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Basic: Why are Slope, Intercept in Regression considered Random Variables?
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Sorry if this is too basic.
In an OLS regression given by
$y=ax+b$
$b$ intercept, $a$ the slope.
Then $a,b$ are not numbers but random variables.
I find this confusing since I start with data points $(x_1,y_1),(x_2,y_2),..(x_n,y_n) ; x_i neq x_j $ when $i neq j $. Then we find
the line of best fit, which would give us actual numbers $a,b$, so that these seem to be constants, not (Random) variables.
Is the reason these are called variables that I am selecting just one
of many possible values $y_j$ for a given variable $x_j$?
regression random-variable
edited 3 hours ago
Glen_b♦
202k22384708
asked 5 hours ago
gary
238210
 |Â
show 4 more comments
up vote
1
down vote
favorite
Sorry if this is too basic.
In an OLS regression given by
$y=ax+b$
$b$ intercept, $a$ the slope.
Then $a,b$ are not numbers but random variables.
I find this confusing since I start with data points $(x_1,y_1),(x_2,y_2),..(x_n,y_n) ; x_i neq x_j $ when $i neq j $. Then we find
the line of best fit, which would give us actual numbers $a,b$, so that these seem to be constants, not (Random) variables.
Is the reason these are called variables that I am selecting just one
of many possible values $y_j$ for a given variable $x_j$?
regression random-variable
edited 3 hours ago
Glen_b♦
202k22384708
asked 5 hours ago
gary
238210
1
Estimators of slope and intercept are random variables.because they're functions of the responses, which are random variables.
– Glen_b♦
4 hours ago
@Glen_b: So the point is that for each data set $(x_i,y_i)$ for the same variables X,Y ( of same size) I would get different values for the slope, the intercept?
– gary
4 hours ago
2
New samples would indeed lead to different estimates (because - even assuming fixed x's - you'd have different realizations of each of the n corresponding y's)
– Glen_b♦
3 hours ago
Thanks, Glen_b , should I delete the question or do you want to answer it. Or should I?
– gary
3 hours ago
I wasn't sure whether that's what you were seeking (which is why I commented, figuring you'd clarify the question if you needed something else), I am happy to post it as an answer (or you can if you prefer).
– Glen_b♦
3 hours ago
 |Â
show 4 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Sorry if this is too basic.
In an OLS regression given by
$y=ax+b$
$b$ intercept, $a$ the slope.
Then $a,b$ are not numbers but random variables.
I find this confusing since I start with data points $(x_1,y_1),(x_2,y_2),..(x_n,y_n) ; x_i neq x_j $ when $i neq j $. Then we find
the line of best fit, which would give us actual numbers $a,b$, so that these seem to be constants, not (Random) variables.
Is the reason these are called variables that I am selecting just one
of many possible values $y_j$ for a given variable $x_j$?
regression random-variable
edited 3 hours ago
Glen_b♦
202k22384708
asked 5 hours ago
gary
238210
Sorry if this is too basic.
In an OLS regression given by
$y=ax+b$
$b$ intercept, $a$ the slope.
Then $a,b$ are not numbers but random variables.
I find this confusing since I start with data points $(x_1,y_1),(x_2,y_2),..(x_n,y_n) ; x_i neq x_j $ when $i neq j $. Then we find
the line of best fit, which would give us actual numbers $a,b$, so that these seem to be constants, not (Random) variables.
Is the reason these are called variables that I am selecting just one
of many possible values $y_j$ for a given variable $x_j$?
regression random-variable
regression random-variable
edited 3 hours ago
Glen_b♦
202k22384708
asked 5 hours ago
gary
238210
edited 3 hours ago
Glen_b♦
202k22384708
asked 5 hours ago
gary
238210
edited 3 hours ago
Glen_b♦
202k22384708
edited 3 hours ago
Glen_b♦
202k22384708
edited 3 hours ago
Glen_b♦
202k22384708
202k22384708
asked 5 hours ago
gary
238210
asked 5 hours ago
gary
238210
asked 5 hours ago
gary
238210
238210
1
Estimators of slope and intercept are random variables.because they're functions of the responses, which are random variables.
– Glen_b♦
4 hours ago
@Glen_b: So the point is that for each data set $(x_i,y_i)$ for the same variables X,Y ( of same size) I would get different values for the slope, the intercept?
– gary
4 hours ago
2
New samples would indeed lead to different estimates (because - even assuming fixed x's - you'd have different realizations of each of the n corresponding y's)
– Glen_b♦
3 hours ago
Thanks, Glen_b , should I delete the question or do you want to answer it. Or should I?
