Good morning everyone! 
First, thank you for your help!
I have 6 variables where my first variable X1 has 5 differentes values in my dataset (X1.1 X1.2 X1.3 X1.4 X1.5).
I have this code in my STATA do-file:
My problem comes from I have the reference in X1.3 and this is what I want; For example, this is what I have from logit regression:
And I interpret this as the probability that an individual will experience Y2 is 6% greater if he belongs to group X1.1 than if he belongs to group X1.3 (group X1.3 being the reference group).
So, Is this right? If this is good, This is what I want to get. After this, here comes the "real problem"
In the multinomial logit, I have my individual reference when X1.3 as I did in logit, but here is the problem, I need the predicted probability of my individual that belongs in X1.3 but I do not have it.
My variable Y1 in the mlogit can have 4 values: Y1.1 Y1.2 Y1.3 Y1.4.
As you can see, I interpret this like the probability that an individual belonging to group X1.1 belongs to Y1.1 is 1.9% greater than that of the individual from group X1.3 belonging to Y1.1. I don't know if this is correct or I'm confusing concepts.
After this (and thinking this is good) I wanna know what is the predicted probability that each different value of X1 (which can have 5 different values) belongs to one of the four different values that Y1 can take, that is, I want to obtain the total predicted probability and not only comparing it with the base outcome (I want both cases).
How can I do it? How can I obtain this predicted probability?
Thank you very much! I appreciate your help!
Regards,
R
First, thank you for your help!I have 6 variables where my first variable X1 has 5 differentes values in my dataset (X1.1 X1.2 X1.3 X1.4 X1.5).
I have this code in my STATA do-file:
mlogit Y1 ib3.X1 i.X2 i.X3 i.X4 i.X5 i.X6, baseoutcome(3) r
margins, dydx(*)
logit Y2 ib3.X1 i.X2 i.X3 i.X4 i.X5 i.X6, r
margins, dydx(*)
margins, dydx(*)
logit Y2 ib3.X1 i.X2 i.X3 i.X4 i.X5 i.X6, r
margins, dydx(*)
My problem comes from I have the reference in X1.3 and this is what I want; For example, this is what I have from logit regression:
dy/dx std. err. z P>z [95% conf. interval]
X1
X1.1 .0600794 .0179045 3.36 0.001 .0249872 .0951717
X1.2 .0673866 .0199228 3.38 0.001 .0283386 .1064346
X1.4 -.0048324 .0201372 -0.24 0.810 -.0443007 .0346358
X1.5 .1126457 .0211299 5.33 0.000 .0712319 .1540594
X1
X1.1 .0600794 .0179045 3.36 0.001 .0249872 .0951717
X1.2 .0673866 .0199228 3.38 0.001 .0283386 .1064346
X1.4 -.0048324 .0201372 -0.24 0.810 -.0443007 .0346358
X1.5 .1126457 .0211299 5.33 0.000 .0712319 .1540594
And I interpret this as the probability that an individual will experience Y2 is 6% greater if he belongs to group X1.1 than if he belongs to group X1.3 (group X1.3 being the reference group).
So, Is this right? If this is good, This is what I want to get. After this, here comes the "real problem"
In the multinomial logit, I have my individual reference when X1.3 as I did in logit, but here is the problem, I need the predicted probability of my individual that belongs in X1.3 but I do not have it.
My variable Y1 in the mlogit can have 4 values: Y1.1 Y1.2 Y1.3 Y1.4.
Delta-method
dy/dx std. err. z P>z [95% conf. interval]
1.X1.1
_predict
1 .0198357 .006931 2.86 0.004 .0062512 .0334202
2 .002201 .0038173 0.58 0.564 -.0052808 .0096827
3 -.0100104 .0035321 -2.83 0.005 -.0169333 -.0030876
4 -.0120262 .0047819 -2.51 0.012 -.0213987 -.0026538
2.X1.2
_predict
1 .0263887 .0067502 3.91 0.000 .0131585 .0396188
2 -.0053821 .0030592 -1.76 0.079 -.0113781 .0006139
3 -.0091103 .0036635 -2.49 0.013 -.0162907 -.0019299
4 -.0118963 .0049251 -2.42 0.016 -.0215494 -.0022432
3.X1.3 (base outcome)
4.X1.4
_predict
1 -.0060001 .006104 -0.98 0.326 -.0179637 .0059635
2 .003009 .0029996 1.00 0.316 -.0028701 .0088881
3 .0120931 .0038648 3.13 0.002 .0045182 .019668
4 -.0091019 .0039111 -2.33 0.020 -.0167676 -.0014363
5.X1.5
_predict
1 .0498023 .0048763 10.21 0.000 .0402449 .0593597
2 -.0099728 .0021525 -4.63 0.000 -.0141917 -.0057539
3 -.0105514 .003214 -3.28 0.001 -.0168507 -.004252
4 -.0292782 .0030971 -9.45 0.000 -.0353484 -.0232079
dy/dx std. err. z P>z [95% conf. interval]
1.X1.1
_predict
1 .0198357 .006931 2.86 0.004 .0062512 .0334202
2 .002201 .0038173 0.58 0.564 -.0052808 .0096827
3 -.0100104 .0035321 -2.83 0.005 -.0169333 -.0030876
4 -.0120262 .0047819 -2.51 0.012 -.0213987 -.0026538
2.X1.2
_predict
1 .0263887 .0067502 3.91 0.000 .0131585 .0396188
2 -.0053821 .0030592 -1.76 0.079 -.0113781 .0006139
3 -.0091103 .0036635 -2.49 0.013 -.0162907 -.0019299
4 -.0118963 .0049251 -2.42 0.016 -.0215494 -.0022432
3.X1.3 (base outcome)
4.X1.4
_predict
1 -.0060001 .006104 -0.98 0.326 -.0179637 .0059635
2 .003009 .0029996 1.00 0.316 -.0028701 .0088881
3 .0120931 .0038648 3.13 0.002 .0045182 .019668
4 -.0091019 .0039111 -2.33 0.020 -.0167676 -.0014363
5.X1.5
_predict
1 .0498023 .0048763 10.21 0.000 .0402449 .0593597
2 -.0099728 .0021525 -4.63 0.000 -.0141917 -.0057539
3 -.0105514 .003214 -3.28 0.001 -.0168507 -.004252
4 -.0292782 .0030971 -9.45 0.000 -.0353484 -.0232079
As you can see, I interpret this like the probability that an individual belonging to group X1.1 belongs to Y1.1 is 1.9% greater than that of the individual from group X1.3 belonging to Y1.1. I don't know if this is correct or I'm confusing concepts.
After this (and thinking this is good) I wanna know what is the predicted probability that each different value of X1 (which can have 5 different values) belongs to one of the four different values that Y1 can take, that is, I want to obtain the total predicted probability and not only comparing it with the base outcome (I want both cases).
How can I do it? How can I obtain this predicted probability?
Thank you very much! I appreciate your help!
Regards,
R
