Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Techniques for analyzing the relationship between one (or more) "dependent" variables and "independent" variables.
0
votes
0
answers
66
views
Herfindahl Hirschman Indices
Now, I wish to run a
regression, say of ${HH}_{t}$ (or the HH index on time $T)$ on a particular
set of world characteristcs, call it $X_{Wt}$. … The regression is:
$$
{HH}_{t}=\alpha+\beta X_{Wt}+\epsilon_{t}
$$
Now, in the above, the problem is that every year, the total number
of countries is changing. …
2
votes
5
answers
174
views
Constant Variance restrictive as a general rule?
Think of a linear regression model:
$$
Y=X\beta+\epsilon
$$
where $\epsilon|X\sim N\left(0,\sigma^{2}\right)$. The vector of
parameters $\beta$ can be consistenly estimated by OLS. …
3
votes
1
answer
2k
views
Linear Regression and Almost Sure Convergence
Consider a linear regression model, wherein:
$$
y_{i}=x_{i}\beta+\epsilon_{i}
$$
where notation is standard and $x$ is a scalar. …
2
votes
0
answers
113
views
Test linearity versus local linearity in regression
Consider the linear regression model
$$y_{i}=x_{i}'\beta+\epsilon_{it}$$
where we have assumed that the model is linear in its parameters. …
0
votes
0
answers
62
views
Notational issues for point estimates
When we estimate a linear regression model, say we obtain an estimate for \beta
as $\hat{\beta}=(x'x)^{-1}(x'y)$
, the standard least standard least squares solution. …
6
votes
1
answer
2k
views
Including both individual and state fixed effects
Consider we have the following regression model:
$$y_{it}=x_{it}'\beta+\alpha_{i}+\upsilon_{it}$$
where we have data on $N$
individuals for $T$
time periods. …
12
votes
2
answers
11k
views
Conditional Mean in Linear Regression
I have a question about linear regression in general. …
1
vote
Difference between the effect and the contribution of a regressor
I think I figured the answer out. Ultimately, I am trying to compare the partial $R^{2}$
to the coefficient. They are in some ways related. Consider we have a model:$$y=x\beta+\epsilon$$
where $x$
…
2
votes
1
answer
177
views
Difference between the effect and the contribution of a regressor
Consider that we have the following time series of observations: $C_{t},I_{t},G_{t},X_{t}-M_{t}$
Now, $Y_{t}$
is defined as:$$Y_{t}=C_{t}+I_{t}+G_{t}+X_{t}-M_{t}$$
If we were to run a regression of …
8
votes
1
answer
820
views
Non Linear Endogeneity
Consider the following Linear Regression Model. …
6
votes
2
answers
7k
views
Conditional variance in OLS regression
Consider the linear regression model: $$y_{it}=x_{it}\beta+\epsilon_{it}$$
where $x$
is single regressor. …
0
votes
1
answer
604
views
Relationship between Conditional Mean and Dummy Variables in the presence of Additional Regr...
Consider the following regression model$$y_{it}=\beta_{1}M_{i}+\beta_{2}F_{i}+x_{it}'\gamma+\epsilon_{it}$$
where the LHS is some individual specific, time varying regressand the RHS variables consist … It is clear that if $x_{it}$
was not present, an OLS regression of the LHS on the RHS would result in the estiamtes representing conditional means by groups. …
1
vote
Using estimated parameters as dependent variables
As the RHS variables are generated by OLS, they are actually "generated regressors". It is perfectly fine to estimate the model by OLS, but you have to adjust standard errors. The classic paper by Pag …
1
vote
1
answer
1k
views
Law of Iterated Expectations in Linear Regression
Consider the following model:$$y=x\beta+u$$
Now, let us take conditional expectations, assuming conditional exogeneity, such that $$E[u|x]=0$$
$$E[y|x]=\beta E[x|x]=\beta x$$
By the Law of Iterated …
0
votes
0
answers
49
views
Linear Regression
Consider the regression model:$$y=x\beta+u$$
Let us take expectations:$$E[y]=E[x\beta+u]=E[x\beta]$$
by linearity of the expectations operator and $E[u]=0$
. …