Trying to understand an inconsistency within basic error analysis equations
$begingroup$
While doing a physics lab, I noticed that the error analysis equation for multiplication $$R = frac{X*Y}{Z}$$ $$ delta R = |R|sqrt{left(frac{delta X}{X}right)^2+left(frac{delta Y}{Y}right)^2+left(frac{delta Z}{Z}right)^2}$$ is not algebraically equivalent to $$R = X^n$$ $$delta R = |n|*frac{delta X}{|X|}*|R|$$ For example, if we take $R = X*X$ then the first equation gives $sqrt{2}*left(frac{delta X}{|X|}right)*X^2$. But if we take $R = X^2$ then the second equation gives $2*left(frac{delta X}{|X|}right)*X^2$.
Why is this the case? I have feeling that I should only use the first equation for differing variables $X$ and $Y$, but I don't understand why.
error-propagation
edited Dec 3 '18 at 3:14
Robert Howard
2,0101826
asked Dec 3 '18 at 2:29
Ryan GreylingRyan Greyling
3449
$endgroup$
add a comment |
$begingroup$
While doing a physics lab, I noticed that the error analysis equation for multiplication $$R = frac{X*Y}{Z}$$ $$ delta R = |R|sqrt{left(frac{delta X}{X}right)^2+left(frac{delta Y}{Y}right)^2+left(frac{delta Z}{Z}right)^2}$$ is not algebraically equivalent to $$R = X^n$$ $$delta R = |n|*frac{delta X}{|X|}*|R|$$ For example, if we take $R = X*X$ then the first equation gives $sqrt{2}*left(frac{delta X}{|X|}right)*X^2$. But if we take $R = X^2$ then the second equation gives $2*left(frac{delta X}{|X|}right)*X^2$.
Why is this the case? I have feeling that I should only use the first equation for differing variables $X$ and $Y$, but I don't understand why.
error-propagation
edited Dec 3 '18 at 3:14
Robert Howard
2,0101826
asked Dec 3 '18 at 2:29
Ryan GreylingRyan Greyling
3449
$endgroup$
add a comment |
$begingroup$
While doing a physics lab, I noticed that the error analysis equation for multiplication $$R = frac{X*Y}{Z}$$ $$ delta R = |R|sqrt{left(frac{delta X}{X}right)^2+left(frac{delta Y}{Y}right)^2+left(frac{delta Z}{Z}right)^2}$$ is not algebraically equivalent to $$R = X^n$$ $$delta R = |n|*frac{delta X}{|X|}*|R|$$ For example, if we take $R = X*X$ then the first equation gives $sqrt{2}*left(frac{delta X}{|X|}right)*X^2$. But if we take $R = X^2$ then the second equation gives $2*left(frac{delta X}{|X|}right)*X^2$.
Why is this the case? I have feeling that I should only use the first equation for differing variables $X$ and $Y$, but I don't understand why.
error-propagation
edited Dec 3 '18 at 3:14
Robert Howard
2,0101826
asked Dec 3 '18 at 2:29
Ryan GreylingRyan Greyling
3449
$endgroup$
While doing a physics lab, I noticed that the error analysis equation for multiplication $$R = frac{X*Y}{Z}$$ $$ delta R = |R|sqrt{left(frac{delta X}{X}right)^2+left(frac{delta Y}{Y}right)^2+left(frac{delta Z}{Z}right)^2}$$ is not algebraically equivalent to $$R = X^n$$ $$delta R = |n|*frac{delta X}{|X|}*|R|$$ For example, if we take $R = X*X$ then the first equation gives $sqrt{2}*left(frac{delta X}{|X|}right)*X^2$. But if we take $R = X^2$ then the second equation gives $2*left(frac{delta X}{|X|}right)*X^2$.
Why is this the case? I have feeling that I should only use the first equation for differing variables $X$ and $Y$, but I don't understand why.
error-propagation
error-propagation
edited Dec 3 '18 at 3:14
Robert Howard
2,0101826
asked Dec 3 '18 at 2:29
Ryan GreylingRyan Greyling
3449
edited Dec 3 '18 at 3:14
Robert Howard
2,0101826
asked Dec 3 '18 at 2:29
Ryan GreylingRyan Greyling
3449
edited Dec 3 '18 at 3:14
Robert Howard
2,0101826
edited Dec 3 '18 at 3:14
Robert Howard
2,0101826
edited Dec 3 '18 at 3:14
Robert Howard
2,0101826
2,0101826
asked Dec 3 '18 at 2:29
Ryan GreylingRyan Greyling
3449
asked Dec 3 '18 at 2:29
Ryan GreylingRyan Greyling
3449
asked Dec 3 '18 at 2:29
Ryan GreylingRyan Greyling
3449
3449
add a comment |
add a comment |
1 Answer
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$begingroup$
This comes down to correlation between $X$ and $Y$, and hence correlation in their errors.
