IQR , standard deviation

Question

there is a a sample such that the standard deviation is significantly smaller than IQR? I think that no but how to prove this?

Answer 1

Chebyshev's inequality is derived by imagining a situation where as much of the distribution as possible is thrown into the outliers—that is, into two equal clusters at equal distances from the center.

The situation you describe is similar, and can be addressed by imagining a similarly extreme distribution: We put $1/4+\varepsilon$ of the distribution at the low end (say, at $-u$), $1/2-2\varepsilon$ in the middle (at $0$), and again $1/4+\varepsilon$ at the high end (at $+u$). In the limit, such a distribution has an IQR of $2u$, and a standard deviation of $u/\sqrt2$; that is, the spread between one SD down and one SD up is $u\sqrt2$. The ratio between twice the SD and the IQR—a ratio of $\sqrt2/2$—is, I believe, as small as that can be.

Answer 2

Let's fix IQR. There are 50% of points that lie inside of it, and 25% of points that lie above and 25% below. While we move the points around and not crossing the boundary of IQR, IQR will remain the same. Then it's easy to show that the smallest SD will be when all of the points from “above” category lie exactly on the top boundary, all of the points from “below” on bottom boundary, and all internal points exactly in the middle.

Thus, the smallest SD is: $$ \text{SD} = \sqrt{0.25\times (-\text{IQR}/2)^2 + 0.5\times 0^2 + 0.25\times (\text{IQR}/2)^2} = \frac{\text{IQR}}{2\sqrt2} = 0.3536\ \text{IQR} $$