Ask for intuitive explanation of an expectation
$begingroup$
If X~HGeom(w,b,n), and Y = $Xchoose2$, then:
$$E(Y) = frac{{w choose 2}{n choose 2}}{{w+b} choose 2}$$
This post:
https://math.stackexchange.com/a/1669384/152816,
has a very detailed mathematical proof.
But since the result is very symmetric, it seems there could be an intuitive explanation of E(Y), i.e. we could story proof this E(Y) formula without complicated mathematical proof, could anyone think of an intuitive explanation of E(Y)?
Thanks.
probability expected-value
asked Dec 4 '18 at 2:57
wangshuaijiewangshuaijie
1788
$endgroup$
add a comment |
$begingroup$
If X~HGeom(w,b,n), and Y = $Xchoose2$, then:
$$E(Y) = frac{{w choose 2}{n choose 2}}{{w+b} choose 2}$$
This post:
https://math.stackexchange.com/a/1669384/152816,
has a very detailed mathematical proof.
But since the result is very symmetric, it seems there could be an intuitive explanation of E(Y), i.e. we could story proof this E(Y) formula without complicated mathematical proof, could anyone think of an intuitive explanation of E(Y)?
Thanks.
probability expected-value
asked Dec 4 '18 at 2:57
wangshuaijiewangshuaijie
1788
$endgroup$
add a comment |
$begingroup$
If X~HGeom(w,b,n), and Y = $Xchoose2$, then:
$$E(Y) = frac{{w choose 2}{n choose 2}}{{w+b} choose 2}$$
This post:
https://math.stackexchange.com/a/1669384/152816,
has a very detailed mathematical proof.
But since the result is very symmetric, it seems there could be an intuitive explanation of E(Y), i.e. we could story proof this E(Y) formula without complicated mathematical proof, could anyone think of an intuitive explanation of E(Y)?
Thanks.
probability expected-value
asked Dec 4 '18 at 2:57
wangshuaijiewangshuaijie
1788
$endgroup$
If X~HGeom(w,b,n), and Y = $Xchoose2$, then:
$$E(Y) = frac{{w choose 2}{n choose 2}}{{w+b} choose 2}$$
This post:
https://math.stackexchange.com/a/1669384/152816,
has a very detailed mathematical proof.
But since the result is very symmetric, it seems there could be an intuitive explanation of E(Y), i.e. we could story proof this E(Y) formula without complicated mathematical proof, could anyone think of an intuitive explanation of E(Y)?
Thanks.
probability expected-value
probability expected-value
asked Dec 4 '18 at 2:57
wangshuaijiewangshuaijie
1788
asked Dec 4 '18 at 2:57
wangshuaijiewangshuaijie
1788
asked Dec 4 '18 at 2:57
wangshuaijiewangshuaijie
1788
asked Dec 4 '18 at 2:57
wangshuaijiewangshuaijie
1788
asked Dec 4 '18 at 2:57
wangshuaijiewangshuaijie
1788
1788
add a comment |
add a comment |
1 Answer
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$begingroup$
$binom{X}2$ is equal to the number of pairs of white balls which are chosen. For each particular pair of white balls, the probability they are chosen is equal to
$$
frac{binom{b+w-2}{n-2}}{binom{b+w}{n}}=frac{n}{b+w}cdotfrac{n-1}{b+w-1}:=p
$$
Since there are $binom{w}2$ pairs of white balls, and each is chosen about $p$ of the time, you would expect there to be $pcdot binom{w}2 $ pairs of white balls chosen. This implies the expected value of $binom{X}2$ is
$$
frac{n}{b+w}cdotfrac{n-1}{b+w-1}cdotbinom{w}2=frac{binom{n}2binom{w}2}{binom{b+w}2}=EBig[binom{X}2Big].