– gary
3 hours ago
I wasn't sure whether that's what you were seeking (which is why I commented, figuring you'd clarify the question if you needed something else), I am happy to post it as an answer (or you can if you prefer).
– Glen_b♦
3 hours ago
 |Â
show 4 more comments
1
Estimators of slope and intercept are random variables.because they're functions of the responses, which are random variables.
– Glen_b♦
4 hours ago
@Glen_b: So the point is that for each data set $(x_i,y_i)$ for the same variables X,Y ( of same size) I would get different values for the slope, the intercept?
– gary
4 hours ago
2
New samples would indeed lead to different estimates (because - even assuming fixed x's - you'd have different realizations of each of the n corresponding y's)
– Glen_b♦
3 hours ago
Thanks, Glen_b , should I delete the question or do you want to answer it. Or should I?
– gary
3 hours ago
I wasn't sure whether that's what you were seeking (which is why I commented, figuring you'd clarify the question if you needed something else), I am happy to post it as an answer (or you can if you prefer).
– Glen_b♦
3 hours ago
1
1
Estimators of slope and intercept are random variables.because they're functions of the responses, which are random variables.
– Glen_b♦
4 hours ago
Estimators of slope and intercept are random variables.because they're functions of the responses, which are random variables.
– Glen_b♦
4 hours ago
@Glen_b: So the point is that for each data set $(x_i,y_i)$ for the same variables X,Y ( of same size) I would get different values for the slope, the intercept?
– gary
4 hours ago
@Glen_b: So the point is that for each data set $(x_i,y_i)$ for the same variables X,Y ( of same size) I would get different values for the slope, the intercept?
– gary
4 hours ago
2
2
New samples would indeed lead to different estimates (because - even assuming fixed x's - you'd have different realizations of each of the n corresponding y's)
– Glen_b♦
3 hours ago
New samples would indeed lead to different estimates (because - even assuming fixed x's - you'd have different realizations of each of the n corresponding y's)
– Glen_b♦
3 hours ago
Thanks, Glen_b , should I delete the question or do you want to answer it. Or should I?
– gary
3 hours ago
Thanks, Glen_b , should I delete the question or do you want to answer it. Or should I?
– gary
3 hours ago
I wasn't sure whether that's what you were seeking (which is why I commented, figuring you'd clarify the question if you needed something else), I am happy to post it as an answer (or you can if you prefer).
– Glen_b♦
3 hours ago
I wasn't sure whether that's what you were seeking (which is why I commented, figuring you'd clarify the question if you needed something else), I am happy to post it as an answer (or you can if you prefer).
– Glen_b♦
3 hours ago
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Estimators of slope and intercept are random variables because they're functions of the responses, which are random variables.
New samples would lead to different estimates (because - even assuming fixed $x$'s - you'd have different realizations of each of the n corresponding sets of $Y$'s)
If we set the situation up to make the variables and their realizations a little more distinct, the situation may become clearer; taking the $x$'s as fixed (for simplicity of exposition), you have $$Y_i = ax_i + b + epsilon_i$$ where $epsilon_i$ is the error term. You draw a sample with that set of $x$'s and you observe a corresponding set of $y$'s, corresponding to a particular realization of the $epsilon_i$. Let's call that set of observed $y$ values $mathbfy^(1)$. We repeat our sampling procedure at the same set of x-values and obtain a new set of responses, $mathbfy^(2)$ and we keep going, up to $mathbfy^(k)$ say. Each realization will have its own slope and intercept, so realization $j$ has $a^(j)$ and intercept $b^(j)$, which will be a function of $mathbfy^(j)$.
answered 3 hours ago
Glen_b♦
202k22384708
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Estimators of slope and intercept are random variables because they're functions of the responses, which are random variables.
New samples would lead to different estimates (because - even assuming fixed $x$'s - you'd have different realizations of each of the n corresponding sets of $Y$'s)
If we set the situation up to make the variables and their realizations a little more distinct, the situation may become clearer; taking the $x$'s as fixed (for simplicity of exposition), you have $$Y_i = ax_i + b + epsilon_i$$ where $epsilon_i$ is the error term. You draw a sample with that set of $x$'s and you observe a corresponding set of $y$'s, corresponding to a particular realization of the $epsilon_i$. Let's call that set of observed $y$ values $mathbfy^(1)$. We repeat our sampling procedure at the same set of x-values and obtain a new set of responses, $mathbfy^(2)$ and we keep going, up to $mathbfy^(k)$ say. Each realization will have its own slope and intercept, so realization $j$ has $a^(j)$ and intercept $b^(j)$, which will be a function of $mathbfy^(j)$.
answered 3 hours ago
Glen_b♦
202k22384708
add a comment |Â
up vote
3
down vote
accepted
Estimators of slope and intercept are random variables because they're functions of the responses, which are random variables.
