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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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.
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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






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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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    2 Answers
    2






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    active

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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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            數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音
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            格利澤436b

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            格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP
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            When can things happen in Etherscan, such as the picture below?

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            2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  | 
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