pdf of sum of two uniform random variables
Why does the cusp in the PDF of $Z_n$ disappear for $n \geq 3$? \end{aligned}$$, $$\begin{aligned} P(X_1=x_1,X_2=x_2,X_3=n-x_1-x_2)=\frac{n!}{x_1! Then you arrive at ($\star$) below. Copy the n-largest files from a certain directory to the current one, Are these quarters notes or just eighth notes? $$h(v)= \frac{1}{20} \int_{-10}^{10} \frac{1}{|y|}\cdot \frac{1}{2}\mathbb{I}_{(0,2)}(v/y)\text{d}y$$(I also corrected the Jacobian by adding the absolute value). 14 0 obj /LastModified (D:20140818172507-05'00') Pdf of the sum of two independent Uniform R.V., but not identical. (Assume that neither a nor b is concentrated at 0.). 107 0 obj Continuing in this way we would find \(P(S_2 = 5) = 4/36, P(S_2 = 6) = 5/36, P(S_2 = 7) = 6/36, P(S_2 = 8) = 5/36, P(S_2 = 9) = 4/36, P(S_2 = 10) = 3/36, P(S_2 = 11) = 2/36,\) and \(P(S_2 = 12) = 1/36\). We would like to determine the distribution function m3(x) of Z. You may receive emails, depending on your. \\&\left. stream N Am Actuar J 11(2):99115, Zhang C-H (2005) Estimation of sums of random variables: examples and information bounds. /Filter /FlateDecode /ModDate (D:20140818172507-05'00') /SaveTransparency false $$h(v) = \int_{y=-\infty}^{y=+\infty}\frac{1}{y}f_Y(y) f_X\left (\frac{v}{y} \right ) dy$$. /BBox [0 0 337.016 8] /XObject << /Fm5 20 0 R >> >> We thank the referees for their constructive comments which helped us to improve the presentation of the manuscript in its current form. /Filter /FlateDecode << endstream Is there such a thing as aspiration harmony? /Subtype /Form Suppose X and Y are two independent random variables, each with the standard normal density (see Example 5.8). xP( Please help. I was still finding this a bit counter intuitive so I just executed this (similar to Xi'an's "simulation"): Hi, Thanks. Plot this distribution. /Subtype /Form Suppose the \(X_i\) are uniformly distributed on the interval [0,1]. Values within (say) $\varepsilon$ of $0$ arise in many ways, including (but not limited to) when (a) one of the factors is less than $\varepsilon$ or (b) both the factors are less than $\sqrt{\varepsilon}$. Suppose X and Y are two independent discrete random variables with distribution functions \(m_1(x)\) and \(m_2(x)\). . \end{aligned}$$, $$\begin{aligned} P(2X_1+X_2=k)= & {} P(X_1=k-n,X_2=2n-k,X_3=0)\\+ & {} P(X_1=k-n+1,X_2=2n-k-2,X_3=1)\\{} & {} +\dots +P(X_1=\frac{k-1}{2},X_2=1,X_3=n-\frac{k+1}{2})\\= & {} \sum _{j=k-n}^{\frac{k-1}{2}}P(X_1=j,X_2=k-2j,X_3=n-k+j)\\= & {} \sum _{j=k-n}^{\frac{k-1}{2}}\frac{n!}{j! In 5e D&D and Grim Hollow, how does the Specter transformation affect a human PC in regards to the 'undead' characteristics and spells? But I'm having some difficulty on choosing my bounds of integration? Learn more about Stack Overflow the company, and our products. /Filter /FlateDecode endobj Well, theoretically, one would expect the solution to be a triangle distribution, with peak at 0, and extremes at -1 and 1. /Type /XObject Sorry, but true. 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 /RoundTrip 1 By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. 0, &\text{otherwise} endobj We explain: first, how to work out the cumulative distribution function of the sum; then, how to compute its probability mass function (if the summands are discrete) or its probability density function (if the summands are continuous). \quad\text{and}\quad stream /Type /Page That square root is enormously larger than $\varepsilon$ itself when $\varepsilon$ is close to $0$. \frac{1}{\lambda([1,2] \cup [4,5])} = \frac{1}{1 + 1} = \frac{1}{2}, &y \in [1,2] \cup [4,5] \\ \\&\left. Legal. /Im0 37 0 R It only takes a minute to sign up. Use MathJax to format equations. \\&\left. \frac{1}{4}z - \frac{1}{2}, &z \in (2,3) \tag{$\star$}\\ /Filter /FlateDecode Statistical Papers So f . Thank you for the link! To do this, it is enough to determine the probability that Z takes on the value z, where z is an arbitrary integer. Consequently. