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OLS effect of X squared ond dependent variable
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I have my OLS regression: $y = beta_1 + beta_2 * X_2 + beta_3 * X_3 +beta_4 * (X_3)^2$
Could anybody please explain to me the effect of a change in $X_3$ on the dependent variable?(Is the effect positive or negative). An example would be greatly appreciated.
regression mathematical-statistics multiple-regression least-squares marginal-effect
edited 19 mins ago
Ferdi
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P.Cay
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up vote
1
down vote
favorite
I have my OLS regression: $y = beta_1 + beta_2 * X_2 + beta_3 * X_3 +beta_4 * (X_3)^2$
Could anybody please explain to me the effect of a change in $X_3$ on the dependent variable?(Is the effect positive or negative). An example would be greatly appreciated.
regression mathematical-statistics multiple-regression least-squares marginal-effect
edited 19 mins ago
Ferdi
3,69042152
asked 1 hour ago
P.Cay
83
New contributor
P.Cay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have my OLS regression: $y = beta_1 + beta_2 * X_2 + beta_3 * X_3 +beta_4 * (X_3)^2$
Could anybody please explain to me the effect of a change in $X_3$ on the dependent variable?(Is the effect positive or negative). An example would be greatly appreciated.
regression mathematical-statistics multiple-regression least-squares marginal-effect
edited 19 mins ago
Ferdi
3,69042152
asked 1 hour ago
P.Cay
83
New contributor
P.Cay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have my OLS regression: $y = beta_1 + beta_2 * X_2 + beta_3 * X_3 +beta_4 * (X_3)^2$
Could anybody please explain to me the effect of a change in $X_3$ on the dependent variable?(Is the effect positive or negative). An example would be greatly appreciated.
regression mathematical-statistics multiple-regression least-squares marginal-effect
regression mathematical-statistics multiple-regression least-squares marginal-effect
edited 19 mins ago
Ferdi
3,69042152
asked 1 hour ago
P.Cay
83
New contributor
P.Cay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 19 mins ago
Ferdi
3,69042152
asked 1 hour ago
P.Cay
83
New contributor
P.Cay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 19 mins ago
Ferdi
3,69042152
edited 19 mins ago
Ferdi
3,69042152
edited 19 mins ago
Ferdi
3,69042152
3,69042152
asked 1 hour ago
P.Cay
83
New contributor
P.Cay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
P.Cay
83
asked 1 hour ago
P.Cay
83
83
New contributor
P.Cay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
P.Cay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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3 Answers
3
active
oldest
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up vote
1
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accepted
As @Tomas already explained greatly in his answer you have to drive to $X_3$.
$$fracpartial ypartial X_3= beta_3 + 2 times beta_4 times X_3$$.
As you want to now whether the derivative is positive you face the inquality:
$$0>beta_3 + 2 times beta_4 times X_3$$.
Depending on $X_3$ you can change the equation to
$$frac-beta_32 * beta_4 > X_3$$
Let us look at an example in which you have real numbers for the Betas:
When you have the equation:
$beta_1 = 5$
$beta_2 = 3$
$beta_3 = 4$
$beta_4 = 2$
$y = 5 + 3 * X_2 + 4 * X_3 + 2 * (X_3)^2$
and $X_3$ will have a positive effect when:
$$frac-42 * 2 > X_3$$
which is simplified:
$$1 > X_3$$
Note that the "change in the conditional mean of outcome y when regressors change by one unit" is also called marginal effect. You might look at the tag description and question tagged with this tag to get a broader understanding. In your case you want to know whether the marginal effect of $X_3$ is positive (or negative).
answered 20 mins ago
Ferdi
3,69042152
add a comment |Â
up vote
1
down vote
You just take the first derivative of the function with respect to $X_3$:
$$fracpartial ypartial X_3= beta_3 + 2 times beta_4 times X_3$$
You may see that the effect of changing $X_3$ on $y$ is not constant but depends on the observed value of $X_3$ (for some, say $i$-th observation).
Also, ceteris paribus (other things fixed) interpretation holds only with respect to $X_2$, but you cannot change $X_3^2$ while holding $X_3$ fixed.
answered 36 mins ago
Tomas
1086
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up vote
0
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The effect that a change in $X_3$ will have on the outcome depend on the coefficients of $beta_3$ and $beta_4$. If both $beta_3$ and $beta_4$ are positive, then any increase in $X_3$ will cause $Y$ to increase. If both $beta_3$ and $beta_4$ are negative, then the an increse in $X_3$ leads to a decrease in $Y$. (Granted, all this is given that $X_3$ is increased without changing the values of $X_2$.)
If the signs of $beta_3$ and $beta_4$ differ, then the net effect of an increase of $X_3$ on $Y$ will depend on the relative magnitude of $beta_3$ and $beta_4$, as well as the increase in $X_3$.
answered 34 mins ago
Phil
36711
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
As @Tomas already explained greatly in his answer you have to drive to $X_3$.
$$fracpartial ypartial X_3= beta_3 + 2 times beta_4 times X_3$$.
