Mixing
Find expected value of a random rectangle in a square grid
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I've tried to solve the following problem. Could you please tell if I did it right.
Problem. Given a n x n square grid. We take a random rectangle so that every ractangle is equally likely.
Find expected value of area of this rectangle.
Solution.
The number of all rectangles is $sumlimits_{i=1}^{n}{i^2} = frac{n(n+1)(2n+1)}{6}$. Let $f((i,j))$ - number of rectangles which consist of square in i-th row and j-th column. $f((i,j)) = i(n-i+1)j(n-j+1)$. Let $X_{i,j} = 1$ if random rectangle consists of square in i-th row and j-th column and $X_{i,j} = 0$ otherwise. Expected value of area of a random rectangle is $E(sumlimits_{i=1}^{n}{sumlimits_{j=1}^{n}{X_{i,j}}}) = sumlimits_{i=1}^{n}{sumlimits_{j=1}^{n}{frac{f((i,j))}{frac{n(n+1)(2n+1)}{6}}}}$
$= sumlimits_{i=1}^{n}{frac{1}{frac{n(n+1)(2n+1)}{6}}sumlimits_{j=1}^{n}{i(n-i+1)j(n-j+1)}} = sumlimits_{i=1}^{n}{i(n-i+1)frac{6}{n(n+1)(2n+1)}sumlimits_{j=1}^{n}{j(n-j+1)}}$
$=sumlimits_{i=1}^{n}{i(n-i+1)frac{6}{n(n+1)(2n+1)}((n+1)sumlimits_{j=1}^{n}{j} - sumlimits_{j=1}^{n}{j^2})} = sumlimits_{i=1}^{n}{i(n-i+1)frac{6}{n(n+1)(2n+1)}(frac{(n+1)(1+n)}{2} - frac{n(n+1)(2n+1)}{6})}$
$=frac{6}{n(n+1)(2n+1)}(frac{(n+1)(1+n)}{2} - frac{n(n+1)(2n+1)}{6})sumlimits_{i=1}^{n}{i(n-i+1)} = frac{6}{n(n+1)(2n+1)}(frac{(n+1)(1+n)}{2} - frac{n(n+1)(2n+1)}{6})^2$
probability expected-value
asked Feb 1 at 23:46
Novak DjokovicNovak Djokovic
45038
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add a comment |
$begingroup$
I've tried to solve the following problem. Could you please tell if I did it right.
Problem. Given a n x n square grid. We take a random rectangle so that every ractangle is equally likely.
Find expected value of area of this rectangle.
Solution.
The number of all rectangles is $sumlimits_{i=1}^{n}{i^2} = frac{n(n+1)(2n+1)}{6}$. Let $f((i,j))$ - number of rectangles which consist of square in i-th row and j-th column. $f((i,j)) = i(n-i+1)j(n-j+1)$. Let $X_{i,j} = 1$ if random rectangle consists of square in i-th row and j-th column and $X_{i,j} = 0$ otherwise. Expected value of area of a random rectangle is $E(sumlimits_{i=1}^{n}{sumlimits_{j=1}^{n}{X_{i,j}}}) = sumlimits_{i=1}^{n}{sumlimits_{j=1}^{n}{frac{f((i,j))}{frac{n(n+1)(2n+1)}{6}}}}$
$= sumlimits_{i=1}^{n}{frac{1}{frac{n(n+1)(2n+1)}{6}}sumlimits_{j=1}^{n}{i(n-i+1)j(n-j+1)}} = sumlimits_{i=1}^{n}{i(n-i+1)frac{6}{n(n+1)(2n+1)}sumlimits_{j=1}^{n}{j(n-j+1)}}$
$=sumlimits_{i=1}^{n}{i(n-i+1)frac{6}{n(n+1)(2n+1)}((n+1)sumlimits_{j=1}^{n}{j} - sumlimits_{j=1}^{n}{j^2})} = sumlimits_{i=1}^{n}{i(n-i+1)frac{6}{n(n+1)(2n+1)}(frac{(n+1)(1+n)}{2} - frac{n(n+1)(2n+1)}{6})}$
$=frac{6}{n(n+1)(2n+1)}(frac{(n+1)(1+n)}{2} - frac{n(n+1)(2n+1)}{6})sumlimits_{i=1}^{n}{i(n-i+1)} = frac{6}{n(n+1)(2n+1)}(frac{(n+1)(1+n)}{2} - frac{n(n+1)(2n+1)}{6})^2$
probability expected-value
asked Feb 1 at 23:46
Novak DjokovicNovak Djokovic
45038
$endgroup$
$begingroup$
If $n=2,$ your formula gives $5$ rectangles, but it seems to me there are $9$. Four $1times1,$ four $2times1,$ and one $2times2$ What am I missing?
$endgroup$
– saulspatz
Feb 2 at 0:02
$begingroup$
@saulspatz That indeed is not the correct formula for the number of all rectangles.
