Oh yes, you are right. It seems that the default distribution used for
sampling is uniform. And whether the resampling generates a random
distribution or not depends on the distribution being sampled.

Thanks everyone for your help! I appreciate your support very much.

Tom

On 10/19/06, Ted Harding <Ted.Harding@nessie.mcc.ac.uk> wrote:
>
> On 19-Oct-06 tom soyer wrote:
> > Hi,
> >
> > I looked up the help file on sample(), but didn't find the
> > info I was looking for.
> >
> > When sample() is used to resample from a distribution, e.g.,
> > bootstrap, how does it do it? Does it use an uniform distribution,
> > e.g., runif(), or something else?
>
> I don't know the details of the algorithm, but since sample()
> has flexible options it may be helpful to describe the effect
> of sample() in different cases.
>
> 1. sample(x,r) where x is a vector of length n
> In effect, the index values (1:n) of x are sampled from
> without replacement (default) with a uniform probability
> distribution over the available elements at all stages.
> Hence, i1 is sampled from (1:n) with probability 1/n for
> each possibility. Then i2 is sampled from the remainder
> with probability 1/(n-1) for each, and so on until r items
> (all distinct) have been sampled. If the resulting indices
> are {i1,i2,...,ir} then the result is x[i1],x[i2],...,x[ir].
> Thus, if some of the values in x[1],...,x[n] are equal,
> you can get 2 or more items in the sample which are equal
> even though the sampling is done without replacement (since
> it is the indices which are sampled).
> [NB I'm describing the *effect* here, not saying that this
> is how the algorithm operates]
>
> 2. sample(x, replace=TRUE)
> Similar to [1], except that the sampled index is returned
> to the pool and is available to be sampled again, so at each
> stage the probability of any value being chosen is 1/n.
>
> 3. sample(x, replace=TRUE, prob=p) where p is a vector of
>   probability weights (which must not all be 0, and none
>   negative).
> First, p is converted into a probability distribution
> (summing to 1) (in effect by dividing by the sum).
> Then an index i1 is sampled from (1:n) with probability
> p[i] that i is chosen. This is repeated (with previously
> sampled i's still available) until r index values have been
> sampled -- i1,...,ir. The result is x[i1],...,x[ir].
>
> 4. sample(x, prob=p) [without replacement]
> First p is scaled to sum to 1, then i1 is sampled as in [3].
> The remaining p-values are rescaled so as to sum to 1,
> and i2 is sampled from the remaining i's; and so on.
>
> These are the essential variants of the use of sample().
>
> runif() can be used to sample i1 from (1:n) with equal
> probabilities by selecting i if runif() is <= i and > (i-1)
> for i = 1:n.
>
> Similarly runif() can be used to sample i1 from (1:n)
> with probabilities p1,...,pn by selecting i if
>
> p[1] + ... + p[i-1] < runif() <= p[1] + ... + p[i]
>
> [LHS=0 if i=0], since the probability of this happening is p[i].
>
> > And, when the help file
> > says:"sample(x) generates a random permutation of the elements
> > of x (or 1:x)",
>
> Since the default value of r (size of sample) is the length
> of x, say n, sample(x) (see [1] above) will sample n elements
> without replacement from the n elements of x with uniform
> probabilities at each stage. In effect, n elements i1,i2,...,in
> will be sampled without replacement from (1:n), giving a
> random permutation of (1:n), so the result x[i1],...,x[in]
> will be a random permutation of x[1],...,x[n] (though
> different random permutations may look identical if there
> are equal values in x[1],...,x[n]).
>
> > would I be correct if I translate the statement
> > as follows: it means that the order of sequence, which was
> > generated from a uniform distribution, would look like a
> > random normal distribution.
>
> No. A normal distribution has nothing to do with it!
>
> *Unless* the values x[1],...,x[n] already loooked like values
> which had already been sampled from a normal distribution (but
> were, say, in increasing order of size). Then sample(x) would
> shuffle them into random order so the result could then look
> like a real sample according ot eh order in which the data
> came in.
>
> Hoping this helps!
> Ted.
>
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> E-Mail: (Ted Harding) <Ted.Harding@nessie.mcc.ac.uk>
> Fax-to-email: +44 (0)870 094 0861
> Date: 19-Oct-06                                       Time: 19:34:13
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