DHS data -Generating the leave-out mean share of a cohort

[Mwenzie Banda]

Hello,

I am trying to run an instrumental variable estimation of child marriage on educational attainment using DHS data. As one of my instruments I want to use the leave-out mean of the past incidence of marriage in the previous decade by measuring the share of women that are currently aged 30-40 that got married at the ages of 12 through to 17 at PSU level. I used the following command:

*v511 is age at first marriage
*v012 is respondents current age
*V001 is the PSU

*Previous Decade PSU leave out mean share women that married at 12 years
gen byte age12=v511==12 if v012>=30&v012<=40
egen pasttotal12=total(pastage12), by(v001)
egen pastcount12= count(pastage12), by(v001)
gen pastlom12=(pasttotal12-pastage12)/(pastcount12-1)
list v001 age12 pasttotal12 pastcount12 pastlom12, sepby(v001)

The issue with this command is that it creates missing values for the sample(those ages 20-29) that I want to estimate on. Is there any other command I can use that will give me leave out mean share of past incidence of marriage at PSU level without generating missing values for my sample.

[Andrew Musau]

Assuming that the variable is constant within PSU:

bys v001 (pastlom12): replace pastlom12= pastlom12[1]

Missing values are sorted last. Therefore, if there is a nonmissing value in a cohort, all missing values will be replaced with it.

[Mwenzie Banda]

Hi Andrew, thanks for this! The variable isn't constant because I calculate the mean to not include the data for the corresponding observation. Here is an example of my output of the same command i used above

. list v001 pastage12 pasttotal12 pastcount12 pastlom12, sepby(v001)      

      +--------------------------------------------------+
      | v001   pasta~12   pastt~12   pastc~12   pastl~12 |
      |--------------------------------------------------|
   1. |    1          0          0          2          0 |
   2. |    1          .          0          2          . |
   3. |    1          .          0          2          . |
   4. |    1          .          0          2          . |
   5. |    1          .          0          2          . |
   6. |    1          .          0          2          . |
   7. |    1          0          0          2          0 |
   8. |    1          .          0          2          . |
   9. |    1          .          0          2          . |
  10. |    1          .          0          2          . |
  11. |    1          .          0          2          . |
  12. |    1          .          0          2          . |
  13. |    1          .          0          2          . |
......
    
      |--------------------------------------------------|
 142. |   15          .          1          6          . |
 143. |   15          0          1          6         .2 |
 144. |   15          .          1          6          . |
 145. |   15          .          1          6          . |
 146. |   15          .          1          6          . |
 147. |   15          .          1          6          . |
 148. |   15          .          1          6          . |
 149. |   15          0          1          6         .2 |
 150. |   15          1          1          6          0 |
 151. |   15          0          1          6         .2 |
 152. |   15          .          1          6          . |
 153. |   15          .          1          6          . |
 154. |   15          0          1          6         .2 |
 155. |   15          0          1          6         .2 |

When I use the code you suggested it changes all the mean values to 0 in PSU 15:

 bys v001 (pastlom12): replace pastlom12= pastlom12[1]
(3,218 real changes made)

. list v001 pastage12 pasttotal12 pastcount12 pastlom12, sepby(v001)      

      +--------------------------------------------------+
      | v001   pasta~12   pastt~12   pastc~12   pastl~12 |
      |--------------------------------------------------|
   1. |    1          0          0          2          0 |
   2. |    1          0          0          2          0 |
   3. |    1          .          0          2          0 |
   4. |    1          .          0          2          0 |
   5. |    1          .          0          2          0 |
   6. |    1          .          0          2          0 |
   7. |    1          .          0          2          0 |
   8. |    1          .          0          2          0 |
   9. |    1          .          0          2          0 |
  10. |    1          .          0          2          0 |
  11. |    1          .          0          2          0 |
  12. |    1          .          0          2          0 |
  13. |    1          .          0          2          0 |
......
      |--------------------------------------------------|
 142. |   15          1          1          6          0 |
 143. |   15          0          1          6          0 |
 144. |   15          0          1          6          0 |
 145. |   15          0          1          6          0 |
 146. |   15          0          1          6          0 |
 147. |   15          0          1          6          0 |
 148. |   15          .          1          6          0 |
 149. |   15          .          1          6          0 |
 150. |   15          .          1          6          0 |
 151. |   15          .          1          6          0 |
 152. |   15          .          1          6          0 |
 153. |   15          .          1          6          0 |
 154. |   15          .          1          6          0 |
 155. |   15          .          1          6          0 |

[Andrew Musau]

Sorry, I misunderstood your question. If you are creating the variable for the subsample of individuals aged 30-40 years, then what values should it take for individuals outside this age range? If you want the value corresponding to pastage12=0, then

bys v001 (pastage12): replace pastlom12= pastlom12[1]

[Mwenzie Banda]

Thank so much!