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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 $g(x)$?

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    $\begingroup$ Since you don't have any $y$, why do you think $x=y$? $z = x^2$ in no way implies the existence of a $y$. $\endgroup$ – jbowman Mar 16 at 15:34
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    $\begingroup$ See stats.stackexchange.com/q/163249/247352. $\endgroup$ – Ed V Mar 16 at 16:52

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