$endgroup$
– Math1000
Feb 2 at 1:40
add a comment |
$begingroup$
I've tried to solve the following problem. Could you please tell if I did it right.
Problem. Given a n x n square grid. We take a random rectangle so that every ractangle is equally likely.
Find expected value of area of this rectangle.
Solution.
The number of all rectangles is $sumlimits_{i=1}^{n}{i^2} = frac{n(n+1)(2n+1)}{6}$. Let $f((i,j))$ - number of rectangles which consist of square in i-th row and j-th column. $f((i,j)) = i(n-i+1)j(n-j+1)$. Let $X_{i,j} = 1$ if random rectangle consists of square in i-th row and j-th column and $X_{i,j} = 0$ otherwise. Expected value of area of a random rectangle is $E(sumlimits_{i=1}^{n}{sumlimits_{j=1}^{n}{X_{i,j}}}) = sumlimits_{i=1}^{n}{sumlimits_{j=1}^{n}{frac{f((i,j))}{frac{n(n+1)(2n+1)}{6}}}}$
$= sumlimits_{i=1}^{n}{frac{1}{frac{n(n+1)(2n+1)}{6}}sumlimits_{j=1}^{n}{i(n-i+1)j(n-j+1)}} = sumlimits_{i=1}^{n}{i(n-i+1)frac{6}{n(n+1)(2n+1)}sumlimits_{j=1}^{n}{j(n-j+1)}}$
$=sumlimits_{i=1}^{n}{i(n-i+1)frac{6}{n(n+1)(2n+1)}((n+1)sumlimits_{j=1}^{n}{j} - sumlimits_{j=1}^{n}{j^2})} = sumlimits_{i=1}^{n}{i(n-i+1)frac{6}{n(n+1)(2n+1)}(frac{(n+1)(1+n)}{2} - frac{n(n+1)(2n+1)}{6})}$
$=frac{6}{n(n+1)(2n+1)}(frac{(n+1)(1+n)}{2} - frac{n(n+1)(2n+1)}{6})sumlimits_{i=1}^{n}{i(n-i+1)} = frac{6}{n(n+1)(2n+1)}(frac{(n+1)(1+n)}{2} - frac{n(n+1)(2n+1)}{6})^2$
probability expected-value
asked Feb 1 at 23:46
Novak DjokovicNovak Djokovic
45038
$endgroup$
I've tried to solve the following problem. Could you please tell if I did it right.
Problem. Given a n x n square grid. We take a random rectangle so that every ractangle is equally likely.
Find expected value of area of this rectangle.
Solution.
The number of all rectangles is $sumlimits_{i=1}^{n}{i^2} = frac{n(n+1)(2n+1)}{6}$. Let $f((i,j))$ - number of rectangles which consist of square in i-th row and j-th column. $f((i,j)) = i(n-i+1)j(n-j+1)$. Let $X_{i,j} = 1$ if random rectangle consists of square in i-th row and j-th column and $X_{i,j} = 0$ otherwise. Expected value of area of a random rectangle is $E(sumlimits_{i=1}^{n}{sumlimits_{j=1}^{n}{X_{i,j}}}) = sumlimits_{i=1}^{n}{sumlimits_{j=1}^{n}{frac{f((i,j))}{frac{n(n+1)(2n+1)}{6}}}}$
$= sumlimits_{i=1}^{n}{frac{1}{frac{n(n+1)(2n+1)}{6}}sumlimits_{j=1}^{n}{i(n-i+1)j(n-j+1)}} = sumlimits_{i=1}^{n}{i(n-i+1)frac{6}{n(n+1)(2n+1)}sumlimits_{j=1}^{n}{j(n-j+1)}}$
$=sumlimits_{i=1}^{n}{i(n-i+1)frac{6}{n(n+1)(2n+1)}((n+1)sumlimits_{j=1}^{n}{j} - sumlimits_{j=1}^{n}{j^2})} = sumlimits_{i=1}^{n}{i(n-i+1)frac{6}{n(n+1)(2n+1)}(frac{(n+1)(1+n)}{2} - frac{n(n+1)(2n+1)}{6})}$
$=frac{6}{n(n+1)(2n+1)}(frac{(n+1)(1+n)}{2} - frac{n(n+1)(2n+1)}{6})sumlimits_{i=1}^{n}{i(n-i+1)} = frac{6}{n(n+1)(2n+1)}(frac{(n+1)(1+n)}{2} - frac{n(n+1)(2n+1)}{6})^2$
probability expected-value
probability expected-value
asked Feb 1 at 23:46
Novak DjokovicNovak Djokovic
45038
asked Feb 1 at 23:46
Novak DjokovicNovak Djokovic
45038
asked Feb 1 at 23:46
Novak DjokovicNovak Djokovic
45038
asked Feb 1 at 23:46
Novak DjokovicNovak Djokovic
45038
asked Feb 1 at 23:46
Novak DjokovicNovak Djokovic
45038
45038
$begingroup$
If $n=2,$ your formula gives $5$ rectangles, but it seems to me there are $9$. Four $1times1,$ four $2times1,$ and one $2times2$ What am I missing?
