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Chebyshev inequality in terms of RMS

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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

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asked 2 hours ago

H. YongH. Yong

112

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2

$begingroup$

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 2 hours ago

H. YongH. Yong

112

New contributor

H. Yong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

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 2 hours ago

H. YongH. Yong

112

New contributor

H. Yong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked 2 hours ago

H. YongH. Yong

112

New contributor

H. Yong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked 2 hours ago

H. YongH. Yong

112

New contributor

H. Yong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 hours ago

H. YongH. Yong

112

New contributor

H. Yong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 hours ago

H. YongH. Yong

112

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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 $vec{x}=(x_1,ldots,x_n)$ such that $||x_i|| geq a > 0$.

Hence $||vec{x}||^2 = sum x_i^2 geq k a^2 + (n - k) times 0$, 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 $rms(vec{x}) = sqrt{frac{||vec{x}||^2}{n}}$, it follows that $rms(vec{x})^2 = frac{||vec{x}||^2}{n} geq frac {k a^2}{n}$.

Therefore, we get the final expression that says
$$
frac {k}{n} leq left( frac{rms(vec{x})}{a} right) ^2
$$

So, following the example, where $a = 5 rms(vec{x})$, we have that $frac {k}{n} leq left( frac{1}{5} right) ^2 = 4 %$, so, the fraction of elements of the vector larger (in absolute value) than $5 rms$ is at most $4%$.

If we chose another number, say $a = 2 rms(vec{x})$, we would have that $frac {k}{n} leq left( frac{1}{2} right) ^2 = 25 %$.

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answered 1 hour ago

ErtxiemErtxiem

31718

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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 $vec{x}=(x_1,ldots,x_n)$ such that $||x_i|| geq a > 0$.

Hence $||vec{x}||^2 = sum x_i^2 geq k a^2 + (n - k) times 0$, 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 $rms(vec{x}) = sqrt{frac{||vec{x}||^2}{n}}$, it follows that $rms(vec{x})^2 = frac{||vec{x}||^2}{n} geq frac {k a^2}{n}$.

Therefore, we get the final expression that says
$$
frac {k}{n} leq left( frac{rms(vec{x})}{a} right) ^2
$$

So, following the example, where $a = 5 rms(vec{x})$, we have that $frac {k}{n} leq left( frac{1}{5} right) ^2 = 4 %$, so, the fraction of elements of the vector larger (in absolute value) than $5 rms$ is at most $4%$.

If we chose another number, say $a = 2 rms(vec{x})$, we would have that $frac {k}{n} leq left( frac{1}{2} right) ^2 = 25 %$.

share|cite|improve this answer

answered 1 hour ago

ErtxiemErtxiem

31718

$endgroup$

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 $vec{x}=(x_1,ldots,x_n)$ such that $||x_i|| geq a > 0$.

Hence $||vec{x}||^2 = sum x_i^2 geq k a^2 + (n - k) times 0$, 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 $rms(vec{x}) = sqrt{frac{||vec{x}||^2}{n}}$, it follows that $rms(vec{x})^2 = frac{||vec{x}||^2}{n} geq frac {k a^2}{n}$.

Therefore, we get the final expression that says
$$
frac {k}{n} leq left( frac{rms(vec{x})}{a} right) ^2
$$

So, following the example, where $a = 5 rms(vec{x})$, we have that $frac {k}{n} leq left( frac{1}{5} right) ^2 = 4 %$, so, the fraction of elements of the vector larger (in absolute value) than $5 rms$ is at most $4%$.

If we chose another number, say $a = 2 rms(vec{x})$, we would have that $frac {k}{n} leq left( frac{1}{2} right) ^2 = 25 %$.

share|cite|improve this answer

answered 1 hour ago

ErtxiemErtxiem

31718

$endgroup$

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 $vec{x}=(x_1,ldots,x_n)$ such that $||x_i|| geq a > 0$.

Hence $||vec{x}||^2 = sum x_i^2 geq k a^2 + (n - k) times 0$, 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 $rms(vec{x}) = sqrt{frac{||vec{x}||^2}{n}}$, it follows that $rms(vec{x})^2 = frac{||vec{x}||^2}{n} geq frac {k a^2}{n}$.

Therefore, we get the final expression that says
$$
frac {k}{n} leq left( frac{rms(vec{x})}{a} right) ^2
$$

So, following the example, where $a = 5 rms(vec{x})$, we have that $frac {k}{n} leq left( frac{1}{5} right) ^2 = 4 %$, so, the fraction of elements of the vector larger (in absolute value) than $5 rms$ is at most $4%$.

If we chose another number, say $a = 2 rms(vec{x})$, we would have that $frac {k}{n} leq left( frac{1}{2} right) ^2 = 25 %$.

share|cite|improve this answer

answered 1 hour ago

ErtxiemErtxiem

31718

$endgroup$

share|cite|improve this answer

answered 1 hour ago

ErtxiemErtxiem

31718

answered 1 hour ago

ErtxiemErtxiem

31718

answered 1 hour ago

ErtxiemErtxiem

31718