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. endobj 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 \((0,\infty )\times (0,\infty )\). What is Wario dropping at the end of Super Mario Land 2 and why? Commun Stat Theory Methods 47(12):29692978, Article 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. Consider the sum of $n$ uniform distributions on $[0,1]$, or $Z_n$. (k-2j)!(n-k+j)!}q_1^jq_2^{k-2j}q_3^{n-k+j}. If a card is dealt at random to a player, then the point count for this card has distribution. That is clearly what we . MathJax reference. First, simple averages . 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. V%H320I !.V The \(X_1\) and \(X_2\) have the common distribution function: \[ m = \bigg( \begin{array}{}1 & 2 & 3 & 4 & 5 & 6 \\ 1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 \end{array} \bigg) .\]. /Shading << /Sh << /ShadingType 3 /ColorSpace /DeviceRGB /Domain [0 1] /Coords [4.00005 4.00005 0.0 4.00005 4.00005 4.00005] /Function << /FunctionType 2 /Domain [0 1] /C0 [0.5 0.5 0.5] /C1 [0 0 0] /N 1 >> /Extend [true false] >> >> Question. /ProcSet [ /PDF ] stream For this to be possible, the density of the product has to become arbitrarily large at $0$. /Length 15 To me, the latter integral seems like the better choice to use. endobj So then why are you using randn, which produces a GAUSSIAN (normal) random variable? endobj \[ p_x = \bigg( \begin{array}{} 0&1 & 2 & 3 & 4 \\ 36/52 & 4/52 & 4/52 & 4/52 & 4/52 \end{array} \bigg) \]. /Matrix [1 0 0 1 0 0] 0, &\text{otherwise} endobj endstream /Matrix [1 0 0 1 0 0] Is this distribution bell-shaped for large values of n? /Matrix [1 0 0 1 0 0] This lecture discusses how to derive the distribution of the sum of two independent random variables. /XObject << /FormType 1 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**************************************. \end{cases}$$. /Group << /S /Transparency /CS /DeviceGray >> &=\frac{\log\{20/|v|\}}{40}\mathbb{I}_{-20\le v\le 20} Stat Probab Lett 79(19):20922097, Frees EW (1994) Estimating densities of functions of observations. \frac{5}{4} - \frac{1}{4}z, &z \in (4,5)\\ A die is rolled twice. /Creator (Adobe Photoshop 7.0) << 21 0 obj Easy Understanding of Convolution The best way to understand convolution is given in the article in the link,using that . Pdf of the sum of two independent Uniform R.V., but not identical. 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 \(p\). /XObject << Doing this we find that, so that about one in four hands should be an opening bid according to this simplified model. (This last step converts a non-negative variate into a symmetric distribution around $0$, both of whose tails look like the original distribution.). /BBox [0 0 362.835 2.657] For certain special distributions it is possible to find an expression for the distribution that results from convoluting the distribution with itself n times. /BBox [0 0 16 16] Suppose X and Y are two independent discrete random variables with distribution functions \(m_1(x)\) and \(m_2(x)\). /FormType 1 xP( \frac{1}{2}z - \frac{3}{2}, &z \in (3,4)\\ xP( $$\begin{aligned}{} & {} {\widehat{F}}_Z(z) - F_{Z_m}(z)\\= & {} \left\{ \frac{1}{2}\sum _{i=0}^{m-1}\left( {\widehat{F}}_X\left( \frac{(i+1) z}{m}\right) -{\widehat{F}}_X\left( \frac{i z}{m}\right) \right) \left( {\widehat{F}}_Y\left( \frac{z (m-i-1)}{m}\right) +{\widehat{F}}_Y\left( \frac{z (m-i)}{m}\right) \right) \right\} \\{} & {} -\left\{ \frac{1}{2}\sum _{i=0}^{m-1}\left( F_X\left( \frac{(i+1) z}{m}\right) -F_X\left( \frac{i z}{m}\right) \right) \left( F_Y\left( \frac{z (m-i-1)}{m}\right) +F_Y\left( \frac{z (m-i)}{m}\right) \right) \right\} \\= & {} \frac{1}{2}\sum _{i=0}^{m-1}\left\{ \left( {\widehat{F}}_X\left( \frac{(i+1) z}{m}\right) -{\widehat{F}}_X\left( \frac{i z}{m}\right) \right) \left( {\widehat{F}}_Y\left( \frac{z (m-i-1)}{m}\right) +{\widehat{F}}_Y\left( \frac{z (m-i)}{m}\right) \right) \right\} \\{} & {} -\frac{1}{2}\sum _{i=0}^{m-1}\left\{ \left( F_X\left( \frac{(i+1) z}{m}\right) -F_X\left( \frac{i z}{m}\right) \right) \left( F_Y\left( \frac{z (m-i-1)}{m}\right) +F_Y\left( \frac{z (m-i)}{m}\right) \right) \right\} \end{aligned}$$, $$\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). Are these quarters notes or just eighth notes? Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. endobj xcbd`g`b``8 "U A)4J@e v
