On the variance in the Bernoulli scheme
I'm trying to solve the following problem.
There are $755$ cards, each of which has a number from $1$ to $755$ (each number occurs exactly once). Of these cards, $20$ pieces are randomly selected and a random value equal to the sum of the numbers written on all $20$ cards is considered. I want to find the variance of this random value.
I think in this problem there is a Bernoulli scheme in which the space of elementary events is a set of numbers from $1$ to $755$. But are there ways to calculate the parameters of such a distribution (e.g. variance)?
probability probability-theory probability-distributions variance
asked Dec 2 '18 at 13:04
Victor
1766
add a comment |
I'm trying to solve the following problem.
There are $755$ cards, each of which has a number from $1$ to $755$ (each number occurs exactly once). Of these cards, $20$ pieces are randomly selected and a random value equal to the sum of the numbers written on all $20$ cards is considered. I want to find the variance of this random value.
I think in this problem there is a Bernoulli scheme in which the space of elementary events is a set of numbers from $1$ to $755$. But are there ways to calculate the parameters of such a distribution (e.g. variance)?
probability probability-theory probability-distributions variance
asked Dec 2 '18 at 13:04
Victor
1766
add a comment |
I'm trying to solve the following problem.
There are $755$ cards, each of which has a number from $1$ to $755$ (each number occurs exactly once). Of these cards, $20$ pieces are randomly selected and a random value equal to the sum of the numbers written on all $20$ cards is considered. I want to find the variance of this random value.
I think in this problem there is a Bernoulli scheme in which the space of elementary events is a set of numbers from $1$ to $755$. But are there ways to calculate the parameters of such a distribution (e.g. variance)?
probability probability-theory probability-distributions variance
asked Dec 2 '18 at 13:04
Victor
1766
I'm trying to solve the following problem.
There are $755$ cards, each of which has a number from $1$ to $755$ (each number occurs exactly once). Of these cards, $20$ pieces are randomly selected and a random value equal to the sum of the numbers written on all $20$ cards is considered. I want to find the variance of this random value.
I think in this problem there is a Bernoulli scheme in which the space of elementary events is a set of numbers from $1$ to $755$. But are there ways to calculate the parameters of such a distribution (e.g. variance)?
