What does it exactly mean if a random variable follows a distribution The 2019 Stack Overflow Developer Survey Results Are InWhat 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?
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What does it exactly mean if a random variable follows a distribution
The 2019 Stack Overflow Developer Survey Results Are InWhat 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;
$begingroup$
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
New contributor
$endgroup$
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
New contributor
$endgroup$
2
$begingroup$
Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
$endgroup$
– Tim♦
yesterday
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
New contributor
$endgroup$
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
New contributor
New contributor
New contributor
asked yesterday
Hello MellowHello Mellow
62
62
New contributor
New contributor
2
$begingroup$
Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
$endgroup$
– Tim♦
yesterday
add a comment |
2
$begingroup$
Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
$endgroup$
– Tim♦
yesterday
2
2
$begingroup$
Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
$endgroup$
– Tim♦
yesterday
$begingroup$
Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
$endgroup$
– Tim♦
yesterday
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, $Y=f(X) + epsilon; epsilon stackreli.i.d.sim N(0,sigma^2)$, 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 $f(X)$.
$endgroup$
add a comment |
$begingroup$
A random variable $varepsilon sim mathrmN(0,sigma^2)$ 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 $mathrmN(0,sigma^2)$ 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.
$endgroup$
$begingroup$
How is it not a random variable? It has a distribution, so it is a random variable.
$endgroup$
– Tim♦
yesterday
$begingroup$
It is a random variable, but not a „variable“ how we typically understand it.
$endgroup$
– Jonas
20 hours ago
$begingroup$
That is..? What do you mean by variable?
$endgroup$
– Tim♦
19 hours 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
19 hours 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♦
17 hours ago
add a comment |
Your Answer
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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, $Y=f(X) + epsilon; epsilon stackreli.i.d.sim N(0,sigma^2)$, 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 $f(X)$.
$endgroup$
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, $Y=f(X) + epsilon; epsilon stackreli.i.d.sim N(0,sigma^2)$, 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 $f(X)$.
$endgroup$
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, $Y=f(X) + epsilon; epsilon stackreli.i.d.sim N(0,sigma^2)$, 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 $f(X)$.
$endgroup$
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 $f(X)$.
answered yesterday
HStamperHStamper
1,114612
1,114612
add a comment |
add a comment |
$begingroup$
A random variable $varepsilon sim mathrmN(0,sigma^2)$ 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 $mathrmN(0,sigma^2)$ 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.
$endgroup$
$begingroup$
How is it not a random variable? It has a distribution, so it is a random variable.
$endgroup$
– Tim♦
yesterday
$begingroup$
It is a random variable, but not a „variable“ how we typically understand it.
$endgroup$
– Jonas
20 hours ago
$begingroup$
That is..? What do you mean by variable?
$endgroup$
– Tim♦
19 hours 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
19 hours 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♦
17 hours ago
add a comment |
$begingroup$
A random variable $varepsilon sim mathrmN(0,sigma^2)$ 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 $mathrmN(0,sigma^2)$ 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.
$endgroup$
$begingroup$
How is it not a random variable? It has a distribution, so it is a random variable.
$endgroup$
– Tim♦
yesterday
$begingroup$
It is a random variable, but not a „variable“ how we typically understand it.
$endgroup$
– Jonas
20 hours ago
$begingroup$
That is..? What do you mean by variable?
$endgroup$
– Tim♦
19 hours 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
19 hours 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♦
17 hours ago
add a comment |
$begingroup$
A random variable $varepsilon sim mathrmN(0,sigma^2)$ 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 $mathrmN(0,sigma^2)$ 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.
$endgroup$
A random variable $varepsilon sim mathrmN(0,sigma^2)$ 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 $mathrmN(0,sigma^2)$ 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.
edited 16 hours ago
answered yesterday
JonasJonas
51211
51211
$begingroup$
How is it not a random variable? It has a distribution, so it is a random variable.
$endgroup$
– Tim♦
yesterday
$begingroup$
It is a random variable, but not a „variable“ how we typically understand it.
$endgroup$
– Jonas
20 hours ago
$begingroup$
That is..? What do you mean by variable?
$endgroup$
– Tim♦
19 hours 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
19 hours 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♦
17 hours ago
add a comment |
$begingroup$
How is it not a random variable? It has a distribution, so it is a random variable.
$endgroup$
– Tim♦
yesterday
$begingroup$
It is a random variable, but not a „variable“ how we typically understand it.
$endgroup$
– Jonas
20 hours ago
$begingroup$
That is..? What do you mean by variable?
$endgroup$
– Tim♦
19 hours 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
19 hours 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♦
17 hours ago
$begingroup$
How is it not a random variable? It has a distribution, so it is a random variable.
$endgroup$
– Tim♦
yesterday
$begingroup$
How is it not a random variable? It has a distribution, so it is a random variable.
$endgroup$
– Tim♦
yesterday
$begingroup$
It is a random variable, but not a „variable“ how we typically understand it.
$endgroup$
– Jonas
20 hours ago
$begingroup$
It is a random variable, but not a „variable“ how we typically understand it.
$endgroup$
– Jonas
20 hours ago
$begingroup$
That is..? What do you mean by variable?
$endgroup$
– Tim♦
19 hours ago
$begingroup$
That is..? What do you mean by variable?
$endgroup$
– Tim♦
19 hours 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
19 hours 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
19 hours 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♦
17 hours 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♦
17 hours ago
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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$begingroup$
Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
$endgroup$
– Tim♦
yesterday