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Chebyshev inequality in terms of RMS

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3

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I'm self studying the book Introduction to Applied Linear Algebra – Vectors, Matrices, and Least Squares

In page 48, the author write: "It says,for example, that no more than 1/25 = 4% of the entries of a vector can exceed its RMS value by more than a factor of 5."

I need more explain about it. Especially about why the factor is 5?

linear-algebra

share|cite|improve this question

asked Apr 18 at 17:23

H. YongH. Yong

183

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add a comment | 

3

$begingroup$

I'm self studying the book Introduction to Applied Linear Algebra – Vectors, Matrices, and Least Squares

In page 48, the author write: "It says,for example, that no more than 1/25 = 4% of the entries of a vector can exceed its RMS value by more than a factor of 5."

I need more explain about it. Especially about why the factor is 5?

linear-algebra

share|cite|improve this question

asked Apr 18 at 17:23

H. YongH. Yong

183

$endgroup$

add a comment | 

$begingroup$

I'm self studying the book Introduction to Applied Linear Algebra – Vectors, Matrices, and Least Squares

In page 48, the author write: "It says,for example, that no more than 1/25 = 4% of the entries of a vector can exceed its RMS value by more than a factor of 5."

I need more explain about it. Especially about why the factor is 5?

linear-algebra

share|cite|improve this question

asked Apr 18 at 17:23

H. YongH. Yong

183

$endgroup$

share|cite|improve this question

asked Apr 18 at 17:23

H. YongH. Yong

183

share|cite|improve this question

asked Apr 18 at 17:23

H. YongH. Yong

183

asked Apr 18 at 17:23

H. YongH. Yong

183

asked Apr 18 at 17:23

H. YongH. Yong

183

1 Answer
1

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oldest

votes

4

$begingroup$

According to Chebyshev's_inequality, the probability of a value to deviate more than $k=5$ standard deviations from the mean is at most $1/k^2$.

When applied to vectors in your specific case, and following the book you cited, let $k$ be the number of elements of the vector $vecx=(x_1,ldots,x_n)$ such that $||x_i|| geq a > 0$.

Hence $|vecx|^2 = sum x_i^2 geq k a^2 + (n - k) times 0$, which means that we have $k$ values larger than $a^2$ and the others $n-k$ values are at least zero.

Since the root mean square value is $operatornamerms(vecx) = sqrtfrac^2n$, it follows that $operatornamerms(vecx)^2 = frac^2n geq frac k a^2n$.

Therefore, we get the final expression that says

$$frac kn leq left( fracoperatornamerms(vecx)a right) ^2$$

So, following the example, where $a = 5 operatornamerms(vecx)$, we have that $frac kn leq left( frac15 right) ^2 = 4 %$, so, the fraction of elements of the vector larger (in absolute value) than $5operatornamerms$ is at most $4%$.

If we chose another number, say $a = 2 operatornamerms(vecx)$, we would have that $frac kn leq left( frac12 right) ^2 = 25 %$.

share|cite|improve this answer

edited Apr 18 at 21:30

StubbornAtom

3,2461536

answered Apr 18 at 18:10

ErtxiemErtxiem

49628

$endgroup$

add a comment | 

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4

$begingroup$

According to Chebyshev's_inequality, the probability of a value to deviate more than $k=5$ standard deviations from the mean is at most $1/k^2$.

When applied to vectors in your specific case, and following the book you cited, let $k$ be the number of elements of the vector $vecx=(x_1,ldots,x_n)$ such that $||x_i|| geq a > 0$.

Hence $|vecx|^2 = sum x_i^2 geq k a^2 + (n - k) times 0$, which means that we have $k$ values larger than $a^2$ and the others $n-k$ values are at least zero.

Since the root mean square value is $operatornamerms(vecx) = sqrtfrac^2n$, it follows that $operatornamerms(vecx)^2 = frac^2n geq frac k a^2n$.

