What does it exactly mean if a random variable follows a distribution





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$begingroup$


Imagine there's a random variable such as $ε$. Then we say that $ε$ is i.i.d and follows a normal distribution with mean $0$ and variance $σ^2$.



What does this mean? Is this not a variable anymore? Is this a function now? I see this in most books and such but I'm still unclear what exactly it means or what it does and etc.



In terms of regression, I know this variable is basically the random errors, but what does it mean if this vector of random errors follows a normal distribution?






regression distributions normal-distribution random-variable







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  • 2




    $begingroup$
    Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
    $endgroup$
    – Tim♦
    Apr 8 at 21:35


















1












$begingroup$


Imagine there's a random variable such as $ε$. Then we say that $ε$ is i.i.d and follows a normal distribution with mean $0$ and variance $σ^2$.



What does this mean? Is this not a variable anymore? Is this a function now? I see this in most books and such but I'm still unclear what exactly it means or what it does and etc.



In terms of regression, I know this variable is basically the random errors, but what does it mean if this vector of random errors follows a normal distribution?






regression distributions normal-distribution random-variable







share|cite|improve this question







New contributor




Hello Mellow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 2




    $begingroup$
    Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
    $endgroup$
    – Tim♦
    Apr 8 at 21:35














1












1








1





$begingroup$


Imagine there's a random variable such as $ε$. Then we say that $ε$ is i.i.d and follows a normal distribution with mean $0$ and variance $σ^2$.



What does this mean? Is this not a variable anymore? Is this a function now? I see this in most books and such but I'm still unclear what exactly it means or what it does and etc.



In terms of regression, I know this variable is basically the random errors, but what does it mean if this vector of random errors follows a normal distribution?






regression distributions normal-distribution random-variable







share|cite|improve this question







New contributor




Hello Mellow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Imagine there's a random variable such as $ε$. Then we say that $ε$ is i.i.d and follows a normal distribution with mean $0$ and variance $σ^2$.



What does this mean? Is this not a variable anymore? Is this a function now? I see this in most books and such but I'm still unclear what exactly it means or what it does and etc.



In terms of regression, I know this variable is basically the random errors, but what does it mean if this vector of random errors follows a normal distribution?






regression distributions normal-distribution random-variable




regression distributions normal-distribution random-variable






share|cite|improve this question







New contributor




Hello Mellow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Hello Mellow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






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Check out our Code of Conduct.









asked Apr 8 at 20:56









Hello MellowHello Mellow

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Hello Mellow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.








  • 2




    $begingroup$
    Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
    $endgroup$
    – Tim♦
    Apr 8 at 21:35














  • 2




    $begingroup$
    Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
    $endgroup$
    – Tim♦
    Apr 8 at 21:35








2




2




$begingroup$
Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
$endgroup$
– Tim♦
Apr 8 at 21:35




$begingroup$
Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
$endgroup$
– Tim♦
Apr 8 at 21:35










2 Answers
2






active

oldest

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2












$begingroup$

I.I.D. means independent and identically distributed, so $epsilon$ is a vector of component random variables with the same distribution.



The meaning of "A follows an X distribution" is equivalent to saying that it "has a distribution," which is to say that it is a random quantity that can be determined only in probability.



In the example of regression that you refer to, $Y=f(X) + epsilon; epsilon stackrel{i.i.d.}{sim} N(0,sigma^2)$, so the response variable $Y$ is equal to some function of the independent $X$ on average, and errors are normally distributed with mean zero, i.e. the observed $Y$ is not exactly $f(X)$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    A random variable $varepsilon sim mathrm{N}(0,sigma^2)$ is not the kind of variable considered when thinking about function arguments or solving equations, but actually represents the outcome of a random experiment. (Mathematically rigorously, but not so important, one would say: it is a function mapping from a sample space into the space in which the random variable lives.)



