<>
"...0 with probability about 1 in 4 billion"
Out of curiosity, how do you calculate that for a continuous random
variable?
HTH
Martin
-----Ursprüngliche Nachricht-----
Von: [email protected]
[mailto:[email protected]] Im Auftrag von Austin Nichols
Gesendet: Donnerstag, 26. Februar 2009 20:59
An: [email protected]
Betreff: Re: st: random draw of single variable, not whole data
Jeph suggests the ceil() solution in a follow-up with the caveat "as
long as uniform() never returns precisely 0, which I've
never seen" --uniform() is 0 with probability about 1 in 4 billion, so
to be safe you could
gen y = x[ceil(_N*(1-uniform()))]
On Thu, Feb 26, 2009 at 2:33 PM, Austin Nichols <[email protected]>
wrote:
> With replacement:
> gen y = x[round(_N*uniform(),1)+1]
>
> should be
> gen y = x[round(_N*uniform()+.5)]
> or
> gen y = x[ceil(_N*uniform())]
>
> I think.... (otherwise you can get _N+1 in the square brackets, for a
> missing y, and the prob of getting y[1] is too low)
>
> On Thu, Feb 26, 2009 at 2:19 PM, Jeph Herrin <[email protected]> wrote:
>>
>> It's not clear if you want sampling with or without replacement.
>>
>> With replacement:
>>
>> gen y = x[round(_N*uniform(),1)+1]
>>
>> without replacement:
>>
>> egen index=rank(uniform())
>> gen y = x[index]
>>
>> hth,
>> Jeph
>
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