$endgroup$
– saulspatz
Feb 2 at 0:02
$begingroup$
@saulspatz That indeed is not the correct formula for the number of all rectangles.
$endgroup$
– Math1000
Feb 2 at 1:40
add a comment |
$begingroup$
If $n=2,$ your formula gives $5$ rectangles, but it seems to me there are $9$. Four $1times1,$ four $2times1,$ and one $2times2$ What am I missing?
$endgroup$
– saulspatz
Feb 2 at 0:02
$begingroup$
@saulspatz That indeed is not the correct formula for the number of all rectangles.
$endgroup$
– Math1000
Feb 2 at 1:40
$begingroup$
If $n=2,$ your formula gives $5$ rectangles, but it seems to me there are $9$. Four $1times1,$ four $2times1,$ and one $2times2$ What am I missing?
$endgroup$
– saulspatz
Feb 2 at 0:02
$begingroup$
If $n=2,$ your formula gives $5$ rectangles, but it seems to me there are $9$. Four $1times1,$ four $2times1,$ and one $2times2$ What am I missing?
$endgroup$
– saulspatz
Feb 2 at 0:02
$begingroup$
@saulspatz That indeed is not the correct formula for the number of all rectangles.
$endgroup$
– Math1000
Feb 2 at 1:40
$begingroup$
@saulspatz That indeed is not the correct formula for the number of all rectangles.
$endgroup$
– Math1000
Feb 2 at 1:40
add a comment |
1 Answer
1
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oldest
votes
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The expected area is the expected base times the expected height. If, for example, $n=10$, there are $10$ bases of length $1$, and $9$ of length $2$, on up to $1$ of length $n$: $sum_{k=1}^n k$ in all. Thus the expected base length is $$b = frac{sum_{k=1}^n k ( n+1-k)}{sum_{k=1}^n k}$$ and the expected height length is the same. So the expected area is $b^2$.
answered Feb 2 at 2:59
kimchi loverkimchi lover
11.8k31229
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add a comment |
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1 Answer
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$begingroup$
The expected area is the expected base times the expected height. If, for example, $n=10$, there are $10$ bases of length $1$, and $9$ of length $2$, on up to $1$ of length $n$: $sum_{k=1}^n k$ in all. Thus the expected base length is $$b = frac{sum_{k=1}^n k ( n+1-k)}{sum_{k=1}^n k}$$ and the expected height length is the same. So the expected area is $b^2$.
answered Feb 2 at 2:59
kimchi loverkimchi lover
11.8k31229
$endgroup$
add a comment |
$begingroup$
The expected area is the expected base times the expected height. If, for example, $n=10$, there are $10$ bases of length $1$, and $9$ of length $2$, on up to $1$ of length $n$: $sum_{k=1}^n k$ in all. Thus the expected base length is $$b = frac{sum_{k=1}^n k ( n+1-k)}{sum_{k=1}^n k}$$ and the expected height length is the same. So the expected area is $b^2$.
answered Feb 2 at 2:59
kimchi loverkimchi lover
11.8k31229
$endgroup$
add a comment |
$begingroup$
The expected area is the expected base times the expected height. If, for example, $n=10$, there are $10$ bases of length $1$, and $9$ of length $2$, on up to $1$ of length $n$: $sum_{k=1}^n k$ in all. Thus the expected base length is $$b = frac{sum_{k=1}^n k ( n+1-k)}{sum_{k=1}^n k}$$ and the expected height length is the same. So the expected area is $b^2$.
answered Feb 2 at 2:59
kimchi loverkimchi lover
11.8k31229
$endgroup$
The expected area is the expected base times the expected height. If, for example, $n=10$, there are $10$ bases of length $1$, and $9$ of length $2$, on up to $1$ of length $n$: $sum_{k=1}^n k$ in all. Thus the expected base length is $$b = frac{sum_{k=1}^n k ( n+1-k)}{sum_{k=1}^n k}$$ and the expected height length is the same. So the expected area is $b^2$.
answered Feb 2 at 2:59
kimchi loverkimchi lover
11.8k31229
answered Feb 2 at 2:59
kimchi loverkimchi lover
11.8k31229
answered Feb 2 at 2:59
kimchi loverkimchi lover
11.8k31229
answered Feb 2 at 2:59
kimchi loverkimchi lover
11.8k31229
11.8k31229
add a comment |
add a comment |
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$begingroup$
If $n=2,$ your formula gives $5$ rectangles, but it seems to me there are $9$. Four $1times1,$ four $2times1,$ and one $2times2$ What am I missing?
$endgroup$
– saulspatz
Feb 2 at 0:02
$begingroup$
@saulspatz That indeed is not the correct formula for the number of all rectangles.
$endgroup$
– Math1000
Feb 2 at 1:40