Chebyshev inequality in terms of RMS Announcing the arrival of Valued Associate #679: Cesar...
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Chebyshev inequality in terms of RMS
Announcing the arrival of Valued Associate #679: Cesar Manara
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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
asked 12 hours ago
H. YongH. Yong
162
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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
asked 12 hours ago
H. YongH. Yong
162
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
asked 12 hours ago
H. YongH. Yong
162
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.
$endgroup$
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
linear-algebra
asked 12 hours ago
H. YongH. Yong
162
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 12 hours ago
H. YongH. Yong
162
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 12 hours ago
H. YongH. Yong
162
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 12 hours ago
H. YongH. Yong
162
asked 12 hours ago
H. YongH. Yong
162
162
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.
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.
H. Yong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
1 Answer
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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 $operatorname{rms}(vec{x}) = sqrt{frac{|vec{x}|^2}{n}}$, it follows that $operatorname{rms}(vec{x})^2 = frac{|vec{x}|^2}{n} geq frac {k a^2}{n}$.
Therefore, we get the final expression that says
$$frac {k}{n} leq left( frac{operatorname{rms}(vec{x})}{a} right) ^2$$
So, following the example, where $a = 5 operatorname{rms}(vec{x})$, 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 $5operatorname{rms}$ is at most $4%$.
If we chose another number, say $a = 2 operatorname{rms}(vec{x})$, we would have that $frac {k}{n} leq left( frac{1}{2} right) ^2 = 25 %$.
edited 8 hours ago
StubbornAtom
3,1511535
answered 12 hours ago
ErtxiemErtxiem
31718
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$begingroup$
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 $operatorname{rms}(vec{x}) = sqrt{frac{|vec{x}|^2}{n}}$, it follows that $operatorname{rms}(vec{x})^2 = frac{|vec{x}|^2}{n} geq frac {k a^2}{n}$.
Therefore, we get the final expression that says
$$frac {k}{n} leq left( frac{operatorname{rms}(vec{x})}{a} right) ^2$$
So, following the example, where $a = 5 operatorname{rms}(vec{x})$, 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 $5operatorname{rms}$ is at most $4%$.
If we chose another number, say $a = 2 operatorname{rms}(vec{x})$, we would have that $frac {k}{n} leq left( frac{1}{2} right) ^2 = 25 %$.
edited 8 hours ago
StubbornAtom
3,1511535
answered 12 hours 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 $operatorname{rms}(vec{x}) = sqrt{frac{|vec{x}|^2}{n}}$, it follows that $operatorname{rms}(vec{x})^2 = frac{|vec{x}|^2}{n} geq frac {k a^2}{n}$.
Therefore, we get the final expression that says
$$frac {k}{n} leq left( frac{operatorname{rms}(vec{x})}{a} right) ^2$$
So, following the example, where $a = 5 operatorname{rms}(vec{x})$, 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 $5operatorname{rms}$ is at most $4%$.
If we chose another number, say $a = 2 operatorname{rms}(vec{x})$, we would have that $frac {k}{n} leq left( frac{1}{2} right) ^2 = 25 %$.
edited 8 hours ago
StubbornAtom
3,1511535
answered 12 hours 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 $operatorname{rms}(vec{x}) = sqrt{frac{|vec{x}|^2}{n}}$, it follows that $operatorname{rms}(vec{x})^2 = frac{|vec{x}|^2}{n} geq frac {k a^2}{n}$.
Therefore, we get the final expression that says
$$frac {k}{n} leq left( frac{operatorname{rms}(vec{x})}{a} right) ^2$$
So, following the example, where $a = 5 operatorname{rms}(vec{x})$, 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 $5operatorname{rms}$ is at most $4%$.
If we chose another number, say $a = 2 operatorname{rms}(vec{x})$, we would have that $frac {k}{n} leq left( frac{1}{2} right) ^2 = 25 %$.
edited 8 hours ago
StubbornAtom
3,1511535
answered 12 hours ago
ErtxiemErtxiem
31718
$endgroup$
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 $operatorname{rms}(vec{x}) = sqrt{frac{|vec{x}|^2}{n}}$, it follows that $operatorname{rms}(vec{x})^2 = frac{|vec{x}|^2}{n} geq frac {k a^2}{n}$.
Therefore, we get the final expression that says
$$frac {k}{n} leq left( frac{operatorname{rms}(vec{x})}{a} right) ^2$$
So, following the example, where $a = 5 operatorname{rms}(vec{x})$, 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 $5operatorname{rms}$ is at most $4%$.
If we chose another number, say $a = 2 operatorname{rms}(vec{x})$, we would have that $frac {k}{n} leq left( frac{1}{2} right) ^2 = 25 %$.
edited 8 hours ago
StubbornAtom
3,1511535
answered 12 hours ago
ErtxiemErtxiem
31718
edited 8 hours ago
StubbornAtom
3,1511535
edited 8 hours ago
StubbornAtom
3,1511535
edited 8 hours ago
StubbornAtom
3,1511535
3,1511535
answered 12 hours ago
ErtxiemErtxiem
31718
answered 12 hours ago
ErtxiemErtxiem
31718
answered 12 hours ago
ErtxiemErtxiem
31718
31718
add a comment |
add a comment |
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