    How can this be understood? A probability measure, like $mathrm{N}(0,sigma^2)$ assigns values to sets, so-called events. In this case, the probability of $varepsilon$ ending up in a set $A$ has probability
    $$
    mathrm{N}(0,sigma^2)(A) = int_A frac{1}{sqrt{2pisigma^2}}expleft(-frac{1}{2sigma^2} |x |^2 right) mathrm{d}x.
    $$
    That means, if you repeatedly saw i.i.d. (independent and identically distributed) $varepsilon$'s, they would (in the large data limit) on average end up in $A$, precisely $mathrm{N}(0,sigma^2)(A)cdot 100 %$ of the time.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      How is it not a random variable? It has a distribution, so it is a random variable.
      $endgroup$
      – Tim♦
      Apr 8 at 21:28










    • $begingroup$
      It is a random variable, but not a „variable“ how we typically understand it.
      $endgroup$
      – Jonas
      2 days ago










    • $begingroup$
      That is..? What do you mean by variable?
      $endgroup$
      – Tim♦
      2 days ago










    • $begingroup$
      Like an unknown value with respect to which we want to solve an equation, or the argument of a function taking values in a given set. A random variable is a measurable function from a probability space to a measurable space. To me, those are completely different concepts; even if they appear similar.
      $endgroup$
      – Jonas
      2 days ago






    • 1




      $begingroup$
      When you see "variable" mentioned in probability theory text, they usually mean "random variable". Also "variable" in statistics means something else then in algebra. So it would be best if you could edit and make it more precise what kind of "variable" the random variable is not.
      $endgroup$
      – Tim♦
      2 days ago














    Your Answer





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    2 Answers
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    active

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    2 Answers
    2






    active

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    active

    oldest

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    active

    oldest

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    2












    $begingroup$

    I.I.D. means independent and identically distributed, so $epsilon$ is a vector of component random variables with the same distribution.



    The meaning of "A follows an X distribution" is equivalent to saying that it "has a distribution," which is to say that it is a random quantity that can be determined only in probability.



    In the example of regression that you refer to, $Y=f(X) + epsilon; epsilon stackrel{i.i.d.}{sim} N(0,sigma^2)$, so the response variable $Y$ is equal to some function of the independent $X$ on average, and errors are normally distributed with mean zero, i.e. the observed $Y$ is not exactly $f(X)$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      I.I.D. means independent and identically distributed, so $epsilon$ is a vector of component random variables with the same distribution.



      The meaning of "A follows an X distribution" is equivalent to saying that it "has a distribution," which is to say that it is a random quantity that can be determined only in probability.



      In the example of regression that you refer to, $Y=f(X) + epsilon; epsilon stackrel{i.i.d.}{sim} N(0,sigma^2)$, so the response variable $Y$ is equal to some function of the independent $X$ on average, and errors are normally distributed with mean zero, i.e. the observed $Y$ is not exactly $f(X)$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        I.I.D. means independent and identically distributed, so $epsilon$ is a vector of component random variables with the same distribution.



        The meaning of "A follows an X distribution" is equivalent to saying that it "has a distribution," which is to say that it is a random quantity that can be determined only in probability.



        In the example of regression that you refer to, $Y=f(X) + epsilon; epsilon stackrel{i.i.d.}{sim} N(0,sigma^2)$, so the response variable $Y$ is equal to some function of the independent $X$ on average, and errors are normally distributed with mean zero, i.e. the observed $Y$ is not exactly $f(X)$.






        share|cite|improve this answer









        $endgroup$



        I.I.D. means independent and identically distributed, so $epsilon$ is a vector of component random variables with the same distribution.



        The meaning of "A follows an X distribution" is equivalent to saying that it "has a distribution," which is to say that it is a random quantity that can be determined only in probability.



        In the example of regression that you refer to, $Y=f(X) + epsilon; epsilon stackrel{i.i.d.}{sim} N(0,sigma^2)$, so the response variable $Y$ is equal to some function of the independent $X$ on average, and errors are normally distributed with mean zero, i.e. the observed $Y$ is not exactly $f(X)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 8 at 21:11









        HStamperHStamper

        1,124612




        1,124612

























            2












            $begingroup$

            A random variable $varepsilon sim mathrm{N}(0,sigma^2)$ is not the kind of variable considered when thinking about function arguments or solving equations, but actually represents the outcome of a random experiment. (Mathematically rigorously, but not so important, one would say: it is a function mapping from a sample space into the space in which the random variable lives.)