o u 2 Choose a web site to get translated content where available and see local events and /Type /XObject 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. Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. We see that, as in the case of Bernoulli trials, the distributions become bell-shaped. Connect and share knowledge within a single location that is structured and easy to search. 2023 Springer Nature Switzerland AG. Indian Statistical Institute, New Delhi, India, Indian Statistical Institute, Chennai, India, You can also search for this author in 15 0 obj \end{cases} Let \(X\) and \(Y\) be two independent integer-valued random variables, with distribution functions \(m_1(x)\) and \(m_2(x)\) respectively. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. 104 0 obj ), (Lvy\(^2\) ) Assume that n is an integer, not prime. You can also select a web site from the following list: Select the China site (in Chinese or English) for best site performance. /BBox [0 0 8 87.073] \frac{1}{4}z - \frac{5}{4}, &z \in (5,6)\\ Find the treasures in MATLAB Central and discover how the community can help you! /Subtype /Form Let \(C_r\) be the number of customers arriving in the first r minutes. Why condition on either the r.v. $$f_Z(z) = Note that, Then, it is observed that, \((C_1,C_2,C_3)\) is distributed as multinomial distribution with parameters \(\left( n_1 n_2,q_1,q_2,q_3\right) ,\) where \(q_1,\,q_2\) and \(q_3\) are as specified in the statement of the theorem. }\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. Learn more about Institutional subscriptions, Atkinson KE (2008) An introduction to numerical analysis. https://doi.org/10.1007/s00362-023-01413-4, DOI: https://doi.org/10.1007/s00362-023-01413-4. 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 $0$. Let us regard the total hand of 13 cards as 13 independent trials with this common distribution. /LastModified (D:20140818172507-05'00') The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. endstream /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] >> >> A die is rolled three times. . >> of \(\frac{2X_1+X_2-\mu }{\sigma }\) converges to \(e^{\frac{t^2}{2}},\) which is the m.g.f. Using the comment by @whuber, I believe I arrived at a more efficient method to reach the solution. For terms and use, please refer to our Terms and Conditions >> This leads to the following definition. Are there any constraint on these terms? \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-1}{2},X_2=1,X_3=n-\frac{k+1}{2})\\= & {} \sum _{j=0}^{\frac{k-1}{2}}P(X_1=j,X_2=k-2j,X_3=n-k+j)\\ {}{} & {} =\sum _{j=0}^{\frac{k-1}{2}}\frac{n!}{j! Then %PDF-1.5 Marcel Dekker Inc., New York, Moschopoulos PG (1985) The distribution of the sum of independent gamma random variables. Learn more about Stack Overflow the company, and our products. We also know that $f_Y(y) = \frac{1}{20}$, $$h(v)= \frac{1}{20} \int_{y=-10}^{y=10} \frac{1}{y}\cdot \frac{1}{2}dy$$ Much can be accomplished by focusing on the forms of the component distributions: $X$ is twice a $U(0,1)$ random variable. i.e. a society or other partner) holds exclusive rights to this article under a publishing agreement with the author(s) or other rightsholder(s); author self-archiving of the accepted manuscript version of this article is solely governed by the terms of such publishing agreement and applicable law. Use this find the distribution of \(Y_3\). Example 7.5), \[f_{X_i}(x) = \frac{1}{\sqrt{2pi}} e^{-x^2/2}, \nonumber \], \[f_{S_n}(x) = \frac{1}{\sqrt{2\pi n}}e^{-x^2/2n} \nonumber \]. 