Therefore, we get the final expression that says

$$frac kn leq left( fracoperatornamerms(vecx)a right) ^2$$

So, following the example, where $a = 5 operatornamerms(vecx)$, we have that $frac kn leq left( frac15 right) ^2 = 4 %$, so, the fraction of elements of the vector larger (in absolute value) than $5operatornamerms$ is at most $4%$.

If we chose another number, say $a = 2 operatornamerms(vecx)$, we would have that $frac kn leq left( frac12 right) ^2 = 25 %$.

share|cite|improve this answer

edited Apr 18 at 21:30

StubbornAtom

3,2461536

answered Apr 18 at 18:10

ErtxiemErtxiem

49628

$endgroup$

add a comment | 

4

$begingroup$

According to Chebyshev's_inequality, the probability of a value to deviate more than $k=5$ standard deviations from the mean is at most $1/k^2$.

When applied to vectors in your specific case, and following the book you cited, let $k$ be the number of elements of the vector $vecx=(x_1,ldots,x_n)$ such that $||x_i|| geq a > 0$.

Hence $|vecx|^2 = sum x_i^2 geq k a^2 + (n - k) times 0$, which means that we have $k$ values larger than $a^2$ and the others $n-k$ values are at least zero.

Since the root mean square value is $operatornamerms(vecx) = sqrtfrac^2n$, it follows that $operatornamerms(vecx)^2 = frac^2n geq frac k a^2n$.

Therefore, we get the final expression that says

$$frac kn leq left( fracoperatornamerms(vecx)a right) ^2$$

So, following the example, where $a = 5 operatornamerms(vecx)$, we have that $frac kn leq left( frac15 right) ^2 = 4 %$, so, the fraction of elements of the vector larger (in absolute value) than $5operatornamerms$ is at most $4%$.

If we chose another number, say $a = 2 operatornamerms(vecx)$, we would have that $frac kn leq left( frac12 right) ^2 = 25 %$.

share|cite|improve this answer

edited Apr 18 at 21:30

StubbornAtom

3,2461536

answered Apr 18 at 18:10

ErtxiemErtxiem

49628

$endgroup$

add a comment | 

$begingroup$

According to Chebyshev's_inequality, the probability of a value to deviate more than $k=5$ standard deviations from the mean is at most $1/k^2$.

When applied to vectors in your specific case, and following the book you cited, let $k$ be the number of elements of the vector $vecx=(x_1,ldots,x_n)$ such that $||x_i|| geq a > 0$.

Hence $|vecx|^2 = sum x_i^2 geq k a^2 + (n - k) times 0$, which means that we have $k$ values larger than $a^2$ and the others $n-k$ values are at least zero.

Since the root mean square value is $operatornamerms(vecx) = sqrtfrac^2n$, it follows that $operatornamerms(vecx)^2 = frac^2n geq frac k a^2n$.

Therefore, we get the final expression that says

$$frac kn leq left( fracoperatornamerms(vecx)a right) ^2$$

So, following the example, where $a = 5 operatornamerms(vecx)$, we have that $frac kn leq left( frac15 right) ^2 = 4 %$, so, the fraction of elements of the vector larger (in absolute value) than $5operatornamerms$ is at most $4%$.

If we chose another number, say $a = 2 operatornamerms(vecx)$, we would have that $frac kn leq left( frac12 right) ^2 = 25 %$.

share|cite|improve this answer

edited Apr 18 at 21:30

StubbornAtom

3,2461536

answered Apr 18 at 18:10

ErtxiemErtxiem

49628

$endgroup$

share|cite|improve this answer

edited Apr 18 at 21:30

StubbornAtom

3,2461536

answered Apr 18 at 18:10

ErtxiemErtxiem

49628

edited Apr 18 at 21:30

StubbornAtom

3,2461536

edited Apr 18 at 21:30

StubbornAtom

3,2461536

answered Apr 18 at 18:10

ErtxiemErtxiem

49628

answered Apr 18 at 18:10

ErtxiemErtxiem

49628