probability probability-theory probability-distributions variance
probability probability-theory probability-distributions variance
asked Dec 2 '18 at 13:04
Victor
1766
asked Dec 2 '18 at 13:04
Victor
1766
asked Dec 2 '18 at 13:04
Victor
1766
asked Dec 2 '18 at 13:04
Victor
1766
asked Dec 2 '18 at 13:04
Victor
1766
1766
add a comment |
add a comment |
1 Answer
1
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oldest
votes
With $n$ the number of cards and $k$ the number drawn we have the
OGF
$$[u^k] prod_{q=1}^n (1+ux^q).$$
Differentiate once for the expectation
$$[u^k] prod_{q=1}^n (1+ux^q)
sum_{q=1}^n frac{q u x^{q-1}}{1+ux^q}$$
and set $x=1$:
$$[u^k] prod_{q=1}^n (1+u)
sum_{q=1}^n frac{q u}{1+u}
= [u^k] (1+u)^{n-1} u frac{1}{2} n (n+1)
\ = frac{1}{2} n (n+1) {n-1choose k-1}.$$
This gives for the expectation
$$frac{1}{2} n (n+1) {n-1choose k-1} {nchoose k}^{-1}
= frac{1}{2} n (n+1) {n-1choose k-1}
frac{k}{n} {n-1choose k-1}^{-1}$$
or
$$bbox[5px,border:2px solid #00A000]{
mathrm{E}[X] = frac{1}{2} k (n+1).}$$
Differentiate again for the next factorial moment
$$[u^k] prod_{q=1}^n (1+ux^q)
left(sum_{q=1}^n frac{q u x^{q-1}}{1+ux^q}right)^2
\ + [u^k] prod_{q=1}^n (1+ux^q)
sum_{q=1}^n frac{q (q-1) u x^{q-2}}{1+ux^q}
- [u^k] prod_{q=1}^n (1+ux^q)
sum_{q=1}^n frac{q u x^{q-1}}{(1+ux^q)^2} q u x^{q-1}$$
and once more set $x=1$:
$$[u^k] (1+u)^n
left(sum_{q=1}^n frac{q u}{1+u}right)^2
\ + [u^k] (1+u)^n
sum_{q=1}^n frac{q (q-1) u}{1+u}
- [u^k] (1+u)^n
sum_{q=1}^n frac{q u}{(1+u)^2} q u
\ = {n-2choose k-2} frac{1}{4} n^2 (n+1)^2
\ + {n-1choose k-1} frac{1}{3} (n-1) n (n+1)
- {n-2choose k-2} frac{1}{6} n (n+1) (2n+1).$$
This gives for the second factorial moment
$$ frac{k(k-1)}{n(n-1)} frac{1}{4} n^2 (n+1)^2
\ + frac{k}{n} frac{1}{3} (n-1) n (n+1)
- frac{k(k-1)}{n(n-1)} frac{1}{6} n (n+1) (2n+1)
\ = frac{k(k-1)}{n-1} frac{1}{4} n (n+1)^2
+ k frac{1}{3} (n-1) (n+1)
- frac{k(k-1)}{n-1} frac{1}{6} (n+1) (2n+1).$$
or
$$bbox[5px,border:2px solid #00A000]{
mathrm{E}[X(X-1)] = frac{1}{12} k (n+1)
(3 k n + 2 k + n - 6).}$$
Finally recall that
$$mathrm{Var}[X] =
mathrm{E}[X(X-1)] + mathrm{E}[X] - mathrm{E}[X]^2$$
so that
$$bbox[5px,border:2px solid #00A000]{
mathrm{Var}[X] = frac{1}{12} k (n+1) (n-k).}$$
answered Dec 3 '18 at 15:34
Marko Riedel
39.2k339107
add a comment |
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With $n$ the number of cards and $k$ the number drawn we have the
OGF
$$[u^k] prod_{q=1}^n (1+ux^q).$$
