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
asked Apr 8 at 20:56
Hello MellowHello Mellow
62
New contributor
Hello Mellow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
asked Apr 8 at 20:56
Hello MellowHello Mellow
62
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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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
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
asked Apr 8 at 20:56
Hello MellowHello Mellow
62
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
asked Apr 8 at 20:56
Hello MellowHello Mellow
62
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 Apr 8 at 20:56
Hello MellowHello Mellow
62
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 Apr 8 at 20:56
Hello MellowHello Mellow
62
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 Apr 8 at 20:56
Hello MellowHello Mellow
62
asked Apr 8 at 20:56
Hello MellowHello Mellow
62
62
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.
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.
Hello Mellow is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
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Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
$endgroup$
– Tim♦
Apr 8 at 21:35
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $f(X)$.
answered Apr 8 at 21:11
HStamperHStamper
1,149612
$endgroup$
add a comment |
$begingroup$
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.
answered Apr 8 at 21:15
JonasJonas
51211
$endgroup$
$begingroup$
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
Apr 9 at 4:31
$begingroup$
That is..? What do you mean by variable?
$endgroup$
– Tim♦
Apr 9 at 4:51
$begingroup$
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
Apr 9 at 5:31
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♦
Apr 9 at 6:53
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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 $f(X)$.
answered Apr 8 at 21:11
HStamperHStamper
1,149612
$endgroup$
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 $f(X)$.
answered Apr 8 at 21:11
HStamperHStamper
1,149612
$endgroup$
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 $f(X)$.
answered Apr 8 at 21:11
HStamperHStamper
1,149612
$endgroup$
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 $f(X)$.
answered Apr 8 at 21:11
HStamperHStamper
1,149612
answered Apr 8 at 21:11
HStamperHStamper
1,149612
answered Apr 8 at 21:11
HStamperHStamper
1,149612
answered Apr 8 at 21:11
HStamperHStamper
1,149612
1,149612
add a comment |
add a comment |
$begingroup$
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.
answered Apr 8 at 21:15
JonasJonas
51211
$endgroup$
$begingroup$
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
Apr 9 at 4:31
$begingroup$
That is..? What do you mean by variable?
$endgroup$
– Tim♦
Apr 9 at 4:51
$begingroup$
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
Apr 9 at 5:31
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♦
Apr 9 at 6:53
add a comment |
$begingroup$
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.
answered Apr 8 at 21:15
JonasJonas
51211
$endgroup$
$begingroup$
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
Apr 9 at 4:31
$begingroup$
That is..? What do you mean by variable?
$endgroup$
– Tim♦
Apr 9 at 4:51
$begingroup$
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
Apr 9 at 5:31
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♦
Apr 9 at 6:53
add a comment |
$begingroup$
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.
answered Apr 8 at 21:15
JonasJonas
51211
$endgroup$
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.
answered Apr 8 at 21:15
JonasJonas
51211
edited Apr 9 at 8:29
answered Apr 8 at 21:15
JonasJonas
51211
answered Apr 8 at 21:15
JonasJonas
51211
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
Apr 9 at 4:31
$begingroup$
That is..? What do you mean by variable?
$endgroup$
– Tim♦
Apr 9 at 4:51
$begingroup$
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
Apr 9 at 5:31
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♦
Apr 9 at 6:53
add a comment |
$begingroup$
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
Apr 9 at 4:31
$begingroup$
That is..? What do you mean by variable?
$endgroup$
– Tim♦
Apr 9 at 4:51
$begingroup$
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
Apr 9 at 5:31
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♦
Apr 9 at 6:53
$begingroup$
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
Apr 9 at 4:31
$begingroup$
It is a random variable, but not a „variable“ how we typically understand it.
$endgroup$
– Jonas
Apr 9 at 4:31
$begingroup$
That is..? What do you mean by variable?
$endgroup$
– Tim♦
Apr 9 at 4:51
$begingroup$
That is..? What do you mean by variable?
$endgroup$
– Tim♦
Apr 9 at 4:51
$begingroup$
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
Apr 9 at 5:31
$begingroup$
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
Apr 9 at 5:31
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♦
Apr 9 at 6:53
$begingroup$
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♦
Apr 9 at 6:53
add a comment |
Hello Mellow is a new contributor. Be nice, and check out our Code of Conduct.
Hello Mellow is a new contributor. Be nice, and check out our Code of Conduct.
Hello Mellow is a new contributor. Be nice, and check out our Code of Conduct.
Hello Mellow is a new contributor. Be nice, and check out our Code of Conduct.
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Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
$endgroup$
– Tim♦
Apr 8 at 21:35