# Is fractional error computed using the actual value or the best estimate?

I'm trying to figure out which is the best way to do error propagation for situations when you have a product.

For instance, for $F(x,y) = xy$, we can use Taylor expansion and keeping 1st order terms:

$$\delta F = \frac{\partial F}{\partial x}(x_0,y_0) \delta x + \frac{\partial F}{\partial y}(x_0,y_0) \delta y$$

where $\delta F = F(x,y)-F(x_0,y_0)$, $\delta x = x-x_0$, and $\delta y = y-y_0$.

After computing the partial derivatives,

$$\delta F = y_0 \delta x + x_0 \delta y$$

Then squaring the above equation and assuming there is no correlation between the error in x and error in y (i.e. $\langle\delta x \delta y\rangle = 0$), $\delta F$ can be redefined to be a definite positive number:

$$\delta F := \sqrt{(y_0 \delta x)^2 + (x_0 \delta y)^2}$$

Then one would write quantities which are calculated from measured quantities as: $F = F_0 \pm \delta F$, where $F_0$ is the best estimate of $F$, and $\delta F$ is the propagated error.

So if we were given $x = 20 \pm 1$ and $y = 40 \pm 3$, then I can say $x_0 = 20$, $\delta x = 1$, $y_0 = 40$, and $\delta y = 3$. Then $F_0 = x_0 y_0 = 800$ and $\delta F = \sqrt{40^2 + 60^2} = 10\sqrt{52} \approx 70$. Therefore $F = 800 \pm 70$. This definition defines a clear procedure.

However, in one of my lab manuals, a different and more confusing approached was used to deal with error propagation in products. It uses fractional errors and more ambiguous notation:

$$\delta F = y \delta x + x \delta y$$

Then dividing by $F=xy$:

$$\frac{\delta F}{F} = \frac{\delta x}{x} + \frac{\delta y}{y}$$

Squaring and assuming no correlation in errors of independent variables:

$$\frac{\delta F}{F} = \sqrt{\biggl(\frac{\delta x}{x}\biggr)^2 + \biggl(\frac{\delta y}{y}\biggr)^2}$$

The reason this confuses me is that I'd expect fractional error be such that the $\frac{\delta F}{F}$ is the error in F divided by the actual value of $F$, and not, for instance, the best estimate, $F_0$. Thus in dividing by $F = xy$, the values $x$ and $y$ on the right hand side of the equations would also be the actual values of their respective quantities. In experiments, we don't always know the actual values.

But let's say we try to use what we do know. Given some $x = x_0 \pm \delta x$ and $y = y_0 \pm \delta y$, this version of the propagated error would be very burdensome to calculate. In fact, to be explicit:

$$\delta F = F - F_0 = F \sqrt{\biggl(\frac{\delta x}{x_0 \pm \delta x}\biggr)^2 + \biggl(\frac{\delta y}{y_0 \pm \delta y}\biggr)^2}$$

thus

$$F = F_0 \left[\sqrt{\biggl(\frac{\delta x}{x_0 \pm \delta x}\biggr)^2 + \biggl(\frac{\delta y}{y_0 \pm \delta y}\biggr)^2}\right]^{-1}$$

and you would have $(x \pm \delta x)^2$ and $(y \pm \delta y)^2$ terms to deal with. This would lead to having to consider all sorts of cases for different combinations of signs.

Assuming instead that $\frac{\delta F}{F}$ is actually the propagated error divided by the best estimate in F, we return to the first result.

• You've defined $F$ as $xy$ so why do you later add $x_0$ and $y_0$? Shouldn't $F_0=x_0y_0=800$? Or at least of that order... – Your Majesty Mar 29 '16 at 11:05
• You're right, x is not necessarily the same as x0 + dx. – user279043 Mar 29 '16 at 11:07
• Yes, my mistake.... :( – user279043 Mar 29 '16 at 11:08
• Ok but then the error is order $10\%$ Or am I wrong? – Your Majesty Mar 29 '16 at 11:09
• Yes the error is just under 10%, and the error in x and y are likewise are in the order just under 10%, so this looks right to me. – user279043 Mar 29 '16 at 11:11

You do it this way because you don't have the actual value of $F$, so it is impossible to divide by it. In the second part of your question, when you discuss the procedure from your lab manual, $F$ refers to the best estimate, i.e. the measured value. That's what you divide by.
Bear in mind that something like $F \pm \delta F$ (or, if you prefer, $F_0 \pm \delta F$) represents a probability distribution, not a number. Typically, the best estimate $F$ (or $F_0$) is the mean of the distribution, and $\delta F$ is its standard deviation. The numerical value you get by adding $F + \delta F$ (or $F_0 + \delta F$) is fairly meaningless; it's only one of many possible values that the true value of the measurement could be.