Differentiate once for the expectation
$$[u^k] prod_{q=1}^n (1+ux^q)
sum_{q=1}^n frac{q u x^{q-1}}{1+ux^q}$$
and set $x=1$:
$$[u^k] prod_{q=1}^n (1+u)
sum_{q=1}^n frac{q u}{1+u}
= [u^k] (1+u)^{n-1} u frac{1}{2} n (n+1)
\ = frac{1}{2} n (n+1) {n-1choose k-1}.$$
This gives for the expectation
$$frac{1}{2} n (n+1) {n-1choose k-1} {nchoose k}^{-1}
= frac{1}{2} n (n+1) {n-1choose k-1}
frac{k}{n} {n-1choose k-1}^{-1}$$
or
$$bbox[5px,border:2px solid #00A000]{
mathrm{E}[X] = frac{1}{2} k (n+1).}$$
Differentiate again for the next factorial moment
$$[u^k] prod_{q=1}^n (1+ux^q)
left(sum_{q=1}^n frac{q u x^{q-1}}{1+ux^q}right)^2
\ + [u^k] prod_{q=1}^n (1+ux^q)
sum_{q=1}^n frac{q (q-1) u x^{q-2}}{1+ux^q}
- [u^k] prod_{q=1}^n (1+ux^q)
sum_{q=1}^n frac{q u x^{q-1}}{(1+ux^q)^2} q u x^{q-1}$$
and once more set $x=1$:
$$[u^k] (1+u)^n
left(sum_{q=1}^n frac{q u}{1+u}right)^2
\ + [u^k] (1+u)^n
sum_{q=1}^n frac{q (q-1) u}{1+u}
- [u^k] (1+u)^n
sum_{q=1}^n frac{q u}{(1+u)^2} q u
\ = {n-2choose k-2} frac{1}{4} n^2 (n+1)^2
\ + {n-1choose k-1} frac{1}{3} (n-1) n (n+1)
- {n-2choose k-2} frac{1}{6} n (n+1) (2n+1).$$
This gives for the second factorial moment
$$ frac{k(k-1)}{n(n-1)} frac{1}{4} n^2 (n+1)^2
\ + frac{k}{n} frac{1}{3} (n-1) n (n+1)
- frac{k(k-1)}{n(n-1)} frac{1}{6} n (n+1) (2n+1)
\ = frac{k(k-1)}{n-1} frac{1}{4} n (n+1)^2
+ k frac{1}{3} (n-1) (n+1)
- frac{k(k-1)}{n-1} frac{1}{6} (n+1) (2n+1).$$
or
$$bbox[5px,border:2px solid #00A000]{
mathrm{E}[X(X-1)] = frac{1}{12} k (n+1)
(3 k n + 2 k + n - 6).}$$
Finally recall that
$$mathrm{Var}[X] =
mathrm{E}[X(X-1)] + mathrm{E}[X] - mathrm{E}[X]^2$$
so that
$$bbox[5px,border:2px solid #00A000]{
mathrm{Var}[X] = frac{1}{12} k (n+1) (n-k).}$$
answered Dec 3 '18 at 15:34
Marko Riedel
39.2k339107
add a comment |
With $n$ the number of cards and $k$ the number drawn we have the
OGF
$$[u^k] prod_{q=1}^n (1+ux^q).$$
Differentiate once for the expectation
$$[u^k] prod_{q=1}^n (1+ux^q)
sum_{q=1}^n frac{q u x^{q-1}}{1+ux^q}$$
and set $x=1$:
$$[u^k] prod_{q=1}^n (1+u)
sum_{q=1}^n frac{q u}{1+u}
= [u^k] (1+u)^{n-1} u frac{1}{2} n (n+1)
\ = frac{1}{2} n (n+1) {n-1choose k-1}.$$
This gives for the expectation
$$frac{1}{2} n (n+1) {n-1choose k-1} {nchoose k}^{-1}
= frac{1}{2} n (n+1) {n-1choose k-1}
frac{k}{n} {n-1choose k-1}^{-1}$$
or
$$bbox[5px,border:2px solid #00A000]{
mathrm{E}[X] = frac{1}{2} k (n+1).}$$
Differentiate again for the next factorial moment
$$[u^k] prod_{q=1}^n (1+ux^q)
left(sum_{q=1}^n frac{q u x^{q-1}}{1+ux^q}right)^2
\ + [u^k] prod_{q=1}^n (1+ux^q)
