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From "jian wang (JIRA)" <j...@apache.org>
Subject [jira] [Comment Edited] (DATAFU-21) Probability weighted sampling without reservoir
Date Tue, 01 Apr 2014 00:25:15 GMT

    [ https://issues.apache.org/jira/browse/DATAFU-21?page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel&focusedCommentId=13944325#comment-13944325
] 

jian wang edited comment on DATAFU-21 at 4/1/14 12:24 AM:
----------------------------------------------------------

Some investigation updates:

Based on the theories from paper: http://utopia.duth.gr/~pefraimi/research/data/2007EncOfAlg.pdf,
I plan to associate each item with a key X(j) = 1 - pow(U, 1/w(j)), U is a random variable
between (0,1). Then follow the thought of Random Sort, we sort the items in ascending order
based on X(j) and select the smallest k = p * n items.

Also as simple random sampling algorithm, we could also consider the possibility of rejecting
items applying Maurer's lemma and accepting items applying Bernstein's lemma.

Apply Maurer's lemma:

we would like to find 0<q1<1, so that we reject items whose  key is greater than q1.

let Y(j) = 1 if (X(j) < q1)
           =  0 otherwise

{Y(j), j = 1 to n} are independent random variables.

E(Y(j)) = Pr(X(j) < q1) * 1 + Pr(X(j) >= q1) * 0
            = Pr(1 - pow( U, 1/w(j) ) < q1)
            = Pr(1 - q1 < pow( U, 1/w(j) ))
            = Pr(pow(1 - q1, w(j) ) < U) = 1 - pow( 1 - q1, w(j) )

E(Y(j) ^ 2) = E(Y(j)) = 1 - pow(1 - q1, w(j) )

set Y = sum(Y(j), j = 1 to n),  

      Q1 = sum( pow( 1-q1, w(j) ) , j = 1 to n)

E(Y) = sum(E(Y(j))) = n - sum( pow( 1-q1, w(j) ) ) = n - Q1

apply Maurer's lemma with t = (1 - p) * n - sum( pow(1 - q1, w(j) ) ) = (1 - p) * n - Q1,
I get  
        Q1  >= n * (1 - p) + log(err) + sqrt( log(err) * (log(err) - 2 * n * p) )     (1)

we could get q1 by solving (1)

Apply Berstein's lemma:

similar to applying Maurer's lemma, we could get a q2 so that we could accept item whose key
is smaller than q2, 0 <= q2 <= 1.

let  Z(j) = 1 if X(j) < q2, 
            = 0 if X(j) >= q2

{Z(j), j = 1 to n} are independent random variables.

E(Z(j)) = Pr(X(j) < q2) * 1 + Pr(X(j) >= q2) *  0
           = Pr(1 - pow( U, 1/w(j) ) < q2)
           = 1 - pow(1 - q2, w(j) )

E(Z(j) ^ 2) = E(Z(j))

Z(j) - E(Z(j)) <= 1 - E(Z(j)) = pow(1 - q2, w(j) ) <= 1 = M

theta(j) ^ 2 = E(Z(j) ^ 2) - E(Z(j)) ^ 2 <= E(Z(j) ^ 2) = 1 - pow(1 - q2, w(j) )

set Z = sum(Z(j),  j = 1 to n)

      Q2 = sum( pow(1 - q2, w(j) ) )

      E(Z) = sum(E(Z(j)), j = 1 to n) = n - sum( pow(1 - q2, w(j) ) ) = n - Q2

apply Berstein's lemma with t = sum( pow(1 - q2, w(j) ) ) - (1 - p) * n = Q2 -  (1 - p) *
n, I get  
    Q2  >= n * (1 - p) + 2 / 3 * log(err)  + 2 / 3 * sqrt( log(err) * (log(err) - 9 * n
* p / 2 ) )     (2)

we could get q2 by solving (2)

Questions:

(1) Please help comment on the above approach. Do they overall make sense? 

(2) I am stuck in getting q1 and q2 by solving (1) and (2) respectively. Would like to seek
some advice on it. 


was (Author: king821221):
Some investigation updates:

Based on the theories from paper: http://utopia.duth.gr/~pefraimi/research/data/2007EncOfAlg.pdf,
I plan to associate each item with a key X(j) = 1 - pow(U, 1/w(j)), U is a random variable
between (0,1). Then follow the thought of Random Sort, we sort the items in ascending order
based on X(j) and select the smallest k = p * n items.

