Quadratic regression: adding CI to lpoly graph

[Aicha SA]

Hello all,

I am running a non-linear regression that looks like this:

vcemway reg ln_y x x^2 i.year FE1 FE 2 FE3, cluster(region year)

I then use the regression coefficients to predict E(Y | X) as follows:

gen beta_S = _b[x]
gen beta_S_sq = _b[x^2]
gen beta_cons = _b[_cons]

gen FE1_mem = _b[hh_members]
gen FE2_KasP =_b[region1]
gen FE3_KusP =_b[region2]

gen share_gen = exp(beta_cons + rand_S*beta_S + rand_S*rand_S*beta_S_sq + FE1_mem + FE2_KasP + FE3_KusP)

Here, rand_S is a variable containing random values of my random variable.

*Output graph*
graph twoway (lpoly share_gen rand_S)

My question is, how can I compute the upper and lower bounds of the confidence intervals to include them in the graph? I am looking to create new variables and plotting them using lpoly.

Any other suggestions would also be welcome! Thank you very much in advance!

[Leonardo Guizzetti]

If you just want the CI overlaid on the plot, there is a corresponding twoway graph called -lpolyci- that does just that.

[Aicha SA]

Thank you for your answer! I'm not sure how I can use lpolyci with my code as simply replacing lpoly with lpolyci doesn't work. Would it be possible to have more details please? Thanks.

[Leonardo Guizzetti]

Are you certain you typed it correctly? This (silly example) works for me.

sysuse auto
graph twoway lpolyci price mpg

[Aicha SA]

Thank you. Sorry, perhaps I wasn't clear with my question. The problem is that I am running the lpoly command on a new variable that I generate from the regression coefficients:

gen share_gen = exp(beta_cons + rand_S*beta_S + rand_S*rand_S*beta_S_sq + FE1_mem + FE2_KasP + FE3_KusP)

That's why lpolyci won't work as I need to add the confidence intervals manually. As this is a quadratic regression, I am not sure how to do that.

Many thanks in advance!

[Leonardo Guizzetti]

It would still work. Generate your -share_gen- and then with -rand_S-, you can make the graph using

graph twoway (lpolyci share_gen rand_S)

For what you are trying to do, generating new variables from regression coefficients, it is probably easier to learn to use -margins- in the long run, but this should get you the graph you want.

[Aicha SA]

Thank you so much for your answers and patience! Unfortunately, this still doesn't work (I ran the exact same code) and I still get a graph without CI... I'm not sure what I'm doing wrong.

Thanks a lot again.

[Leonardo Guizzetti]

still doesn't work

I can't diagnose what is wrong without seeing the exact code you typed and having a reproducible data example. To make a reproducible data example, type -dataex- to generate it, and when you paste the code and data here, be sure to use the CODE delimiters.

What I can see is that the following two lines of code are not legal syntax, because you cannot specify a quadratic in the way that you have.

vcemway reg ln_y x x^2 i.year FE1 FE 2 FE3, cluster(region year)
gen beta_S_sq = _b[x^2]

At a minimum, you should fix these two lines to match below. See -help factor variables- for more information on how this works. The upside is that Stata will now recognize that x and x^2 are correlated.

vcemway reg ln_y c.x##c.x i.year FE1 FE 2 FE3, cluster(region year)
gen beta_S_sq = _b[c.x#c.x]