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Suppose we have the formula

$Y = \frac{2WV^{2}}{\pi D^2}\hspace{1cm}$(1)

where:

w = weight variable, v = velocity variable, d = diameter variable

we want to find $SD(Y)$

the solution which was proposed to me was (exactly):

$SD(Y) = \sqrt{(\frac{2v^2}{\pi d^2})^2SD(W)^2 + (\frac{4wv}{\pi d^2})^2SD(V)^2 + (\frac{-4wv^2}{\pi d^3})^2SD(D)^2 }\hspace{1cm}$(2)

and we assume independence of W, D, and V...

The answer provided does not look correct to me...I can see what they did (albeit I'm not sure where the 4s came from), but it doesn't sit well with me, so I decided to do some simulations in R. I set W, D, and V to be normal distributions with different means and standard deviations and simulated values plugging them into (1) and then estimated the SD of Y. I plugged the means into the lower case values of the variables in (2) and the standard deviations in the respective areas. The results were sort of close but not close enough (446 vs 436) (I did enough simulations that the simulation variance was negligible). I trust the SD from my simulation more than the result from (2). Any thoughts on if (2) is correct, if not any solutions?

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2 Answers 2

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The origin of that formula is

$ s_f = \sqrt{ \left(\frac{\partial f}{\partial x}\right)^2 s_x^2 + \left(\frac{\partial f}{\partial y} \right)^2 s_y^2 + \left(\frac{\partial f}{\partial z} \right)^2 s_z^2 + \cdots} $

(see https://en.wikipedia.org/wiki/Propagation_of_uncertainty). Those 4's come just from taking the derivative of variables that are squared in your eq. (1).

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    $\begingroup$ You input an estimation of their value and use the uncertainty in that estimation. If the estimation that you have is the average, you can use that together with the error in that average. $\endgroup$
    – rasmodius
    Commented Feb 22, 2021 at 18:38
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    $\begingroup$ You should make sure about the correlation between the parameters. If it's not zero, you should use a more general formula that can take it into account. $\endgroup$
    – rasmodius
    Commented Feb 22, 2021 at 18:39
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    $\begingroup$ First of all, if you have 1000 observations {m,d,v}, you can compute 1000 values of Y. From that you can compute EVERYTHING. The average of Y, its SD and even it's PDF. The other method is used when you have just one value for m, d and v (for example you only know their average) and the uncertainty in that value. $\endgroup$
    – rasmodius
    Commented Feb 22, 2021 at 19:14
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    $\begingroup$ But if you choose to do it the way you said, then you can compute the average <m> and SD(m). Then, the SD(<m>) = SD(m)/sqrt(N), where N is the number of points. Then in the formula you can substitute the averages <...> and the error in the averages SD(<...>) to obtain SD(<Y>), where <Y> is approximated by the formula substituting the <...>. But definitely computing 1000 values of Y is preferred, error propagation is just an approximation while the other way is as good as you can get. $\endgroup$
    – rasmodius
    Commented Feb 22, 2021 at 19:17
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    $\begingroup$ Yes. But it's crucial that if you have several data points you make sure what SD you are using, that of w, v and d or that of their average, with that sqrt(N) factor. If you only have one, then it's the same and it's as you said. $\endgroup$
    – rasmodius
    Commented Feb 22, 2021 at 19:42
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multivariate Delta method
The gradient is the vector consisting of the terms which are the three fractions under the square root. You are assuming the three random variables are independent, so $\Sigma$ is a diagonal matrix and that is what you will get when you multiply the terms in the formula. It is intended to be used for functions of estimators from a sample of size $n$ and where $n\rightarrow \infty$. It does not necessarily mean it will be a good approximation for the variance of some function of three random variables.

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  • $\begingroup$ Thank you, I have been reading about issues with the approximation especially when ratios are present. I think that's likely why my sim values were off a bit...I cranked up the mean so they were very far away from 0 and now they are converging fairly well. $\endgroup$
    – DanE
    Commented Feb 22, 2021 at 18:25

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