Also as simple random sampling algorithm, we could also consider the possibility of rejecting
items applying Maurer's lemma and accepting items applying Bernstein's lemma.

Apply Maurer's lemma:

we would like to find 0<q1<1, so that we reject items whose  key is greater than q1.

let Y(j) = 1 if (X(j) < q1)
           =  0 otherwise

{Y(j), j = 1 to n} are independent random variables.

E(Y(j)) = Pr(X(j) < q1) * 1 + Pr(X(j) >= q1) * 0
            = Pr(1 - pow( U, 1/w(j) ) < q1)
            = Pr(1 - q1 < pow( U, 1/w(j) ))
            = Pr(pow(1 - q1, w(j) ) < U) = 1 - pow( 1 - q1, w(j) )

E(Y(j) ^ 2) = E(Y(j)) = 1 - pow(1 - q1, w(j) )

set Y = sum(Y(j), j = 1 to n),  

      Q1 = sum( pow( 1-q1, w(j) ) , j = 1 to n)

E(Y) = sum(E(Y(j))) = n - sum( pow( 81-q1, w(j) ) ) = n - Q1

apply Maurer's lemma with t = (1 - p) * n - sum( pow(1 - q1, w(j) ) ) = (1 - p) * n - Q1,
I get  
        Q1  >= n * (1 - p) + log(err) + sqrt( log(err) * (log(err) - 2 * n * p) )     (1)

we could get q1 by solving (1)

Apply Berstein's lemma:

similar to applying Maurer's lemma, we could get a q2 so that we could accept item whose key
is smaller than q2, 0 <= q2 <= 1.

let  Z(j) = 1 if X(j) < q2, 
            = 0 if X(j) >= q2

{Z(j), j = 1 to n} are independent random variables.

E(Z(j)) = Pr(X(j) < q2) * 1 + Pr(X(j) >= q2) *  0
           = Pr(1 - pow( U, 1/w(j) ) < q2)
           = 1 - pow(1 - q2, w(j) )

E(Z(j) ^ 2) = E(Z(j))

Z(j) - E(Z(j)) <= 1 - E(Z(j)) = pow(1 - q2, w(j) ) <= 1 = M

theta(j) ^ 2 = E(Z(j) ^ 2) - E(Z(j)) ^ 2 <= E(Z(j) ^ 2) = 1 - pow(1 - q2, w(j) )

set Z = sum(Z(j),  j = 1 to n)

      Q2 = sum( pow(1 - q2, w(j) ) )

      E(Z) = sum(E(Z(j)), j = 1 to n) = n - sum( pow(1 - q2, w(j) ) ) = n - Q2

apply Berstein's lemma with t = sum( pow(1 - q2, w(j) ) ) - (1 - p) * n = Q2 -  (1 - p) *
n, I get  
    Q2  >= n * (1 - p) + 2 / 3 * log(err)  + 2 / 3 * sqrt( log(err) * (log(err) - 9 * n
* p / 2 ) )     (2)

we could get q2 by solving (2)

Questions:

(1) Please help comment on the above approach. Do they overall make sense? 

(2) I am stuck in getting q1 and q2 by solving (1) and (2) respectively. Would like to seek
some advice on it. 

> Probability weighted sampling without reservoir
> -----------------------------------------------
>
>                 Key: DATAFU-21
>                 URL: https://issues.apache.org/jira/browse/DATAFU-21
>             Project: DataFu
>          Issue Type: New Feature
>         Environment: Mac OS, Linux
>            Reporter: jian wang
>            Assignee: jian wang
>
> This issue is used to track investigation on finding a weighted sampler without using
internal reservoir. 
> At present, the SimpleRandomSample has implemented a good acceptance-rejection sampling
algo on probability random sampling. The weighted sampler could utilize the simple random
sample with slight modification.
> One slight modification is:  the present simple random sample generates a uniform random
number lies between (0, 1) as the random variable to accept or reject an item. The weighted
sample may generate this random variable based on the item's weight and this random number
still lies between (0, 1) and each item's random variable remain independent between each
other.
> Need further think and experiment the correctness of this solution and how to implement
it in an effective way.



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