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

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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

asked 4 hours ago

Hello MellowHello Mellow

61

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♦
  4 hours ago
  

  
  
  
  

add a comment | 

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

asked 4 hours ago

Hello MellowHello Mellow

61

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♦
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

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

asked 4 hours ago

Hello MellowHello Mellow

61

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.

$endgroup$

share|cite|improve this question

asked 4 hours ago

Hello MellowHello Mellow

61

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

asked 4 hours ago

Hello MellowHello Mellow

61

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 4 hours ago

Hello MellowHello Mellow

61

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 4 hours ago

Hello MellowHello Mellow

61

2 Answers
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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

answered 4 hours ago

HStamperHStamper

1,114612

$endgroup$

add a comment | 

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

answered 4 hours ago

JonasJonas

51211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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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

answered 4 hours ago

HStamperHStamper

1,114612

$endgroup$

add a comment | 

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

answered 4 hours ago

HStamperHStamper

1,114612

$endgroup$

add a comment | 

$begingroup$

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

answered 4 hours ago

HStamperHStamper

1,114612

$endgroup$

share|cite|improve this answer

answered 4 hours ago

HStamperHStamper

1,114612

answered 4 hours ago

HStamperHStamper

1,114612

answered 4 hours ago

HStamperHStamper

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

answered 4 hours ago

JonasJonas

51211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

answered 4 hours ago

JonasJonas

51211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

answered 4 hours ago

JonasJonas

51211

$endgroup$

share|cite|improve this answer

answered 4 hours ago

JonasJonas

51211

answered 4 hours ago

JonasJonas

51211

answered 4 hours ago

JonasJonas

51211