sum_{q=1}^n frac{q (q-1) u x^{q-2}}{1+ux^q}
- [u^k] prod_{q=1}^n (1+ux^q)
sum_{q=1}^n frac{q u x^{q-1}}{(1+ux^q)^2} q u x^{q-1}$$
and once more set $x=1$:
$$[u^k] (1+u)^n
left(sum_{q=1}^n frac{q u}{1+u}right)^2
\ + [u^k] (1+u)^n
sum_{q=1}^n frac{q (q-1) u}{1+u}
- [u^k] (1+u)^n
sum_{q=1}^n frac{q u}{(1+u)^2} q u
\ = {n-2choose k-2} frac{1}{4} n^2 (n+1)^2
\ + {n-1choose k-1} frac{1}{3} (n-1) n (n+1)
- {n-2choose k-2} frac{1}{6} n (n+1) (2n+1).$$
This gives for the second factorial moment
$$ frac{k(k-1)}{n(n-1)} frac{1}{4} n^2 (n+1)^2
\ + frac{k}{n} frac{1}{3} (n-1) n (n+1)
- frac{k(k-1)}{n(n-1)} frac{1}{6} n (n+1) (2n+1)
\ = frac{k(k-1)}{n-1} frac{1}{4} n (n+1)^2
+ k frac{1}{3} (n-1) (n+1)
- frac{k(k-1)}{n-1} frac{1}{6} (n+1) (2n+1).$$
or
$$bbox[5px,border:2px solid #00A000]{
mathrm{E}[X(X-1)] = frac{1}{12} k (n+1)
(3 k n + 2 k + n - 6).}$$
Finally recall that
$$mathrm{Var}[X] =
mathrm{E}[X(X-1)] + mathrm{E}[X] - mathrm{E}[X]^2$$
so that
$$bbox[5px,border:2px solid #00A000]{
mathrm{Var}[X] = frac{1}{12} k (n+1) (n-k).}$$
answered Dec 3 '18 at 15:34
Marko Riedel
39.2k339107
add a comment |
With $n$ the number of cards and $k$ the number drawn we have the
OGF
$$[u^k] prod_{q=1}^n (1+ux^q).$$
Differentiate once for the expectation
$$[u^k] prod_{q=1}^n (1+ux^q)
sum_{q=1}^n frac{q u x^{q-1}}{1+ux^q}$$
and set $x=1$:
$$[u^k] prod_{q=1}^n (1+u)
sum_{q=1}^n frac{q u}{1+u}
= [u^k] (1+u)^{n-1} u frac{1}{2} n (n+1)
\ = frac{1}{2} n (n+1) {n-1choose k-1}.$$
This gives for the expectation
$$frac{1}{2} n (n+1) {n-1choose k-1} {nchoose k}^{-1}
= frac{1}{2} n (n+1) {n-1choose k-1}
frac{k}{n} {n-1choose k-1}^{-1}$$
or
$$bbox[5px,border:2px solid #00A000]{
mathrm{E}[X] = frac{1}{2} k (n+1).}$$
Differentiate again for the next factorial moment
$$[u^k] prod_{q=1}^n (1+ux^q)
left(sum_{q=1}^n frac{q u x^{q-1}}{1+ux^q}right)^2
\ + [u^k] prod_{q=1}^n (1+ux^q)
sum_{q=1}^n frac{q (q-1) u x^{q-2}}{1+ux^q}
- [u^k] prod_{q=1}^n (1+ux^q)
sum_{q=1}^n frac{q u x^{q-1}}{(1+ux^q)^2} q u x^{q-1}$$
and once more set $x=1$:
$$[u^k] (1+u)^n
left(sum_{q=1}^n frac{q u}{1+u}right)^2
\ + [u^k] (1+u)^n
sum_{q=1}^n frac{q (q-1) u}{1+u}
- [u^k] (1+u)^n
sum_{q=1}^n frac{q u}{(1+u)^2} q u
\ = {n-2choose k-2} frac{1}{4} n^2 (n+1)^2
\ + {n-1choose k-1} frac{1}{3} (n-1) n (n+1)
- {n-2choose k-2} frac{1}{6} n (n+1) (2n+1).$$
This gives for the second factorial moment
$$ frac{k(k-1)}{n(n-1)} frac{1}{4} n^2 (n+1)^2
\ + frac{k}{n} frac{1}{3} (n-1) n (n+1)
- frac{k(k-1)}{n(n-1)} frac{1}{6} n (n+1) (2n+1)
\ = frac{k(k-1)}{n-1} frac{1}{4} n (n+1)^2
+ k frac{1}{3} (n-1) (n+1)
