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


What is meant by a “random variable”?What is meant by using a probability distribution to model the output data for a regression problem?What does truncated distribution mean?What does “chi” mean and come from in “chi-squared distribution”?What exactly is a distribution?If $X$ and $Y$ are normally distributed random variables, what kind of distribution their sum follows?“Let random variables $X_1,dots, X_n$ be a iid random sample from $f(x)$” - what does it mean?What does it mean to have a probability as random variable?What does it mean by error has a Gaussian Distribution?what exactly does it mean when we say “Let $X_1, X_2 …$ be iid random variables”Mean and S.D of Normal distributionWhat does it mean to generate a random variable from a distribution when random variable is a function?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








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.







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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♦
    6 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♦
    6 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






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.









asked 7 hours ago









Hello MellowHello Mellow

62




62




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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.






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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♦
    6 hours ago












  • 1




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







1




1




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




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










2 Answers
2






active

oldest

votes


















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, $Y=f(X) + epsilon; epsilon stackreli.i.d.sim N(0,sigma^2)$, 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 $varepsilon sim mathrmN(0,sigma^2)$ 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 $mathrmN(0,sigma^2)$ assigns values to sets, so-called events. In this case, the probability of $varepsilon$ ending up in a set $A$ has probability
    $$
    mathrmN(0,sigma^2)(A) = int_A frac1sqrt2pisigma^2expleft(-frac12sigma^2 |x |^2 right) mathrmdx.
    $$
    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 $mathrmN(0,sigma^2)(A)cdot 100 %$ 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♦
      7 hours ago











    Your Answer





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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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, $Y=f(X) + epsilon; epsilon stackreli.i.d.sim N(0,sigma^2)$, 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, $Y=f(X) + epsilon; epsilon stackreli.i.d.sim N(0,sigma^2)$, 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, $Y=f(X) + epsilon; epsilon stackreli.i.d.sim N(0,sigma^2)$, 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, $Y=f(X) + epsilon; epsilon stackreli.i.d.sim N(0,sigma^2)$, 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 7 hours ago









        HStamperHStamper

        1,114612




        1,114612























            2












            $begingroup$

            A random variable $varepsilon sim mathrmN(0,sigma^2)$ 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 $mathrmN(0,sigma^2)$ assigns values to sets, so-called events. In this case, the probability of $varepsilon$ ending up in a set $A$ has probability
            $$
            mathrmN(0,sigma^2)(A) = int_A frac1sqrt2pisigma^2expleft(-frac12sigma^2 |x |^2 right) mathrmdx.
            $$
            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 $mathrmN(0,sigma^2)(A)cdot 100 %$ 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♦
              7 hours ago















            2












            $begingroup$

            A random variable $varepsilon sim mathrmN(0,sigma^2)$ 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 $mathrmN(0,sigma^2)$ assigns values to sets, so-called events. In this case, the probability of $varepsilon$ ending up in a set $A$ has probability
            $$
            mathrmN(0,sigma^2)(A) = int_A frac1sqrt2pisigma^2expleft(-frac12sigma^2 |x |^2 right) mathrmdx.
            $$
            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 $mathrmN(0,sigma^2)(A)cdot 100 %$ 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♦
              7 hours ago













            2












            2








            2





            $begingroup$

            A random variable $varepsilon sim mathrmN(0,sigma^2)$ 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 $mathrmN(0,sigma^2)$ assigns values to sets, so-called events. In this case, the probability of $varepsilon$ ending up in a set $A$ has probability
            $$
            mathrmN(0,sigma^2)(A) = int_A frac1sqrt2pisigma^2expleft(-frac12sigma^2 |x |^2 right) mathrmdx.
            $$
            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 $mathrmN(0,sigma^2)(A)cdot 100 %$ of the time.






            share|cite|improve this answer









            $endgroup$



            A random variable $varepsilon sim mathrmN(0,sigma^2)$ 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 $mathrmN(0,sigma^2)$ assigns values to sets, so-called events. In this case, the probability of $varepsilon$ ending up in a set $A$ has probability
            $$
            mathrmN(0,sigma^2)(A) = int_A frac1sqrt2pisigma^2expleft(-frac12sigma^2 |x |^2 right) mathrmdx.
            $$
            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 $mathrmN(0,sigma^2)(A)cdot 100 %$ of the time.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 7 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♦
              7 hours ago
















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















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




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










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









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