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In descriptive statistics, continuous variables are often presented using class intervals of uniform width, for instance:

Annual salary (€)     Frequency
[20000,  40000[          10
[40000,  60000[          25
[60000,  80000[           5

and one is told to compute the mean or variance by assuming that all data in a class are at the center of the class. For instance, one would compute the mean annual salary from the table above as

$$ \frac{10 \times 30000 + 25 \times 50000 + 5 \times 70000}{40} = 47500. $$

It is easy to show that the uncertainty (maximal error) when computing the mean this way is half the width of a class interval (in this case, $10000$). In other words, the true mean is in the interval $[37500, 57500[$.

My question is: are there simple bounds for the uncertainty of the variance?

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There are. We can regard the errors introduced by the bunching of the data as a second random variable $Y$ added to the original data $X$. Then the variance of the sum is

$$\sigma_{X+Y}^2=\sigma_X^2+\sigma_Y^2+2\sigma_{XY}\;,$$

where $\sigma_{XY}$ is the covariance of $X$ and $Y$. Since $|\sigma_{XY}|\le\sigma_X\sigma_Y$, we have

$$(\sigma_X-\sigma_Y)^2=\sigma_X^2+\sigma_Y^2-2\sigma_X\sigma_Y\le\sigma_{X+Y}^2\le\sigma_X^2+\sigma_Y^2+2\sigma_X\sigma_Y=(\sigma_X+\sigma_Y)^2\;.$$

Thus, while for uncorrelated variables the variances add, for correlated variables the standard deviation may lie anywhere between the difference and the sum of the standard deviations.

It remains to find the maximal value of $\sigma_Y$. The magnitude of each error is at most the half-width $\Delta$ of the intervals, and the maximal standard deviation is thus attained if half the errors are $\Delta$ and half are $-\Delta$, for a standard deviation of $\sigma_Y=\Delta$. Thus we have the simple result that the standard deviation can be off by up to $\Delta$, just like the mean.

In case you prefer to work with the variances, the corresponding bounds on the exact variance $\sigma^2$ in terms of the approximated variance $\sigma'^2$ are

$$(\sqrt{\sigma'^2}-\Delta)^2\le\sigma^2\le(\sqrt{\sigma'^2}+\Delta)^2\;.$$

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  • $\begingroup$ @user115: You're welcome! $\endgroup$
    – joriki
    Feb 29, 2012 at 0:28

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