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The formula for Chebyshev's inequality for the asymmetric two-sided case is: $$ \mathrm{Pr}( l < X < h ) \ge \frac{ 4 [ ( \mu - l )( h - \mu ) - \sigma^2 ] }{ ( h - l )^2 } . $$ What I don't understand is how it behaves when the interval increases.

To simplify things, let $\mu = 0$, $\sigma = 1$ and $l = -1$. In this case, we get $$\mathrm{Pr}( -1 < X < h ) \ge \frac{ 4 ( h - 1 ) }{ ( h + 1 )^2 } .$$

If we restrict ourselves to the case that $h > 3$, the above is a decreasing function of $h$. For me, it makes no sense, since in this situation the probability should not decrease with $h$ because the interval is getting bigger. Moreover, when $h \to +\infty$ the expression goes to zero!

I acknowledge that the formula is a lower bound on the probability, but it "fells" wrong for large $h$.

Am I missing something?


Edit:

In case we have a generic $l < -1$, and keeping $\mu = 0$ and $\sigma = 1$ for simplicity, we get $$\mathrm{Pr}( l < X < h ) \ge \frac{ 4 ( |l| h - 1 ) }{ ( h + |l| )^2 } .$$

The conclusion is similar to the previous example, since this function is decreasing for $$h > |l| + \frac{2}{|l|} \ .$$

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  • $\begingroup$ The Chebyshev inequality does not provide useful answers when either boundary lies at distance $\sigma$ or less from the mean. Try your example with $l$ strictly smaller than $-1$ and $h$ strictly larger than $+1$. $\endgroup$ Apr 9, 2020 at 0:28
  • $\begingroup$ @DilipSarwate thank you for your comment. I've added a generic example considering $l < -1$. The conclusion is similar, although requiring a possibly larger $h$ in order to observe the decrease. $\endgroup$ Apr 9, 2020 at 0:56

2 Answers 2

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What you are observing here is an idiosyncracy of the general Chebyshev inequality. Generally speaking, the inequality gets better as the midpoint of the interval gets closer to the mean $\mu$ and it also gets better as the length of the interval increases. However, if you hold one of the bounds constant and move the other one out to expand the interval, eventually you pass a point where the midpoint of the interval is now far from the mean, and the effect of further movement of the midpoint away from the mean outweighs the effect of expanding the length of the interval. As such, the probability bound gets worse rather than getting better.


Describing the phenomenon in greater generality: A simpler and more general way to frame this phenomenon is in terms of the standardised part-lengths of the interval, which I will denote by:

$$k_- = \frac{\mu-l}{\sigma} \quad \quad \quad \quad \quad k_+ = \frac{u-\mu}{\sigma}.$$

The lower probability bound given by the Chebyshev inequatity can be written as:

$$B(k_-, k_+) = 4 \cdot \frac{k_- k_+ - 1}{(k_- + k_+)^2},$$

and the bound is "binding" (i.e., greater than zero) if and only if the interval contains the mean $\mu$ in its interior and we also have $k_- k_+>1$. If you hold one of these arguments constant, it can easily be shown that this function is strictly quasi-concave in the other argument. In particular, holding $k_-$ constant and varying $k_+$ gives the maximiser:

$$\underset{k_+}{\text{arg max}} \ B(k_-, k_+) = \hat{k}_+ = k_- + \frac{2}{k_-} \quad \quad \quad \quad \quad \underset{k_+}{\text{max}} \ B(k_-, k_+) = B(\hat{k}_+) = \frac{k_-^2}{k_-^2+1}.$$

The bound function is increasing up to $k_+ = \hat{k}_+$ and then after this it decreases. As stated above, this occurs because after we get past this point, the negative effect of moving the midpoint of the interval away from the mean outweighs the positive effect of making the interval wider.

Of course, the true probability of the interval cannot be getting smaller as you move a boundary point outward to make the interval larger. Thus, you may legitimately use the probability bound at $k_+ = \hat{k}_+$ whenever you have $k_+ > \hat{k}_+$ (and it is desirable to do this, since that lower bound is larger). Indeed, this is what is done in adjust versions of the interval. In the adjusted version, we take the generalised Chebyshev interval to be given by the formula you have written, but adjusted so that it doesn't get smaller as you move outward.

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  • $\begingroup$ Thank you @Ben. You posted your answer while I was writing mine and what you said is coherent with Ferentino's 1982 paper that points to Selberg's inequalty. As far as I understood, what you posted is equivalent to the "second" branch of Selberg's inequalty - and I've found your explanation easier to follow. $\endgroup$ Apr 10, 2020 at 2:40
  • $\begingroup$ Yes, it looks like my formula is equivalent to Selberg's inequality, as presented in Ferantino. $\endgroup$
    – Ben
    Apr 10, 2020 at 3:14
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The reference for the formula for Chebyshev's inequality for the asymmetric two-sided case, $$ \mathrm{Pr}( l < X < h ) \ge \frac{ 4 [ ( \mu - l )( h - \mu ) - \sigma^2 ] }{ ( h - l )^2 } , $$ points to the paper by Steliga and Szynal (2010). I've done some further research and Steliga and Szynal cite Ferentinos (1982). And it turns out that Ferentinos states that the above formula is part of Selberg's inequalty (1940) which I could not access (and it's written in German).

Nevertheless, Ferentinos provides enough information to understand what is happening. Ferentinos states that: $$ \mathrm{Pr}( l < X < h ) \ge \begin{cases} \displaystyle \frac{ 4 [ ( \mu - l )( h - \mu ) - \sigma^2 ] }{ ( h - l )^2 } & \text{if } (\mu - l) (h - \mu) \geq \sigma^2 \text{ and } (\mu - l) (h - \mu) - k^2 \leq 2 \sigma^2 \\ \displaystyle \frac{k^2}{\sigma + k^2} & \text{if } (\mu - l) (h - \mu) - k^2 \geq 2 \sigma^2 \end{cases} \ , $$ where $k = \min(\mu-l, h-\mu)$ and $l < \mu < h$ .

Therefore, provided $\mu-l$ and $h-\mu$ are large enough, but not too large, the top formula is applied but when either $\mu-l$ or $h-\mu$ are too large, the second expression should be used.

Hence, this second expression will be the relevant one for the examples in the question. Specifically, for the first example we will get $$\mathrm{Pr}( -1 < X < h ) \ge \frac{1}{2} \ , $$ and for the second example we will get $$\mathrm{Pr}( l < X < h ) \ge \frac{l^2}{l^2+1} \ . $$


As I was writing this text, the answer by Ben was posted which provided me further insight.

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