Error Propagation

Topics: Numerical Analysis - Error

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Let $p$ and $q$ be two exact values, and $\hat{p}$ and $\hat{q}$, their respective approximate values; $\hat{p}$ and $\hat{q}$ have the errors ${\epsilon }_p$ and ${\epsilon }_q$.

That is:

$$p=\hat{p}+{\epsilon }_p$$

$$q=\hat{q}+{\epsilon }_q$$

Effect on Addition and Subtraction §

The effect of the errors on addition and subtraction is:

$$p\pm q=(\hat{p}+{\epsilon }_p)\pm (\hat{q}+{\epsilon }_q)=(\hat{p}\pm \hat{q})+({\epsilon }_p+{\epsilon }_q)$$

Notice that, effectively, the errors are added when both adding and subtracting $p$ and $q$ with their approximate values.

Effect on Product and Division §

In the case of the product, we have that:

$$pq=(\hat{p}+{\epsilon }_p)(\hat{q}+{\epsilon }_q)=\hat{p}\hat{q}+\hat{p}{\epsilon }_q+\hat{q}{\epsilon }_p+{\epsilon }_p{\epsilon }_q$$

If the approximate values $\hat{p}$ and $\hat{q}$ are greater that $1$, then the error will grow in a significative way.

Given $p\ne 0$ and $\hat{p},\hat{q}$ good approximations, we can show that the relative error of the product of two numbers is approximately the sum of the relative errors of the individual approximations:

$$E_{r_{pq}}=\frac{pq-\hat{p}\hat{q}}{pq}\approx \frac{{\epsilon }_p}{p}+\frac{{\epsilon }_q}{q}+0=E_{r_p}+E_{r_p}$$