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





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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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    $begingroup$
    Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
    $endgroup$
    – Tim♦
    5 hours ago


















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.







$endgroup$








  • 1




    $begingroup$
    Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
    $endgroup$
    – Tim♦
    5 hours ago














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 6 hours ago









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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  • 1




    $begingroup$
    Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
    $endgroup$
    – Tim♦
    5 hours ago














  • 1




    $begingroup$
    Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
    $endgroup$
    – Tim♦
    5 hours ago








1




1




$begingroup$
Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
$endgroup$
– Tim♦
5 hours ago




$begingroup$
Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
$endgroup$
– Tim♦
5 hours ago










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 really a variable, 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♦
      5 hours ago












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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

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    active

    oldest

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    active

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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 6 hours ago









        HStamperHStamper

        1,114612




        1,114612

























            2












            $begingroup$

            A random variable $varepsilon sim mathrm{N}(0,sigma^2)$ is not really a variable, 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♦
              5 hours ago
















            2












            $begingroup$

            A random variable $varepsilon sim mathrm{N}(0,sigma^2)$ is not really a variable, 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♦
              5 hours ago














            2












            2








            2





            $begingroup$

            A random variable $varepsilon sim mathrm{N}(0,sigma^2)$ is not really a variable, 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 really a variable, 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










            answered 5 hours ago









            JonasJonas

            51211




            51211












            • $begingroup$
              How is it not a random variable? It has a distribution, so it is a random variable.
              $endgroup$
              – Tim♦
              5 hours ago


















            • $begingroup$
              How is it not a random variable? It has a distribution, so it is a random variable.
              $endgroup$
              – Tim♦
              5 hours ago
















            $begingroup$
            How is it not a random variable? It has a distribution, so it is a random variable.
            $endgroup$
            – Tim♦
            5 hours ago




            $begingroup$
            How is it not a random variable? It has a distribution, so it is a random variable.
            $endgroup$
            – Tim♦
            5 hours ago










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