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I'm stuck on a seemingly simple question about propagation of error. Say we repeatedly measure the speed of a particle, and we estimate the uncertainty in the measured speed to be 10 percent. What is the uncertainty in the kinetic energy of the particle? (Uncertainty in mass is negligible.)

(This was a physics GRE practice question. The answer is 20 percent.)

The formula I know for propagation of error is, for $f = AB$,

$$\left( \frac{\sigma_f}{f}\right)^2 = \left( \frac{\sigma_A}{A}\right)^2 + \left( \frac{\sigma_B}{B}\right)^2 .$$

So, if I'm correct that uncertainty is defined $\left( \frac{\sigma_f}{f}\right)$, it follows that

$$\Delta KE = \sqrt{2} \Delta v$$

where $\Delta KE$ is uncertainty in kinetic energy and $\Delta v$ is uncertainty in speed. But this gives $14$ percent. Could somebody point out my error?

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    $\begingroup$ That's only true if $A$ and $B$ are uncorrelated. Presumably you're letting $A$ and $B$ both represent speed, in which case they are perfectly correlated. $\endgroup$ – J. Murray Apr 3 '18 at 17:41
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To expand on my comment, your equation holds only if $A$ and $B$ are uncorrelated. More generally, if $F$ is a function of uncorrelated variables $A$ and $B$, then

$$(\delta F)^2 = \left(\frac{\partial F}{\partial A}\delta A\right)^2 + \left(\frac{\partial F}{\partial B} \delta B\right)^2$$

If you plug in $F=AB$, then you find that $$ \left(\delta F\right)^2 = (B\cdot \delta A)^2 + (A \cdot \delta B)^2$$ or $$ \left(\frac{\delta F}{F}\right)^2 = \left(\frac{\delta A}{A}\right)^2 + \left(\frac{\delta B}{B}\right)^2 $$

In your case, $A$ and $B$ are the same, so they are the exact opposite of uncorrelated. In your case, the appropriate thing to do would be to consider $F$ to be a function of a single variable, and simply let

$$\delta F = \left|\frac{\partial F}{\partial A}\right| \delta A$$ $$\frac{\delta F}{F} = \left|\frac{1}{F}\frac{\partial F}{\partial A}\right| \delta A$$


To answer the question at a deeper level, we model experimentally measured quantities as continuous random variables, with our experimental uncertainties corresponding to their standard deviations.

Let $X$ be a random variable with expected value $\mathbb{E}[X]=\mu_X$ and variance $Var(X)\equiv \mathbb{E}\big[ (X-\mu_X)^2\big]=\sigma_X^2$, and let $g$ be a function of $X$. We can expand $g$ in a Taylor series around $\mu_X$:

$$g(X) = g(\mu_x) + g'(\mu_X)\cdot (X-\mu_X) + \frac{1}{2}g''(\mu_X) (X-\mu_X)^2 + \ldots $$

Truncating after the linear term, we write

$$g(x) = g(\mu_X) + g'(\mu_X)\cdot (X-\mu_X)$$

We can now calculate the mean and variance of $g$:

$$\mathbb{E}[g(X)] = g(\mu_X) + g'(\mu_X)\mathbb{E}[X-\mu_X] = g(\mu_X) + g'(\mu_X)\cdot (\mu_X-\mu_X) = g(\mu_X)$$ $$Var(g(X)) = \mathbb{E}\big[\big(g(X)-g(\mu_X)\big)^2\big] = \big(g'(\mu_X)\big)^2 \cdot \mathbb{E}\big[(X-\mu_X)^2\big] = \left(\frac{dg}{dX}\cdot \sigma_X\right)^2$$

This is where the single-variable error propagation formula comes from. But now, consider a function $F$ of two variables $X$ and $Y$, with respective means $\mu_X,\mu_Y$ and variances $\sigma_X^2,\sigma_Y^2$, and covariance

$$Cov(X,Y) = \mathbb{E}\big[(X-\mu_X)\cdot(Y-\mu_Y)\big]$$

We Taylor expand $F$ to linear order:

$$F(X,Y) = F(\mu_X,\mu_Y) + \frac{\partial F}{\partial X}(X-\mu_X) + \frac{\partial F}{\partial Y} (Y-\mu_Y) $$

Repeating the earlier steps, the mean of $F$ is

$$\mathbb{E}\big[F(X,Y)\big] = F(\mu_X,\mu_Y) $$

The variance, however, develops a slight subtlety. Notice that

$$\left(F(X,Y)-F(\mu_X,\mu_Y)\right)^2 = \left(\frac{\partial F}{\partial X}\right)^2(X-\mu_X)^2 + \left(\frac{\partial F}{\partial Y}\right)^2(Y-\mu_Y)^2 + 2\frac{\partial F}{\partial X}\frac{\partial F}{\partial Y}(X-\mu_X)(Y-\mu_Y)$$

It follows that

$$Var\big(F(X,Y)\big) = \left(\frac{\partial F}{\partial X} \sigma_X\right)^2 + \left(\frac{\partial F}{\partial Y}\sigma_Y\right)^2 + 2\frac{\partial F}{\partial X} \frac{\partial F}{\partial Y} Cov(X,Y)$$

If $X$ and $Y$ are uncorrelated, then $Cov(X,Y)=0$, and so we get our nice simple formula again. However, if $Cov(X,Y)\neq 0$, we need to take it into account.

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Your rule works for multiplication and division of independent uncertainties not raised to any interesting powers. The more general rule is

$$ (\mathrm d f)^2 = \left( \frac{\partial f}{\partial A} \mathrm dA \right)^2 + \left( \frac{\partial f}{\partial B} \mathrm dB \right)^2 + \cdots $$

If the terms in the sum were linear rather than added in quadrature, this would just be the total derivative of $f$; adding in quadrature accounts for the fact that uncorrelated errors are not likely to all contribute in the same direction.

You have, with the usual kinetic energy $K=\frac12 mv^2$,

\begin{align} (\mathrm dK)^2 &= \left( \frac{\partial K}{\partial v} \mathrm dv \right)^2 + \text{negligible}^2 \\ \mathrm dK &= mv\ \mathrm dv = \frac{2K}{v} \mathrm dv \\ \frac{\mathrm dK}{K} &= 2\frac{\mathrm dv}{v} \end{align}

The factor of two arises because $v$ is squared.

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  • $\begingroup$ Why did you write negligible? I see that that there is no B term, so instead of writing zero you wrote negligible. Can you please elaborate on that part? $\endgroup$ – Ankur Singh Apr 3 '18 at 19:41
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    $\begingroup$ There is a "B" term: the mass. You said the uncertainty in the mass was negligible. So the answerer is following your lead. $\endgroup$ – Paul T. Apr 3 '18 at 19:56

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