These formulas for calculating $delta X$ are in some sense talking about the standard deviation of the measurements of $X$.
If $X$ and $Y$ are independent measurements, then we don't expect the error in $X+Y$ to be just the sum of errors (it is too pessimistic) because there's some chance a positive error in $X$ will be cancelled out by a negative error in $Y$. Statistically you can then recover $sqrt{left(delta Xright)^2+left(delta Yright)^2}$ as the expected (average) error.
This reasoning completely fails when $X$ and $X$ are added. Whenever $X$ has a positive error, obviously $X$ (the second copy) also has positive error. There is no cancelling whatsoever, and the error of $2X$ is just twice the error of $X$.
answered Dec 3 '18 at 2:41
obscuransobscurans
1,152311
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
This comes down to correlation between $X$ and $Y$, and hence correlation in their errors.
These formulas for calculating $delta X$ are in some sense talking about the standard deviation of the measurements of $X$.
If $X$ and $Y$ are independent measurements, then we don't expect the error in $X+Y$ to be just the sum of errors (it is too pessimistic) because there's some chance a positive error in $X$ will be cancelled out by a negative error in $Y$. Statistically you can then recover $sqrt{left(delta Xright)^2+left(delta Yright)^2}$ as the expected (average) error.
This reasoning completely fails when $X$ and $X$ are added. Whenever $X$ has a positive error, obviously $X$ (the second copy) also has positive error. There is no cancelling whatsoever, and the error of $2X$ is just twice the error of $X$.
answered Dec 3 '18 at 2:41
obscuransobscurans
1,152311
$endgroup$
add a comment |
$begingroup$
This comes down to correlation between $X$ and $Y$, and hence correlation in their errors.
These formulas for calculating $delta X$ are in some sense talking about the standard deviation of the measurements of $X$.
If $X$ and $Y$ are independent measurements, then we don't expect the error in $X+Y$ to be just the sum of errors (it is too pessimistic) because there's some chance a positive error in $X$ will be cancelled out by a negative error in $Y$. Statistically you can then recover $sqrt{left(delta Xright)^2+left(delta Yright)^2}$ as the expected (average) error.
This reasoning completely fails when $X$ and $X$ are added. Whenever $X$ has a positive error, obviously $X$ (the second copy) also has positive error. There is no cancelling whatsoever, and the error of $2X$ is just twice the error of $X$.
answered Dec 3 '18 at 2:41
obscuransobscurans
1,152311
$endgroup$
add a comment |
$begingroup$
This comes down to correlation between $X$ and $Y$, and hence correlation in their errors.
These formulas for calculating $delta X$ are in some sense talking about the standard deviation of the measurements of $X$.
If $X$ and $Y$ are independent measurements, then we don't expect the error in $X+Y$ to be just the sum of errors (it is too pessimistic) because there's some chance a positive error in $X$ will be cancelled out by a negative error in $Y$. Statistically you can then recover $sqrt{left(delta Xright)^2+left(delta Yright)^2}$ as the expected (average) error.
This reasoning completely fails when $X$ and $X$ are added. Whenever $X$ has a positive error, obviously $X$ (the second copy) also has positive error. There is no cancelling whatsoever, and the error of $2X$ is just twice the error of $X$.
answered Dec 3 '18 at 2:41
obscuransobscurans
1,152311
$endgroup$
This comes down to correlation between $X$ and $Y$, and hence correlation in their errors.
These formulas for calculating $delta X$ are in some sense talking about the standard deviation of the measurements of $X$.
If $X$ and $Y$ are independent measurements, then we don't expect the error in $X+Y$ to be just the sum of errors (it is too pessimistic) because there's some chance a positive error in $X$ will be cancelled out by a negative error in $Y$. Statistically you can then recover $sqrt{left(delta Xright)^2+left(delta Yright)^2}$ as the expected (average) error.
This reasoning completely fails when $X$ and $X$ are added. Whenever $X$ has a positive error, obviously $X$ (the second copy) also has positive error. There is no cancelling whatsoever, and the error of $2X$ is just twice the error of $X$.
answered Dec 3 '18 at 2:41
obscuransobscurans
1,152311
edited Dec 3 '18 at 2:46
answered Dec 3 '18 at 2:41
obscuransobscurans
1,152311
answered Dec 3 '18 at 2:41
obscuransobscurans
1,152311
answered Dec 3 '18 at 2:41
obscuransobscurans
1,152311
1,152311
add a comment |
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