What does it exactly mean if a random variable follows a distributionWhat is meant by a “random...
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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;
}
$begingroup$
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 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.
$endgroup$
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 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
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.
$endgroup$
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 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
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
asked 4 hours ago
Hello MellowHello Mellow
61
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.
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.
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
$begingroup$
Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
$endgroup$
– Tim♦
4 hours ago
1
1
$begingroup$
Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
$endgroup$
– Tim♦
4 hours ago
$begingroup$
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 |
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 4 hours ago
HStamperHStamper
1,114612
$endgroup$
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.
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 |
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, $Y=f(X) + epsilon; epsilon stackrel{i.i.d.}{sim} N(0,sigma^2)$, 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 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 $f(X)$.
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 $f(X)$.
answered 4 hours ago
HStamperHStamper
1,114612
$endgroup$
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 4 hours ago
HStamperHStamper
1,114612
answered 4 hours ago
HStamperHStamper
1,114612
answered 4 hours ago
HStamperHStamper
1,114612
answered 4 hours ago
HStamperHStamper
1,114612
1,114612
add a comment |
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.
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.
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.
answered 4 hours ago
JonasJonas
51211
$endgroup$
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.
answered 4 hours ago
JonasJonas
51211
answered 4 hours ago
JonasJonas
51211
answered 4 hours ago
JonasJonas
51211
answered 4 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♦
4 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♦
4 hours ago
$begingroup$
How is it not a random variable? It has a distribution, so it is a random variable.
$endgroup$
– Tim♦
4 hours ago
$begingroup$
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 |
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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1
$begingroup$
Does this help stats.stackexchange.com/a/54894/35989? Or maybe this stats.stackexchange.com/questions/194558/… ?
$endgroup$
– Tim♦
4 hours ago