As you want to now whether the derivative is positive you face the inquality:
$$0>beta_3 + 2 times beta_4 times X_3$$.
Depending on $X_3$ you can change the equation to
$$frac-beta_32 * beta_4 > X_3$$
Let us look at an example in which you have real numbers for the Betas:
When you have the equation:
$beta_1 = 5$
$beta_2 = 3$
$beta_3 = 4$
$beta_4 = 2$
$y = 5 + 3 * X_2 + 4 * X_3 + 2 * (X_3)^2$
and $X_3$ will have a positive effect when:
$$frac-42 * 2 > X_3$$
which is simplified:
$$1 > X_3$$
Note that the "change in the conditional mean of outcome y when regressors change by one unit" is also called marginal effect. You might look at the tag description and question tagged with this tag to get a broader understanding. In your case you want to know whether the marginal effect of $X_3$ is positive (or negative).
answered 20 mins ago
Ferdi
3,69042152
add a comment |Â
up vote
1
down vote
accepted
As @Tomas already explained greatly in his answer you have to drive to $X_3$.
$$fracpartial ypartial X_3= beta_3 + 2 times beta_4 times X_3$$.
As you want to now whether the derivative is positive you face the inquality:
$$0>beta_3 + 2 times beta_4 times X_3$$.
Depending on $X_3$ you can change the equation to
$$frac-beta_32 * beta_4 > X_3$$
Let us look at an example in which you have real numbers for the Betas:
When you have the equation:
$beta_1 = 5$
$beta_2 = 3$
$beta_3 = 4$
$beta_4 = 2$
$y = 5 + 3 * X_2 + 4 * X_3 + 2 * (X_3)^2$
and $X_3$ will have a positive effect when:
$$frac-42 * 2 > X_3$$
which is simplified:
$$1 > X_3$$
Note that the "change in the conditional mean of outcome y when regressors change by one unit" is also called marginal effect. You might look at the tag description and question tagged with this tag to get a broader understanding. In your case you want to know whether the marginal effect of $X_3$ is positive (or negative).
answered 20 mins ago
Ferdi
3,69042152
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
As @Tomas already explained greatly in his answer you have to drive to $X_3$.
$$fracpartial ypartial X_3= beta_3 + 2 times beta_4 times X_3$$.
As you want to now whether the derivative is positive you face the inquality:
$$0>beta_3 + 2 times beta_4 times X_3$$.
Depending on $X_3$ you can change the equation to
$$frac-beta_32 * beta_4 > X_3$$
Let us look at an example in which you have real numbers for the Betas:
When you have the equation:
$beta_1 = 5$
$beta_2 = 3$
$beta_3 = 4$
$beta_4 = 2$
$y = 5 + 3 * X_2 + 4 * X_3 + 2 * (X_3)^2$
and $X_3$ will have a positive effect when:
$$frac-42 * 2 > X_3$$
which is simplified:
$$1 > X_3$$
Note that the "change in the conditional mean of outcome y when regressors change by one unit" is also called marginal effect. You might look at the tag description and question tagged with this tag to get a broader understanding. In your case you want to know whether the marginal effect of $X_3$ is positive (or negative).
answered 20 mins ago
Ferdi
3,69042152
As @Tomas already explained greatly in his answer you have to drive to $X_3$.
$$fracpartial ypartial X_3= beta_3 + 2 times beta_4 times X_3$$.
As you want to now whether the derivative is positive you face the inquality:
$$0>beta_3 + 2 times beta_4 times X_3$$.
Depending on $X_3$ you can change the equation to
$$frac-beta_32 * beta_4 > X_3$$
Let us look at an example in which you have real numbers for the Betas:
When you have the equation:
$beta_1 = 5$
$beta_2 = 3$
$beta_3 = 4$
$beta_4 = 2$
$y = 5 + 3 * X_2 + 4 * X_3 + 2 * (X_3)^2$
and $X_3$ will have a positive effect when:
$$frac-42 * 2 > X_3$$
which is simplified:
$$1 > X_3$$
Note that the "change in the conditional mean of outcome y when regressors change by one unit" is also called marginal effect. You might look at the tag description and question tagged with this tag to get a broader understanding. In your case you want to know whether the marginal effect of $X_3$ is positive (or negative).
answered 20 mins ago
Ferdi
3,69042152
answered 20 mins ago
Ferdi
3,69042152
answered 20 mins ago
Ferdi
3,69042152
answered 20 mins ago
Ferdi
3,69042152
3,69042152
add a comment |Â
add a comment |Â
up vote
1
down vote
You just take the first derivative of the function with respect to $X_3$:
$$fracpartial ypartial X_3= beta_3 + 2 times beta_4 times X_3$$
You may see that the effect of changing $X_3$ on $y$ is not constant but depends on the observed value of $X_3$ (for some, say $i$-th observation).
