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Chebyshev inequality in terms of RMS
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Inequality for trace of product of matrices given norms of the matricesCoding for Regression AnalysisQR factorization and linear regressionWhat is the “expressive power” of the composition function in a Recursive Neural Tensor Network?How can one design a polynomial function that really does require higher order terms to approximate it well?Matrix Orthogonal to Vector: why take transpose?
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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 2 hours ago
H. YongH. Yong
112
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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 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
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.
$endgroup$
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 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
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
asked 2 hours ago
H. YongH. Yong
112
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.
New contributor
H. Yong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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 $vecx=(x_1,ldots,x_n)$ 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 $rms(vecx) = sqrtfracn$, it follows that $rms(vecx)^2 = fracn geq frac k a^2n$.
Therefore, we get the final expression that says
$$
frac kn leq left( fracrms(vecx)a right) ^2
$$
So, following the example, where $a = 5 rms(vecx)$, we have that $frac kn leq left( frac15 right) ^2 = 4 %$, so, the fraction of elements of the vector larger (in absolute value) than $5 rms$ is at most $4%$.
If we chose another number, say $a = 2 rms(vecx)$, we would have that $frac kn leq left( frac12 right) ^2 = 25 %$.
answered 1 hour ago
ErtxiemErtxiem
30718
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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 $vecx=(x_1,ldots,x_n)$ 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 $rms(vecx) = sqrtfracn$, it follows that $rms(vecx)^2 = fracn geq frac k a^2n$.
Therefore, we get the final expression that says
$$
frac kn leq left( fracrms(vecx)a right) ^2
$$
So, following the example, where $a = 5 rms(vecx)$, we have that $frac kn leq left( frac15 right) ^2 = 4 %$, so, the fraction of elements of the vector larger (in absolute value) than $5 rms$ is at most $4%$.
If we chose another number, say $a = 2 rms(vecx)$, we would have that $frac kn leq left( frac12 right) ^2 = 25 %$.
answered 1 hour ago
ErtxiemErtxiem
30718
$endgroup$
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 $rms(vecx) = sqrtfracn$, it follows that $rms(vecx)^2 = fracn geq frac k a^2n$.
Therefore, we get the final expression that says
$$
frac kn leq left( fracrms(vecx)a right) ^2
$$
So, following the example, where $a = 5 rms(vecx)$, we have that $frac kn leq left( frac15 right) ^2 = 4 %$, so, the fraction of elements of the vector larger (in absolute value) than $5 rms$ is at most $4%$.
If we chose another number, say $a = 2 rms(vecx)$, we would have that $frac kn leq left( frac12 right) ^2 = 25 %$.
answered 1 hour ago
ErtxiemErtxiem
30718
$endgroup$
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 $rms(vecx) = sqrtfracn$, it follows that $rms(vecx)^2 = fracn geq frac k a^2n$.
Therefore, we get the final expression that says
$$
frac kn leq left( fracrms(vecx)a right) ^2
$$
So, following the example, where $a = 5 rms(vecx)$, we have that $frac kn leq left( frac15 right) ^2 = 4 %$, so, the fraction of elements of the vector larger (in absolute value) than $5 rms$ is at most $4%$.
If we chose another number, say $a = 2 rms(vecx)$, we would have that $frac kn leq left( frac12 right) ^2 = 25 %$.
answered 1 hour ago
ErtxiemErtxiem
30718
$endgroup$
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 $vecx=(x_1,ldots,x_n)$ 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 $rms(vecx) = sqrtfracn$, it follows that $rms(vecx)^2 = fracn geq frac k a^2n$.
Therefore, we get the final expression that says
$$
frac kn leq left( fracrms(vecx)a right) ^2
$$
So, following the example, where $a = 5 rms(vecx)$, we have that $frac kn leq left( frac15 right) ^2 = 4 %$, so, the fraction of elements of the vector larger (in absolute value) than $5 rms$ is at most $4%$.
If we chose another number, say $a = 2 rms(vecx)$, we would have that $frac kn leq left( frac12 right) ^2 = 25 %$.
answered 1 hour ago
ErtxiemErtxiem
30718
answered 1 hour ago
ErtxiemErtxiem
30718
answered 1 hour ago
ErtxiemErtxiem
30718
answered 1 hour ago
ErtxiemErtxiem
30718
30718
add a comment |
add a comment |
H. Yong is a new contributor. Be nice, and check out our Code of Conduct.
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