            How can this be understood? A probability measure, like $mathrm{N}(0,sigma^2)$ assigns values to sets, so-called events. In this case, the probability of $varepsilon$ ending up in a set $A$ has probability
            $$
            mathrm{N}(0,sigma^2)(A) = int_A frac{1}{sqrt{2pisigma^2}}expleft(-frac{1}{2sigma^2} |x |^2 right) mathrm{d}x.
            $$
            That means, if you repeatedly saw i.i.d. (independent and identically distributed) $varepsilon$'s, they would (in the large data limit) on average end up in $A$, precisely $mathrm{N}(0,sigma^2)(A)cdot 100 %$ of the time.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              How is it not a random variable? It has a distribution, so it is a random variable.
              $endgroup$
              – Tim♦
              Apr 8 at 21:28










            • $begingroup$
              It is a random variable, but not a „variable“ how we typically understand it.
              $endgroup$
              – Jonas
              2 days ago










            • $begingroup$
              That is..? What do you mean by variable?
              $endgroup$
              – Tim♦
              2 days ago










            • $begingroup$
              Like an unknown value with respect to which we want to solve an equation, or the argument of a function taking values in a given set. A random variable is a measurable function from a probability space to a measurable space. To me, those are completely different concepts; even if they appear similar.
              $endgroup$
              – Jonas
              2 days ago






            • 1




              $begingroup$
              When you see "variable" mentioned in probability theory text, they usually mean "random variable". Also "variable" in statistics means something else then in algebra. So it would be best if you could edit and make it more precise what kind of "variable" the random variable is not.
              $endgroup$
              – Tim♦
              2 days ago


















            2












            $begingroup$

            A random variable $varepsilon sim mathrm{N}(0,sigma^2)$ is not the kind of variable considered when thinking about function arguments or solving equations, but actually represents the outcome of a random experiment. (Mathematically rigorously, but not so important, one would say: it is a function mapping from a sample space into the space in which the random variable lives.)



            How can this be understood? A probability measure, like $mathrm{N}(0,sigma^2)$ assigns values to sets, so-called events. In this case, the probability of $varepsilon$ ending up in a set $A$ has probability
            $$
            mathrm{N}(0,sigma^2)(A) = int_A frac{1}{sqrt{2pisigma^2}}expleft(-frac{1}{2sigma^2} |x |^2 right) mathrm{d}x.
            $$
            That means, if you repeatedly saw i.i.d. (independent and identically distributed) $varepsilon$'s, they would (in the large data limit) on average end up in $A$, precisely $mathrm{N}(0,sigma^2)(A)cdot 100 %$ of the time.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              How is it not a random variable? It has a distribution, so it is a random variable.
              $endgroup$
              – Tim♦
              Apr 8 at 21:28










            • $begingroup$
              It is a random variable, but not a „variable“ how we typically understand it.
              $endgroup$
              – Jonas
              2 days ago










            • $begingroup$
              That is..? What do you mean by variable?
              $endgroup$
              – Tim♦
              2 days ago










            • $begingroup$
              Like an unknown value with respect to which we want to solve an equation, or the argument of a function taking values in a given set. A random variable is a measurable function from a probability space to a measurable space. To me, those are completely different concepts; even if they appear similar.
              $endgroup$
              – Jonas
              2 days ago






            • 1




              $begingroup$
              When you see "variable" mentioned in probability theory text, they usually mean "random variable". Also "variable" in statistics means something else then in algebra. So it would be best if you could edit and make it more precise what kind of "variable" the random variable is not.
              $endgroup$
              – Tim♦
              2 days ago
















            2












            2








            2





            $begingroup$

            A random variable $varepsilon sim mathrm{N}(0,sigma^2)$ is not the kind of variable considered when thinking about function arguments or solving equations, but actually represents the outcome of a random experiment. (Mathematically rigorously, but not so important, one would say: it is a function mapping from a sample space into the space in which the random variable lives.)



            How can this be understood? A probability measure, like $mathrm{N}(0,sigma^2)$ assigns values to sets, so-called events. In this case, the probability of $varepsilon$ ending up in a set $A$ has probability
            $$
            mathrm{N}(0,sigma^2)(A) = int_A frac{1}{sqrt{2pisigma^2}}expleft(-frac{1}{2sigma^2} |x |^2 right) mathrm{d}x.
            $$
            That means, if you repeatedly saw i.i.d. (independent and identically distributed) $varepsilon$'s, they would (in the large data limit) on average end up in $A$, precisely $mathrm{N}(0,sigma^2)(A)cdot 100 %$ of the time.






            share|cite|improve this answer











            $endgroup$



            A random variable $varepsilon sim mathrm{N}(0,sigma^2)$ is not the kind of variable considered when thinking about function arguments or solving equations, but actually represents the outcome of a random experiment. (Mathematically rigorously, but not so important, one would say: it is a function mapping from a sample space into the space in which the random variable lives.)