Also, ceteris paribus (other things fixed) interpretation holds only with respect to $X_2$, but you cannot change $X_3^2$ while holding $X_3$ fixed.
answered 36 mins ago
Tomas
1086
add a comment |Â
up vote
1
down vote
You just take the first derivative of the function with respect to $X_3$:
$$fracpartial ypartial X_3= beta_3 + 2 times beta_4 times X_3$$
You may see that the effect of changing $X_3$ on $y$ is not constant but depends on the observed value of $X_3$ (for some, say $i$-th observation).
Also, ceteris paribus (other things fixed) interpretation holds only with respect to $X_2$, but you cannot change $X_3^2$ while holding $X_3$ fixed.
answered 36 mins ago
Tomas
1086
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You just take the first derivative of the function with respect to $X_3$:
$$fracpartial ypartial X_3= beta_3 + 2 times beta_4 times X_3$$
You may see that the effect of changing $X_3$ on $y$ is not constant but depends on the observed value of $X_3$ (for some, say $i$-th observation).
Also, ceteris paribus (other things fixed) interpretation holds only with respect to $X_2$, but you cannot change $X_3^2$ while holding $X_3$ fixed.
answered 36 mins ago
Tomas
1086
You just take the first derivative of the function with respect to $X_3$:
$$fracpartial ypartial X_3= beta_3 + 2 times beta_4 times X_3$$
You may see that the effect of changing $X_3$ on $y$ is not constant but depends on the observed value of $X_3$ (for some, say $i$-th observation).
Also, ceteris paribus (other things fixed) interpretation holds only with respect to $X_2$, but you cannot change $X_3^2$ while holding $X_3$ fixed.
answered 36 mins ago
Tomas
1086
edited 30 mins ago
answered 36 mins ago
Tomas
1086
answered 36 mins ago
Tomas
1086
answered 36 mins ago
Tomas
1086
1086
add a comment |Â
add a comment |Â
up vote
0
down vote
The effect that a change in $X_3$ will have on the outcome depend on the coefficients of $beta_3$ and $beta_4$. If both $beta_3$ and $beta_4$ are positive, then any increase in $X_3$ will cause $Y$ to increase. If both $beta_3$ and $beta_4$ are negative, then the an increse in $X_3$ leads to a decrease in $Y$. (Granted, all this is given that $X_3$ is increased without changing the values of $X_2$.)
If the signs of $beta_3$ and $beta_4$ differ, then the net effect of an increase of $X_3$ on $Y$ will depend on the relative magnitude of $beta_3$ and $beta_4$, as well as the increase in $X_3$.
answered 34 mins ago
Phil
36711
add a comment |Â
up vote
0
down vote
The effect that a change in $X_3$ will have on the outcome depend on the coefficients of $beta_3$ and $beta_4$. If both $beta_3$ and $beta_4$ are positive, then any increase in $X_3$ will cause $Y$ to increase. If both $beta_3$ and $beta_4$ are negative, then the an increse in $X_3$ leads to a decrease in $Y$. (Granted, all this is given that $X_3$ is increased without changing the values of $X_2$.)
If the signs of $beta_3$ and $beta_4$ differ, then the net effect of an increase of $X_3$ on $Y$ will depend on the relative magnitude of $beta_3$ and $beta_4$, as well as the increase in $X_3$.
answered 34 mins ago
Phil
36711
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The effect that a change in $X_3$ will have on the outcome depend on the coefficients of $beta_3$ and $beta_4$. If both $beta_3$ and $beta_4$ are positive, then any increase in $X_3$ will cause $Y$ to increase. If both $beta_3$ and $beta_4$ are negative, then the an increse in $X_3$ leads to a decrease in $Y$. (Granted, all this is given that $X_3$ is increased without changing the values of $X_2$.)
If the signs of $beta_3$ and $beta_4$ differ, then the net effect of an increase of $X_3$ on $Y$ will depend on the relative magnitude of $beta_3$ and $beta_4$, as well as the increase in $X_3$.
answered 34 mins ago
Phil
36711
The effect that a change in $X_3$ will have on the outcome depend on the coefficients of $beta_3$ and $beta_4$. If both $beta_3$ and $beta_4$ are positive, then any increase in $X_3$ will cause $Y$ to increase. If both $beta_3$ and $beta_4$ are negative, then the an increse in $X_3$ leads to a decrease in $Y$. (Granted, all this is given that $X_3$ is increased without changing the values of $X_2$.)
If the signs of $beta_3$ and $beta_4$ differ, then the net effect of an increase of $X_3$ on $Y$ will depend on the relative magnitude of $beta_3$ and $beta_4$, as well as the increase in $X_3$.
answered 34 mins ago
Phil
36711
answered 34 mins ago
Phil
36711
answered 34 mins ago
Phil
36711
answered 34 mins ago
Phil
36711
36711
add a comment |Â
add a comment |Â
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