# Error propagation of squared function, insert or not the self-covariance?

As my question suggests, I'm estimating the error of $$z=x^2$$. The general error propagation formula for two measures $$x$$, $$y$$ with errors $$\sigma_x$$, $$\sigma_y$$, is:

$$\sigma^2_z = (\frac{\partial z}{\partial x})^2\sigma^2_x + (\frac{\partial z}{\partial y})^2\sigma^2_y + 2 (\frac{\partial z}{\partial x})(\frac{\partial z}{\partial y})cov(x,y)$$

Since my $$z=x^2$$, i.e., $$x=y$$, $$\sigma_x=\sigma_y$$, the previous relation reduces to:

$$\sigma^2_z = z^2 \cdot(2\sigma^2_x/z+2cov(x,x)/z)$$

Since $$cov(x,x)\equiv\sigma^2_x$$, a further simplification is:

$$\sigma^2_z = z^2 \cdot(2(\sigma_x/x)^2+2(\sigma_x/x)^2)$$

Thus, since $$xy=x^2=y^2=z$$, the squared error on $$z$$ including the self covariance is:

$$\sigma^2_z = 4 z \sigma^2_x$$

While, removing the self covariance:

$$\sigma^2_z = 2 z \sigma^2_x$$

implying a factor $$2$$ between the two versions. Is this reason corrected? Which version should I use? Why?

EDIT Thanks for the comments, my question hiddens another question: if my $$z$$ has a form like this: $$z = f(x, y)$$, where $$y=f(x)$$, should I insert the covariance between $$x$$ and $$g(x)$$?

• Since you don't have any $y$, why do you think $x=y$? $z = x^2$ in no way implies the existence of a $y$. – jbowman Mar 16 at 15:34
• – Ed V Mar 16 at 16:52