            How can this be understood? A probability measure, like $mathrm{N}(0,sigma^2)$ assigns values to sets, so-called events. In this case, the probability of $varepsilon$ ending up in a set $A$ has probability
            $$
            mathrm{N}(0,sigma^2)(A) = int_A frac{1}{sqrt{2pisigma^2}}expleft(-frac{1}{2sigma^2} |x |^2 right) mathrm{d}x.
            $$
            That means, if you repeatedly saw i.i.d. (independent and identically distributed) $varepsilon$'s, they would (in the large data limit) on average end up in $A$, precisely $mathrm{N}(0,sigma^2)(A)cdot 100 %$ of the time.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 days ago

























            answered Apr 8 at 21:15









            JonasJonas

            51211




            51211












            • $begingroup$
              How is it not a random variable? It has a distribution, so it is a random variable.
              $endgroup$
              – Tim♦
              Apr 8 at 21:28










            • $begingroup$
              It is a random variable, but not a „variable“ how we typically understand it.
              $endgroup$
              – Jonas
              2 days ago










            • $begingroup$
              That is..? What do you mean by variable?
              $endgroup$
              – Tim♦
              2 days ago










            • $begingroup$
              Like an unknown value with respect to which we want to solve an equation, or the argument of a function taking values in a given set. A random variable is a measurable function from a probability space to a measurable space. To me, those are completely different concepts; even if they appear similar.
              $endgroup$
              – Jonas
              2 days ago






            • 1




              $begingroup$
              When you see "variable" mentioned in probability theory text, they usually mean "random variable". Also "variable" in statistics means something else then in algebra. So it would be best if you could edit and make it more precise what kind of "variable" the random variable is not.
              $endgroup$
              – Tim♦
              2 days ago




















            • $begingroup$
              How is it not a random variable? It has a distribution, so it is a random variable.
              $endgroup$
              – Tim♦
              Apr 8 at 21:28










            • $begingroup$
              It is a random variable, but not a „variable“ how we typically understand it.
              $endgroup$
              – Jonas
              2 days ago










            • $begingroup$
              That is..? What do you mean by variable?
              $endgroup$
              – Tim♦
              2 days ago










            • $begingroup$
              Like an unknown value with respect to which we want to solve an equation, or the argument of a function taking values in a given set. A random variable is a measurable function from a probability space to a measurable space. To me, those are completely different concepts; even if they appear similar.
              $endgroup$
              – Jonas
              2 days ago






            • 1




              $begingroup$
              When you see "variable" mentioned in probability theory text, they usually mean "random variable". Also "variable" in statistics means something else then in algebra. So it would be best if you could edit and make it more precise what kind of "variable" the random variable is not.
              $endgroup$
              – Tim♦
              2 days ago


















            $begingroup$
            How is it not a random variable? It has a distribution, so it is a random variable.
            $endgroup$
            – Tim♦
            Apr 8 at 21:28




            $begingroup$
            How is it not a random variable? It has a distribution, so it is a random variable.
            $endgroup$
            – Tim♦
            Apr 8 at 21:28












            $begingroup$
            It is a random variable, but not a „variable“ how we typically understand it.
            $endgroup$
            – Jonas
            2 days ago




            $begingroup$
            It is a random variable, but not a „variable“ how we typically understand it.
            $endgroup$
            – Jonas
            2 days ago












            $begingroup$
            That is..? What do you mean by variable?
            $endgroup$
            – Tim♦
            2 days ago




            $begingroup$
            That is..? What do you mean by variable?
            $endgroup$
            – Tim♦
            2 days ago












            $begingroup$
            Like an unknown value with respect to which we want to solve an equation, or the argument of a function taking values in a given set. A random variable is a measurable function from a probability space to a measurable space. To me, those are completely different concepts; even if they appear similar.
            $endgroup$
            – Jonas
            2 days ago




            $begingroup$
            Like an unknown value with respect to which we want to solve an equation, or the argument of a function taking values in a given set. A random variable is a measurable function from a probability space to a measurable space. To me, those are completely different concepts; even if they appear similar.
            $endgroup$
            – Jonas
            2 days ago




            1




            1




            $begingroup$
            When you see "variable" mentioned in probability theory text, they usually mean "random variable". Also "variable" in statistics means something else then in algebra. So it would be best if you could edit and make it more precise what kind of "variable" the random variable is not.
            $endgroup$
            – Tim♦
            2 days ago






            $begingroup$
            When you see "variable" mentioned in probability theory text, they usually mean "random variable". Also "variable" in statistics means something else then in algebra. So it would be best if you could edit and make it more precise what kind of "variable" the random variable is not.
            $endgroup$
            – Tim♦
            2 days ago












            Hello Mellow is a new contributor. Be nice, and check out our Code of Conduct.