New samples would lead to different estimates (because - even assuming fixed $x$'s - you'd have different realizations of each of the n corresponding sets of $Y$'s)
If we set the situation up to make the variables and their realizations a little more distinct, the situation may become clearer; taking the $x$'s as fixed (for simplicity of exposition), you have $$Y_i = ax_i + b + epsilon_i$$ where $epsilon_i$ is the error term. You draw a sample with that set of $x$'s and you observe a corresponding set of $y$'s, corresponding to a particular realization of the $epsilon_i$. Let's call that set of observed $y$ values $mathbfy^(1)$. We repeat our sampling procedure at the same set of x-values and obtain a new set of responses, $mathbfy^(2)$ and we keep going, up to $mathbfy^(k)$ say. Each realization will have its own slope and intercept, so realization $j$ has $a^(j)$ and intercept $b^(j)$, which will be a function of $mathbfy^(j)$.
answered 3 hours ago
Glen_b♦
202k22384708
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Estimators of slope and intercept are random variables because they're functions of the responses, which are random variables.
New samples would lead to different estimates (because - even assuming fixed $x$'s - you'd have different realizations of each of the n corresponding sets of $Y$'s)
If we set the situation up to make the variables and their realizations a little more distinct, the situation may become clearer; taking the $x$'s as fixed (for simplicity of exposition), you have $$Y_i = ax_i + b + epsilon_i$$ where $epsilon_i$ is the error term. You draw a sample with that set of $x$'s and you observe a corresponding set of $y$'s, corresponding to a particular realization of the $epsilon_i$. Let's call that set of observed $y$ values $mathbfy^(1)$. We repeat our sampling procedure at the same set of x-values and obtain a new set of responses, $mathbfy^(2)$ and we keep going, up to $mathbfy^(k)$ say. Each realization will have its own slope and intercept, so realization $j$ has $a^(j)$ and intercept $b^(j)$, which will be a function of $mathbfy^(j)$.
answered 3 hours ago
Glen_b♦
202k22384708
Estimators of slope and intercept are random variables because they're functions of the responses, which are random variables.
New samples would lead to different estimates (because - even assuming fixed $x$'s - you'd have different realizations of each of the n corresponding sets of $Y$'s)
If we set the situation up to make the variables and their realizations a little more distinct, the situation may become clearer; taking the $x$'s as fixed (for simplicity of exposition), you have $$Y_i = ax_i + b + epsilon_i$$ where $epsilon_i$ is the error term. You draw a sample with that set of $x$'s and you observe a corresponding set of $y$'s, corresponding to a particular realization of the $epsilon_i$. Let's call that set of observed $y$ values $mathbfy^(1)$. We repeat our sampling procedure at the same set of x-values and obtain a new set of responses, $mathbfy^(2)$ and we keep going, up to $mathbfy^(k)$ say. Each realization will have its own slope and intercept, so realization $j$ has $a^(j)$ and intercept $b^(j)$, which will be a function of $mathbfy^(j)$.
answered 3 hours ago
Glen_b♦
202k22384708
edited 3 hours ago
answered 3 hours ago
Glen_b♦
202k22384708
answered 3 hours ago
Glen_b♦
202k22384708
answered 3 hours ago
Glen_b♦
202k22384708
202k22384708
add a comment |Â
add a comment |Â
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1
Estimators of slope and intercept are random variables.because they're functions of the responses, which are random variables.
– Glen_b♦
4 hours ago
@Glen_b: So the point is that for each data set $(x_i,y_i)$ for the same variables X,Y ( of same size) I would get different values for the slope, the intercept?
– gary
4 hours ago
2
New samples would indeed lead to different estimates (because - even assuming fixed x's - you'd have different realizations of each of the n corresponding y's)
– Glen_b♦
3 hours ago
Thanks, Glen_b , should I delete the question or do you want to answer it. Or should I?
– gary
3 hours ago
I wasn't sure whether that's what you were seeking (which is why I commented, figuring you'd clarify the question if you needed something else), I am happy to post it as an answer (or you can if you prefer).
– Glen_b♦
3 hours ago