How does error propagate in a complex equation involving averages and variances?

[tjosan]

Homework Statement

Hello,

I have the following operation that I want to perform:

f=\frac{\bar{X}}{100-\sum \bar{Y}_j}*K
\bar{X} and \bar{Y} are averages with variances S_{X}^2 and S_{Y_j}^2 and K is a constant.

How will the error propagate?

Homework Equations

According to Wikipedia:
(1) f=a\bar{A} \Rightarrow S_f^2=a^2S_f^2 where a is a constant.
(2) f=\bar{A}\bar{B} \Rightarrow S_f^2=S_A^2+S_B^2
(3) f=\frac{\bar{A}}{\bar{B}} \Rightarrow S_f^2=f^2\left(\frac{S_A^2}{A^2}+\frac{S_B^2}{B^2}\right)

The Attempt at a Solution

So then the error of the nominator will be S_{X}^2
Only looking at the denominator i will have: 100-\sum S_{Y_j}^2
Using the third and first equation will then yield:

S_f^2=f^2 \left(\frac{S_{X}^2}{\bar{X}^2} + \frac{100-\sum S_{Y_j}^2}{(100-\sum \bar{Y}_j)^2} \right)K^2

Where K^2 comes from the first equation.

I am a little bit confused though. Is this correct?

Thanks.

Edit: Covariance=0

 

[Ray Vickson]

Homework Statement

Hello,

I have the following operation that I want to perform:

f=\frac{\bar{X}}{100-\sum \bar{Y}_j}*K
\bar{X} and \bar{Y} are averages with variances S_{X}^2 and S_{Y_j}^2 and K is a constant.

How will the error propagate?

Homework Equations

According to Wikipedia:
(1) f=a\bar{A} \Rightarrow S_f^2=a^2S_f^2 where a is a constant.
(2) f=\bar{A}\bar{B} \Rightarrow S_f^2=S_A^2+S_B^2
(3) f=\frac{\bar{A}}{\bar{B}} \Rightarrow S_f^2=f^2\left(\frac{S_A^2}{A^2}+\frac{S_B^2}{B^2}\right)

The Attempt at a Solution

So then the error of the nominator will be S_{X}^2
Only looking at the denominator i will have: 100-\sum S_{Y_j}^2
Using the third and first equation will then yield:

S_f^2=f^2 \left(\frac{S_{X}^2}{\bar{X}^2} + \frac{100-\sum S_{Y_j}^2}{(100-\sum \bar{Y}_j)^2} \right)K^2

Where K^2 comes from the first equation.

I am a little bit confused though. Is this correct?

Thanks.

Edit: Covariance=0

No, it is not correct: the squared error in $100 - \sum Y_j$ is not $100 - \sum S_{Y_j}^2$. For one thing, the '100' is a constant that has no error; for another thing, the $Y_i$ squared errors should not be subtracted from anything.

 

[tjosan]

Thank you for your answer. Would this be correct?

S_f^2=f^2 \left(\frac{S_{X}^2}{\bar{X}^2} + \frac{\sum S_{Y_j}^2}{(100-\sum \bar{Y}_j)^2} \right)K^2

Thanks.

 

[tjosan]

To answer my own question:

S_f^2=\left(\frac{\bar{X}}{100-\sum \bar{Y}_j}\right)^2 \left(\frac{S_{X}^2}{\bar{X}^2} + \frac{\sum S_{Y_j}^2}{(100-\sum \bar{Y}_j)^2} \right)K^2

It should be clarified that K is an exakt number.