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            迪纳利峰 位于迪纳利国家公园和保留区的迪纳利峰（旧名麦金利峰） 最高点 海拔 20,237英尺（6,168米）   NGVD 29 . [1] [2] 相對高度 20,073英尺（6,118米） 列表 各洲最高峰列表 特独立山峰 各国最高点列表 美国最高点  排名第一 [3] 坐标 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley ) 坐标： 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley )   地理 位置   美國阿拉斯加州迪纳利自治市镇 迪纳利国家公园和保留区 山脈 阿拉斯加山脉 攀山 首次登頂 （ 英语 ： First ascent ） 1913年6月7日 哈德逊·史塔克 哈里·卡斯坦斯 沃尔特·哈珀 罗伯特·塔特姆 最簡路線 西坡路线（攀登冰川/雪） 迪纳利 （北阿萨巴斯卡语支、 英语： Denali ， / d ᵻ ˈ n ɑː l i / ，直译“高山”），1917年至2015年官方名称为 麦金利山 （另译作 麦金利峰 ， 英语： Mount McKinley ），位于阿拉斯加州东南部、阿拉斯加山脉中段，海拔6194米，是北美洲最高峰，也是美国的最高峰。它的俄语名称是 Большая Гора ， 罗马化：Bolshaya Gora ，意思是“大山”。 迪纳利山地区拥有变幻莫测的高山风、典型的北极植被以及野生动植物。这里大部分地区终年积雪，山间经常浓雾不断，雾气在皑皑白雪中缭绕弥漫时，几百米之外的景物便不可见。夏季，迪纳利山的青青山坡上鲜花盛开，紫色的杜鹃和精巧的铃状石南花随处可见。迪纳利山还是世界登山爱好者的汇集之地。每年5月到7月有数百人登山，只有一半不到的人能登顶成功。一次登山约花3个星期。 2015年8月31日，美國政府宣佈將山名由“麦金利山”改回原名“迪纳利山”。美國內政部發表聲
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            南乌拉尔铁路局

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            南乌拉尔铁路局 Южно-Уральская железная дорога 成立 1934年1月4日 国家   蘇聯（1934年—1991年）   俄羅斯（1991年—） 总部 车里雅宾斯克 隶属 俄罗斯铁路股份公司 路局代码 80 铁路里程 4,934.9公里 奖项 网址 http://yuzd.rzd.ru/ 查 论 编 南乌拉尔铁路局 （俄语： Южно-Уральская железная дорога ， 羅馬化： Yuzhno-Uralskaya zheleznaya doroga ，简称 ЮУЖД ）是俄罗斯铁路股份公司属下的铁路管理局之一，总部位于车里雅宾斯克，负责管辖俄罗斯联邦乌拉尔南部地区的铁路系统，范围包括奥伦堡州、车里雅宾斯克州、库尔干州的大部分铁路，以及萨马拉州、斯维尔德洛夫斯克州、巴什科尔托斯坦共和国的少部分铁路，并有部分铁路延伸至或穿过哈萨克斯坦境内。南乌拉尔铁路局的营业里程为4,934.9公里，与伏尔加铁路局、古比雪夫铁路局、斯维尔德洛夫斯克铁路局、西西伯利亚铁路局相邻，并与南方的哈萨克斯坦铁路连接。 外部链接 （俄文） Южно-Уральская железная дорога — филиал ОАО РЖД （俄文） 南乌拉尔铁路局管辖路线图（一） （俄文） 南乌拉尔铁路局管辖路线图（二） 查 论 编 俄罗斯铁路（集团）股份公司 十月铁路局 · 莫斯科铁路局 · 北高加索铁路局 · 高尔基铁路局 · 古比雪夫铁路局 · 北方铁路局 · 伏尔加铁路局 · 东南铁路局 · 斯维尔德洛夫斯克铁路局 · 南乌拉尔铁路局 · 西西伯利亚铁路局 · 克拉斯诺亚尔斯克铁路局 · 东西伯利亚铁路局 · 后贝加尔铁路局 · 远东铁路局 · 加里宁格勒铁路局 This page is only for reference, If you need detailed information, please check here
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