12 0 obj Viewed 132 times 2 $\begingroup$ . /Resources 19 0 R Why does Acts not mention the deaths of Peter and Paul? What does 'They're at four. Sums of independent random variables. So, we have that $f_X(t -y)f_Y(y)$ is either $0$ or $\frac{1}{4}$. /PTEX.FileName (../TeX/PurdueLogo.pdf) Interpreting non-statistically significant results: Do we have "no evidence" or "insufficient evidence" to reject the null? \(\square \). the PDF of W=X+Y /CreationDate (D:20140818172507-05'00') << /Filter /FlateDecode /S 100 /O 156 /Length 146 >> 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 . EE 178/278A: Multiple Random Variables Page 3-11 Two Continuous Random variables - Joint PDFs Two continuous r.v.s dened over the same experiment are jointly continuous if they take on a continuum of values each with probability 0. That is clearly what we see. 20 0 obj >> /Length 15 endobj /Subtype /Form Then the distribution function of \(S_1\) is m. We can write. So how might you plot the pdf of a difference of two uniform variables? Sum of two independent uniform random variables in different regions. << Find the distribution of, \[ \begin{array}{} (a) & Y+X \\ (b) & Y-X \end{array}\]. I'm learning and will appreciate any help. Legal. What more terms would be added to make the pdf of the sum look normal? << As I understand the LLN, it makes statements about the convergence of the sample mean, but not about the distribution of the sample mean. Does \(Y_3\) have a bell-shaped distribution? f_Y(y) = Other MathWorks country endstream Let Z = X + Y. Using the program NFoldConvolution find the distribution for your total winnings after ten (independent) plays. (14), we can write, As \(n_1,n_2\rightarrow \infty \), the right hand side of the above expression converges to zero a.s. \(\square \), The p.m.f. maybe something with log? 103 0 obj Thanks, The answer looks correct, cgo. Thus \(P(S_3 = 3) = P(S_2 = 2)P(X_3 = 1)\). Thanks for contributing an answer to Cross Validated! stream Something tells me, there is something weird here since it is discontinuous at 0. >> Which was the first Sci-Fi story to predict obnoxious "robo calls"? endstream >>/ProcSet [ /PDF /ImageC ] + X_n \) be the sum of n independent random variables of an independent trials process with common distribution function m defined on the integers. Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Find the distribution for change in stock price after two (independent) trading days. [1Sti2 k(VjRX=U `9T[%fbz~_5&%d7s`Z:=]ZxBcvHvH-;YkD'}F1xNY?6\\- \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) . \nonumber \]. Ruodu Wang (wang@uwaterloo.ca) Sum of two uniform random variables 18/25. \end{aligned}$$, $$\begin{aligned} {\widehat{F}}_Z(z)&=\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) \frac{\left( {\widehat{F}}_Y\left( \frac{z (m-i-1)}{m}\right) +{\widehat{F}}_Y\left( \frac{z (m-i)}{m}\right) \right) }{2} \right] \\&=\frac{1}{2}\sum _{i=0}^{m-1}\left[ \left( \frac{\#X_v's\le \frac{(i+1) z}{m}}{n_1}-\frac{\#X_v's\le \frac{iz}{m}}{n_1}\right) \left( \frac{\#Y_w's\le \frac{(m-i) z}{m}}{n_2}+\frac{\#Y_w's\le \frac{(m-i-1) z}{m}}{n_2}\right) \right] ,\\&\,\,\,\,\,\,\, \quad v=1,2\dots n_1,\,w=1,2\dots n_2\\ {}&=\frac{1}{2}\sum _{i=0}^{m-1}\left[ \left( \frac{\#X_v's \text { between } \frac{iz}{m} \text { and } \frac{(i+1) z}{m}}{n_1}\right) \right. /Matrix [1 0 0 1 0 0] Let \(X_1\) and \(X_2\) be independent random variables with common distribution. << Stat Probab Lett 34(1):4351, Modarres M, Kaminskiy M, Krivtsov V (1999) Reliability engineering and risk analysis. << 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. }q_1^jq_2^{k-2j}q_3^{n-k+j}, &{} \text{ if } k> n. \end{array}\right. } Since $X\sim\mathcal{U}(0,2)$, $$f_X(x) = \frac{1}{2}\mathbb{I}_{(0,2)}(x)$$so in your convolution formula 106 0 obj That singularity first appeared when we considered the exponential of (the negative of) a $\Gamma(2,1)$ distribution, corresponding to multiplying one $U(0,1)$ variate by another one. Why did DOS-based Windows require HIMEM.SYS to boot? The purpose of this one is to derive the same result in a way that may be a little more revealing of the underlying structure of $XY$. Unable to complete the action because of changes made to the page. endobj Book: Introductory Probability (Grinstead and Snell), { "7.01:_Sums_of_Discrete_Random_Variables" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.
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