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I'm trying to do propagation of error using the linearized variance method (assuming independent variables, thus no need for the covariance terms):

$$\sigma^2_f = \sum^n_{k=0} \left(\frac{\partial f}{\partial x_k}\right)^2 \sigma^2_{x_k}$$

However, I have a nasty function that doesn't give me a clear-cut explicit definition of my variable. For simplicity, I will just give a simple example of what I'm trying to accomplish. Take the equation

$$x - y = e^f + e^{2f} + e^{2x}$$

This is algebraicly impossible to solving explicitly for $f = f(x,y)$. So, if I wanted to find the variance of f, I had two ideas (one of them backfired...). First, make a new function equal to zero,

$$g(x,y,f) = e^f + e^{2f} + e^{2x} + y - x = 0$$

That way I could easily find the partial derivatives, and the variance of this new function would be zero, since it's value always equaled zero.

$$\sigma^2_g = 0$$

Unfortunately, this backfired on me (after I did the 26 partial derivatives, ouch) as you can see with

$$\sigma^2_g = \left(\frac{\partial g}{\partial x}\right)^2 \sigma^2_x + \left(\frac{\partial g}{\partial y}\right)^2 \sigma^2_y + \left(\frac{\partial g}{\partial f}\right)^2 \sigma^2_f$$

If you set the variance of g to zero, then you could solve for the variance of f, and be a happy camper! Wrong. Because all the terms are squared, there is NO WAY they can add up to be zero unless they are all identically zero. That really messed me up.

The other idea I had just take the original equation and performing the partial differential with respect to x and y to each side of the equation, then solve for the partial diff quantities. That would require me to do a complete overhaul on all my work.

My question then: is there anyway to use the first method I thought of, just modifying my steps? Or maybe a third way? If not, then I will be surprised, since mathematics usually has a way to solve such twisted scenarios. Please advise soon, as I need to finish this up quickly.

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3 Answers

Alas, your second idea is the only correct one to use here. The big (and irrepairable) problem with your first idea is that you tried to use the formula for the variance of a function of several independent variables in the case when they are severely dependent (namely, for $x,y,f=f(x,y)$). So, however sad it is, you've got to redo everything from the beginning.

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Except you can't just naively divide by the partial differentials, you have to multiply by the inverse Jacobian to find the correct equation.

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Actually, I figured it out with a stoke of luck last night. I was so set on using an implicit function that I forgot the rest of the implications that go along with that.

The Implicit Function Theorem provides a method to perform implicit differentiation:

$\frac{\partial f}{\partial x} = -\frac{\displaystyle\frac{\partial g}{\partial x}}{\displaystyle\frac{\partial g}{\partial f}}$

This relationship is often used in manipulating total differentials. I used it a lot in my thermodynamics class, which does this kind of calculation all the time. So instead of re-doing everything, I just had to use the existing partials I had and divide them by the appropriate partial as shown above. Hooray! All hard work was not lost! Thanks anyways for your thoughts and contributions.

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