$$
answered Dec 4 '18 at 6:42
Mike EarnestMike Earnest
23.7k12051
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
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active
oldest
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active
oldest
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$begingroup$
$binom{X}2$ is equal to the number of pairs of white balls which are chosen. For each particular pair of white balls, the probability they are chosen is equal to
$$
frac{binom{b+w-2}{n-2}}{binom{b+w}{n}}=frac{n}{b+w}cdotfrac{n-1}{b+w-1}:=p
$$
Since there are $binom{w}2$ pairs of white balls, and each is chosen about $p$ of the time, you would expect there to be $pcdot binom{w}2 $ pairs of white balls chosen. This implies the expected value of $binom{X}2$ is
$$
frac{n}{b+w}cdotfrac{n-1}{b+w-1}cdotbinom{w}2=frac{binom{n}2binom{w}2}{binom{b+w}2}=EBig[binom{X}2Big].
$$
answered Dec 4 '18 at 6:42
Mike EarnestMike Earnest
23.7k12051
$endgroup$
add a comment |
$begingroup$
$binom{X}2$ is equal to the number of pairs of white balls which are chosen. For each particular pair of white balls, the probability they are chosen is equal to
$$
frac{binom{b+w-2}{n-2}}{binom{b+w}{n}}=frac{n}{b+w}cdotfrac{n-1}{b+w-1}:=p
$$
Since there are $binom{w}2$ pairs of white balls, and each is chosen about $p$ of the time, you would expect there to be $pcdot binom{w}2 $ pairs of white balls chosen. This implies the expected value of $binom{X}2$ is
$$
frac{n}{b+w}cdotfrac{n-1}{b+w-1}cdotbinom{w}2=frac{binom{n}2binom{w}2}{binom{b+w}2}=EBig[binom{X}2Big].
$$
answered Dec 4 '18 at 6:42
Mike EarnestMike Earnest
23.7k12051
$endgroup$
add a comment |
$begingroup$
$binom{X}2$ is equal to the number of pairs of white balls which are chosen. For each particular pair of white balls, the probability they are chosen is equal to
$$
frac{binom{b+w-2}{n-2}}{binom{b+w}{n}}=frac{n}{b+w}cdotfrac{n-1}{b+w-1}:=p
$$
Since there are $binom{w}2$ pairs of white balls, and each is chosen about $p$ of the time, you would expect there to be $pcdot binom{w}2 $ pairs of white balls chosen. This implies the expected value of $binom{X}2$ is
$$
frac{n}{b+w}cdotfrac{n-1}{b+w-1}cdotbinom{w}2=frac{binom{n}2binom{w}2}{binom{b+w}2}=EBig[binom{X}2Big].
$$
answered Dec 4 '18 at 6:42
Mike EarnestMike Earnest
23.7k12051
$endgroup$
$binom{X}2$ is equal to the number of pairs of white balls which are chosen. For each particular pair of white balls, the probability they are chosen is equal to
$$
frac{binom{b+w-2}{n-2}}{binom{b+w}{n}}=frac{n}{b+w}cdotfrac{n-1}{b+w-1}:=p
$$
Since there are $binom{w}2$ pairs of white balls, and each is chosen about $p$ of the time, you would expect there to be $pcdot binom{w}2 $ pairs of white balls chosen. This implies the expected value of $binom{X}2$ is
$$
frac{n}{b+w}cdotfrac{n-1}{b+w-1}cdotbinom{w}2=frac{binom{n}2binom{w}2}{binom{b+w}2}=EBig[binom{X}2Big].
$$
answered Dec 4 '18 at 6:42
Mike EarnestMike Earnest
23.7k12051
answered Dec 4 '18 at 6:42
Mike EarnestMike Earnest
23.7k12051
answered Dec 4 '18 at 6:42
Mike EarnestMike Earnest
23.7k12051
answered Dec 4 '18 at 6:42
Mike EarnestMike Earnest
23.7k12051
23.7k12051
add a comment |
add a comment |
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