In chapter 3, Norm and Distance of Introduction to Applied Linear Algebra by Boyd, an example explaining the Chebyshev inequality for standard deviation is given as:

Consider a time series of return on investment, with a mean return of 8%, and a risk (standard deviation) 3%.

The author states -

By the Chebyshev inequality, the fraction of periods with a loss (i.e., $x_i$ ≤ 0) is no more than $(\frac{3}{8})^2 = 14.1%$. (In fact, the fraction of periods when the return is either a loss, $x_i$ ≤ 0, or very good, $x_i$ ≥ 16%, is together no more than 14.1%.)

Currently, I know that the Chebyshev inequality states that If k is the number of entries of x that satisfy |$x_i$ − avg(x)| ≥ a, then $\frac{k}{n}$$(\frac{std(x)}{a})^2$ .

But I am not able to correlate this to the example. Can someone please explain, how $(\frac{3}{8})^2$ is inferred?


The way to see that is to remember he wants to know the fraction of periods incurred in losses, i.e. $$x_i <= 0$$.

If the mean return is 8% it implies that in order to have a loss, $$a$$ has to be equal or greater than 8. So $$a = 8$$ here. Also the reason he says the fraction of periods is no more than 14.1% for either a loss, $$x_i <= 0$$ or for very good returns, $$x_i >= 16$$.


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