# Chebyshev inequality in terms of RMS

I'm self studying the book Introduction to Applied Linear Algebra – Vectors, Matrices, and Least Squares

In page 48, the author write: "It says,for example, that no more than 1/25 = 4% of the entries of a vector can exceed its RMS value by more than a factor of 5."

I need more explain about it. Especially about why the factor is 5?

According to Chebyshev's_inequality, the probability of a value to deviate more than $$k=5$$ standard deviations from the mean is at most $$1/k^2$$.

When applied to vectors in your specific case, and following the book you cited, let $$k$$ be the number of elements of the vector $$\vec{x}=(x_1,\ldots,x_n)$$ such that $$||x_i|| \geq a > 0$$.

Hence $$\|\vec{x}\|^2 = \sum x_i^2 \geq k a^2 + (n - k) \times 0$$, which means that we have $$k$$ values larger than $$a^2$$ and the others $$n-k$$ values are at least zero.

Since the root mean square value is $$\operatorname{rms}(\vec{x}) = \sqrt{\frac{\|\vec{x}\|^2}{n}}$$, it follows that $$\operatorname{rms}(\vec{x})^2 = \frac{\|\vec{x}\|^2}{n} \geq \frac {k a^2}{n}$$.

Therefore, we get the final expression that says

$$\frac {k}{n} \leq \left( \frac{\operatorname{rms}(\vec{x})}{a} \right) ^2$$

So, following the example, where $$a = 5 \operatorname{rms}(\vec{x})$$, we have that $$\frac {k}{n} \leq \left( \frac{1}{5} \right) ^2 = 4 \%$$, so, the fraction of elements of the vector larger (in absolute value) than $$5\operatorname{rms}$$ is at most $$4\%$$.

If we chose another number, say $$a = 2 \operatorname{rms}(\vec{x})$$, we would have that $$\frac {k}{n} \leq \left( \frac{1}{2} \right) ^2 = 25 \%$$.