0 and (b) its absolute value is 2 10 = 20 times the product of two independent U ( 0, 1) random variables. /Type /XObject endobj /XObject << /Fm5 20 0 R >> /BBox [0 0 362.835 2.657] /Subtype /Form I5I'hR-U&bV&L&xN'uoMaKe!*R'ojYY:`9T+_:8h);-mWaQ9~:|%(Lw. What are the advantages of running a power tool on 240 V vs 120 V? f_{XY}(z)dz &= 0\ \text{otherwise}. 18 0 obj Google Scholar, Buonocore A, Pirozzi E, Caputo L (2009) A note on the sum of uniform random variables. endobj \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 }}. >> /Length 15 /Subtype /Form endstream stream MathSciNet >> Note that when $-20\lt v \lt 20$, $\log(20/|v|)$ is. /Size 4458 Plot this distribution. sites are not optimized for visits from your location. Legal. By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site :) (Hey, what can I say?) >> 14 0 obj 23 0 obj /ProcSet [ /PDF ] 2 - \frac{1}{4}z, &z \in (7,8)\\ Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. \[ p_X = \bigg( \begin{array}{} 0 & 1 & 2 \\ 1/2 & 3/8 & 1/2 \end{array} \bigg) \]. %PDF-1.5 (It is actually more complicated than this, taking into account voids in suits, and so forth, but we consider here this simplified form of the point count.) N Am Actuar J 11(2):99115, Zhang C-H (2005) Estimation of sums of random variables: examples and information bounds. I was hoping for perhaps a cleaner method than strictly plotting. endobj statisticians, and ordinarily not highly technical. 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 . What I was getting at is it is a bit cumbersome to draw a picture for problems where we have disjoint intervals (see my comment above). \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) . Consider a Bernoulli trials process with a success if a person arrives in a unit time and failure if no person arrives in a unit time. In our experience, deriving and working with the pdf for sums of random variables facilitates an understanding of the convergence properties of the density of such sums and motivates consideration of other algebraic manipulation for random variables. h(v) &= \frac{1}{40} \int_{-10}^{0} \frac{1}{|y|} \mathbb{I}_{0\le v/y\le 2}\text{d}y+\frac{1}{40} \int_{0}^{10} \frac{1}{|y|}\mathbb{I}_{0\le v/y\le 2}\text{d}y\\ &= \frac{1}{40} \int_{-10}^{0} \frac{1}{|y|} \mathbb{I}_{0\ge v/2\ge y\ge -10}\text{d}y+\frac{1}{40} \int_{0}^{10} \frac{1}{|y|}\mathbb{I}_{0\le v/2\le y\le 10}\text{d}y\\&= \frac{1}{40} \mathbb{I}_{-20\le v\le 0} \int_{-10}^{v/2} \frac{1}{|y|}\text{d}y+\frac{1}{40} \mathbb{I}_{20\ge v\ge 0} \int_{v/2}^{10} \frac{1}{|y|}\text{d}y\\ \frac{1}{2}z - \frac{3}{2}, &z \in (3,4)\\ What positional accuracy (ie, arc seconds) is necessary to view Saturn, Uranus, beyond? << Pdf of sum of two uniform random variables on $\left[-\frac{1}{2},\frac{1}{2}\right]$ Ask Question Asked 2 years, 6 months ago. Suppose we choose independently two numbers at random from the interval [0, 1] with uniform probability density. You can also select a web site from the following list: Select the China site (in Chinese or English) for best site performance. Why is my arxiv paper not generating an arxiv watermark? >> Making statements based on opinion; back them up with references or personal experience. In this case the density \(f_{S_n}\) for \(n = 2, 4, 6, 8, 10\) is shown in Figure 7.8. Thanks, The answer looks correct, cgo. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. << \end{cases} \end{cases}$$. Ann Stat 33(5):20222041. It's not them. stream Indeed, it is well known that the negative log of a U ( 0, 1) variable has an Exponential distribution (because this is about the simplest way to . John Venier left a comment to a previous post about the following method for generating a standard normal: add 12 uniform random variables and subtract 6.
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