Timeline for Sequence of p draws without replacement with biased probabilities
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 23, 2010 at 8:22 | vote | accept | GuillaumeThomas | ||
| Sep 23, 2010 at 8:22 | vote | accept | GuillaumeThomas | ||
| Sep 23, 2010 at 8:22 | |||||
| Sep 23, 2010 at 8:22 | vote | accept | GuillaumeThomas | ||
| Sep 23, 2010 at 8:22 | |||||
| Sep 23, 2010 at 6:03 | comment | added | Ori Gurel-Gurevich | Perhaps I completely misunderstood your comment. The way I see it is this: Denote the total weight of the remaining black marbles after $k$ draws by $X_k$. This is a random variable. The probability to now draw the white marble is $w/(w+X_k)$. Do you claim that the expected value of $X_k$ is a symmetric polynomial of the weights? I don't think that's true (except for $k=1$) since already the second draw involves division by $w+X_1$, so it seems like you get a some rational symmetric function. | |
| Sep 23, 2010 at 1:14 | comment | added | Bill Thurston | The mean weight after $n$ disjoint picks is a symmetric polynomial of degree $n+1$ in the weights: it's the sum of all products of distinct weights times the their total, times a constant (the total weight of all black marbles). There vector space ofof symmetric degree $ \le n$ polynomials has dimension $n$, so since moments are linearly independent polynomials, they form a basis. There's a dimension shift if normalize sum of weights = 1, to coincide with probabilistic moments. Cf. en.wikipedia.org/wiki/Elementary_symmetric_polynomial | |
| Sep 22, 2010 at 20:41 | comment | added | Ori Gurel-Gurevich | $B_k$ as you define it is a random variable. Its distribution is a symmetric function of the weights, but why does this mean it depends only on the first $k$ moments? | |
| Sep 22, 2010 at 19:44 | history | edited | Ori Gurel-Gurevich | CC BY-SA 2.5 |
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| Sep 22, 2010 at 19:33 | comment | added | Bill Thurston | (Re-entered after TeX error) Just a remark: the probability of drawing the white marble on the kth draw depends on the average weight of black marbles that are left (it's $w/(w + B_k)$, where $B_k$ is the mean weight of black marbles weight after $k$ draws of black marbles without replacement.) Since $B_k$ is a symmetric function of the weights of black marbles, the sequence of $B_k$ depends precisely on the first $k$ moments of the distribution of weights of black marbles, and probably has a nice formula that someone can supply. | |
| Sep 22, 2010 at 17:58 | history | answered | Ori Gurel-Gurevich | CC BY-SA 2.5 |