- frac{k(k-1)}{n-1} frac{1}{6} (n+1) (2n+1).$$
or
$$bbox[5px,border:2px solid #00A000]{
mathrm{E}[X(X-1)] = frac{1}{12} k (n+1)
(3 k n + 2 k + n - 6).}$$
Finally recall that
$$mathrm{Var}[X] =
mathrm{E}[X(X-1)] + mathrm{E}[X] - mathrm{E}[X]^2$$
so that
$$bbox[5px,border:2px solid #00A000]{
mathrm{Var}[X] = frac{1}{12} k (n+1) (n-k).}$$
answered Dec 3 '18 at 15:34
Marko Riedel
39.2k339107
With $n$ the number of cards and $k$ the number drawn we have the
OGF
$$[u^k] prod_{q=1}^n (1+ux^q).$$
Differentiate once for the expectation
$$[u^k] prod_{q=1}^n (1+ux^q)
sum_{q=1}^n frac{q u x^{q-1}}{1+ux^q}$$
and set $x=1$:
$$[u^k] prod_{q=1}^n (1+u)
sum_{q=1}^n frac{q u}{1+u}
= [u^k] (1+u)^{n-1} u frac{1}{2} n (n+1)
\ = frac{1}{2} n (n+1) {n-1choose k-1}.$$
This gives for the expectation
$$frac{1}{2} n (n+1) {n-1choose k-1} {nchoose k}^{-1}
= frac{1}{2} n (n+1) {n-1choose k-1}
frac{k}{n} {n-1choose k-1}^{-1}$$
or
$$bbox[5px,border:2px solid #00A000]{
mathrm{E}[X] = frac{1}{2} k (n+1).}$$
Differentiate again for the next factorial moment
$$[u^k] prod_{q=1}^n (1+ux^q)
left(sum_{q=1}^n frac{q u x^{q-1}}{1+ux^q}right)^2
\ + [u^k] prod_{q=1}^n (1+ux^q)
sum_{q=1}^n frac{q (q-1) u x^{q-2}}{1+ux^q}
- [u^k] prod_{q=1}^n (1+ux^q)
sum_{q=1}^n frac{q u x^{q-1}}{(1+ux^q)^2} q u x^{q-1}$$
and once more set $x=1$:
$$[u^k] (1+u)^n
left(sum_{q=1}^n frac{q u}{1+u}right)^2
\ + [u^k] (1+u)^n
sum_{q=1}^n frac{q (q-1) u}{1+u}
- [u^k] (1+u)^n
sum_{q=1}^n frac{q u}{(1+u)^2} q u
\ = {n-2choose k-2} frac{1}{4} n^2 (n+1)^2
\ + {n-1choose k-1} frac{1}{3} (n-1) n (n+1)
- {n-2choose k-2} frac{1}{6} n (n+1) (2n+1).$$
This gives for the second factorial moment
$$ frac{k(k-1)}{n(n-1)} frac{1}{4} n^2 (n+1)^2
\ + frac{k}{n} frac{1}{3} (n-1) n (n+1)
- frac{k(k-1)}{n(n-1)} frac{1}{6} n (n+1) (2n+1)
\ = frac{k(k-1)}{n-1} frac{1}{4} n (n+1)^2
+ k frac{1}{3} (n-1) (n+1)
- frac{k(k-1)}{n-1} frac{1}{6} (n+1) (2n+1).$$
or
$$bbox[5px,border:2px solid #00A000]{
mathrm{E}[X(X-1)] = frac{1}{12} k (n+1)
(3 k n + 2 k + n - 6).}$$
Finally recall that
$$mathrm{Var}[X] =
mathrm{E}[X(X-1)] + mathrm{E}[X] - mathrm{E}[X]^2$$
so that
$$bbox[5px,border:2px solid #00A000]{
mathrm{Var}[X] = frac{1}{12} k (n+1) (n-k).}$$
answered Dec 3 '18 at 15:34
Marko Riedel
39.2k339107
edited Dec 3 '18 at 15:42
answered Dec 3 '18 at 15:34
Marko Riedel
39.2k339107
answered Dec 3 '18 at 15:34
Marko Riedel
39.2k339107
answered Dec 3 '18 at 15:34
Marko Riedel
39.2k339107
39.2k339107
add a comment |
add a comment |
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