Good morning,
I am implementing Bootstrap manually as explained in Angquist, Stata tip 92: Manual implementation of permutations and bootstraps. The results are vastly different from the results of bsample. Why is that, and am I making some mistake in my code?
***
----------------------------------------------------------------------------------------------------------------
name: <unnamed>
log: C:\Program Files\Stata15\bootLog.log
log type: text
opened on: 29 Aug 2018, 10:47:10
.
. cap program drop boot1 boot2
.
.
.
. program define boot1, rclass
1. args obs
2.
. drop _all
3.
. set obs `obs'
4.
.
. * I generate one sample from N(0,1000), then I calculate the mean
.
. gen x = 1000*rnormal()
5.
. summ x
6.
. return sca meanx1 = r(mean)
7.
. * Then I draw a Bootstrap sample using Bsample and calculate the mean
.
. bsample
8.
. summ x
9.
. return sca meanxboot1 = r(mean)
10.
. end
.
.
.
. program define boot2, rclass
1. args obs
2.
. drop _all
3.
. set obs `obs'
4.
. * I generate one sample from N(0,1000), then I calculate the mean
.
. gen x = 1000*rnormal()
5.
. summ x
6.
. return sca meanx2 = r(mean)
7.
. * Then I draw a Bootstrap sample manually and calculate the mean
.
. generate u=ceil(runiform()*_N) // this generates a uniformly distributed set of integers on 1,2,...,_N
8. replace x =x[u] // this replaces the original x by the value of x at the random univorm integer on 1,2,...,
> _N
9.
. summ x
10.
. return sca meanxboot2 = r(mean)
11.
. end
.
.
.
. simulate meanx1=r(meanx1) meanxboot1=r(meanxboot1), reps(10000) seed(69) saving(boot1temp, replace) nodots: bo
> ot1 999
command: boot1 999
meanx1: r(meanx1)
meanxboot1: r(meanxboot1)
.
.
.
.
. simulate meanx2=r(meanx2) meanxboot2=r(meanxboot2), reps(10000) seed(69) nodots: boot2 999
command: boot2 999
meanx2: r(meanx2)
meanxboot2: r(meanxboot2)
.
.
. merge using boot1temp
(note: you are using old merge syntax; see [D] merge for new syntax)
.
. summ meanx1 meanxboot1 meanx2 meanxboot2
Variable | Obs Mean Std. Dev. Min Max
-------------+---------------------------------------------------------
meanx1 | 10,000 .5236758 31.18312 -131.0225 108.7608
meanxboot1 | 10,000 .3181329 44.58329 -171.1018 175.1777
meanx2 | 10,000 .2532531 31.49465 -126.415 116.299
meanxboot2 | 10,000 .3867981 66.27218 -256.7187 262.4239
.
. twoway (kdensity meanx1) (kdensity meanxboot1) (kdensity meanx2) (kdensity meanxboot2)
.
. log close
name: <unnamed>
log: C:\Program Files\Stata15\bootLog.log
log type: text
closed on: 29 Aug 2018, 10:47:30
----------------------------------------------------------------------------------------------------------------
***
Note that the distribution of the bsample generated bootstrap mean (meanxboot1, std=44.58, min= -171.10, max= 175.17) is vastly different from the distribution of the manual bootstrap generated bootstrap mean (meanxboot2, std=66.27, min=-256.71, max=262.42). The manually generated bootstrap mean has much more spread out distribution.
I am implementing Bootstrap manually as explained in Angquist, Stata tip 92: Manual implementation of permutations and bootstraps. The results are vastly different from the results of bsample. Why is that, and am I making some mistake in my code?
***
----------------------------------------------------------------------------------------------------------------
name: <unnamed>
log: C:\Program Files\Stata15\bootLog.log
log type: text
opened on: 29 Aug 2018, 10:47:10
.
. cap program drop boot1 boot2
.
.
.
. program define boot1, rclass
1. args obs
2.
. drop _all
3.
. set obs `obs'
4.
.
. * I generate one sample from N(0,1000), then I calculate the mean
.
. gen x = 1000*rnormal()
5.
. summ x
6.
. return sca meanx1 = r(mean)
7.
. * Then I draw a Bootstrap sample using Bsample and calculate the mean
.
. bsample
8.
. summ x
9.
. return sca meanxboot1 = r(mean)
10.
. end
.
.
.
. program define boot2, rclass
1. args obs
2.
. drop _all
3.
. set obs `obs'
4.
. * I generate one sample from N(0,1000), then I calculate the mean
.
. gen x = 1000*rnormal()
5.
. summ x
6.
. return sca meanx2 = r(mean)
7.
. * Then I draw a Bootstrap sample manually and calculate the mean
.
. generate u=ceil(runiform()*_N) // this generates a uniformly distributed set of integers on 1,2,...,_N
8. replace x =x[u] // this replaces the original x by the value of x at the random univorm integer on 1,2,...,
> _N
9.
. summ x
10.
. return sca meanxboot2 = r(mean)
11.
. end
.
.
.
. simulate meanx1=r(meanx1) meanxboot1=r(meanxboot1), reps(10000) seed(69) saving(boot1temp, replace) nodots: bo
> ot1 999
command: boot1 999
meanx1: r(meanx1)
meanxboot1: r(meanxboot1)
.
.
.
.
. simulate meanx2=r(meanx2) meanxboot2=r(meanxboot2), reps(10000) seed(69) nodots: boot2 999
command: boot2 999
meanx2: r(meanx2)
meanxboot2: r(meanxboot2)
.
.
. merge using boot1temp
(note: you are using old merge syntax; see [D] merge for new syntax)
.
. summ meanx1 meanxboot1 meanx2 meanxboot2
Variable | Obs Mean Std. Dev. Min Max
-------------+---------------------------------------------------------
meanx1 | 10,000 .5236758 31.18312 -131.0225 108.7608
meanxboot1 | 10,000 .3181329 44.58329 -171.1018 175.1777
meanx2 | 10,000 .2532531 31.49465 -126.415 116.299
meanxboot2 | 10,000 .3867981 66.27218 -256.7187 262.4239
.
. twoway (kdensity meanx1) (kdensity meanxboot1) (kdensity meanx2) (kdensity meanxboot2)
.
. log close
name: <unnamed>
log: C:\Program Files\Stata15\bootLog.log
log type: text
closed on: 29 Aug 2018, 10:47:30
----------------------------------------------------------------------------------------------------------------
***
Note that the distribution of the bsample generated bootstrap mean (meanxboot1, std=44.58, min= -171.10, max= 175.17) is vastly different from the distribution of the manual bootstrap generated bootstrap mean (meanxboot2, std=66.27, min=-256.71, max=262.42). The manually generated bootstrap mean has much more spread out distribution.
