w**a
发帖数: 1024 | 1 a population of 3000 unique numbers, 1, 2, 3, ..., 3000.
randomly draw 3000 times from the above population with replacement (i.e.,
some numbers can be drawn more than once). Now we have 3000 new numbers,
some of which could be identical. The number of distinct elements in the
3000
new numbers is denoted as Y.
what is the expectation value for Y.
computer modeling shows E(Y) ~ 1960. Why is that? any theoretical
explanation?
thanks. | m******e
发帖数: 212 | 2 I read a similar problem somewhere else, and there's a solution like this --
Define the random variables X_i: X_i=1 if the i-th number is picked; X_i=0
if
the i-th number is never touched. The sum S = X_1 + ...+X_n (n=3000) is the
number of distinct element and we want to know its expectation value.
Note that X_i's are iid. The probability that the i-th number is not drawn
for m (m=3000) times is (1-1/n)^m; the probablity that it is picked is
1 - (1-1/n)^m --- this is also E(X_i).
Therefore E(S)
【在 w**a 的大作中提到】
: a population of 3000 unique numbers, 1, 2, 3, ..., 3000.
: randomly draw 3000 times from the above population with replacement (i.e.,
: some numbers can be drawn more than once). Now we have 3000 new numbers,
: some of which could be identical. The number of distinct elements in the
: 3000
: new numbers is denoted as Y.
: what is the expectation value for Y.
: computer modeling shows E(Y) ~ 1960. Why is that? any theoretical
: explanation?
: thanks.
| m******e
发帖数: 212 | 3 i think there's a caveat in this solution though: The avarge number E(S)
includes the case that S=0, i.e., none of the numbers is picked. For the
original problem, the expectation of S should be taken under the condition
that S!=0.
More discussions can be found in this post
http://www.wilmott.com/messageview.cfm?catid=26&threadid=42145
--
0
the
drawn
1896.
【在 m******e 的大作中提到】
: I read a similar problem somewhere else, and there's a solution like this --
: Define the random variables X_i: X_i=1 if the i-th number is picked; X_i=0
: if
: the i-th number is never touched. The sum S = X_1 + ...+X_n (n=3000) is the
: number of distinct element and we want to know its expectation value.
: Note that X_i's are iid. The probability that the i-th number is not drawn
: for m (m=3000) times is (1-1/n)^m; the probablity that it is picked is
: 1 - (1-1/n)^m --- this is also E(X_i).
: Therefore E(S)
| w**a
发帖数: 1024 | 4 in my case, S >= 1.
【在 m******e 的大作中提到】
: i think there's a caveat in this solution though: The avarge number E(S)
: includes the case that S=0, i.e., none of the numbers is picked. For the
: original problem, the expectation of S should be taken under the condition
: that S!=0.
: More discussions can be found in this post
: http://www.wilmott.com/messageview.cfm?catid=26&threadid=42145
:
: --
: 0
: the
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