Source: 1955 Miklós Schweitzer Problem 3

Let the density function $f(x)$ of a random variable $\xi$ be an even function; let further $f(x)$ be monotonically non-increasing for $x > 0$. Suppose that $$D^2= \int_\mathbb{R} x^2 f(x)\;dx$$ exists. Prove that for $\lambda > 0$, $$P(\left|\xi\right| \geq \lambda D)\leq \frac{1}{1+\lambda^2}.$$

Attempt: As $f(\cdot)$ is even, $\mathbb{E}(\xi)=0$. I apply Cantelli's inequality to obtain

$$P(|\xi| \geq \lambda D)\leq \frac{2}{1+\lambda^2}.$$

Well, I guess it wasn't meant to be that easy haha... so I assume I must do something using monotonicity of $f(\cdot)$. Alternatively, I wanted to apply some exponential tail bound - for instance ideally yielding something similar to the following:

$$P(|\xi| \geq \lambda D)\leq^* e^{-g(\lambda)}\leq \frac{1}{1+\lambda^2},$$

where $\leq^*$ would follow by showing $\xi$ is sub-Gamma/Gaussian/etc. but I didn't put much thought into this approach.

Question: Is there a way to just slightly change the initial lazy approach I took - i.e. maybe another concentration inequality in a clever way? Or is the approach totally different? I appreciate any help - thanks!

  • $\begingroup$ If you put 아ppreciate into google you go back to this question. $\endgroup$ Commented Jul 16, 2022 at 23:52
  • $\begingroup$ ㅋㅋㅋㅋ - fixed now thanks ;) $\endgroup$
    – user672552
    Commented Jul 17, 2022 at 0:09
  • $\begingroup$ Perhaps Chebyshev will help $\endgroup$ Commented Jul 17, 2022 at 18:00
  • $\begingroup$ It gives $\frac{1}{\lambda^2}$ as an upper bound which doesn't help. Are you maybe suggesting to relate this to the one-sided version of Chebyshev's? To be honest, I see no way of doing so $\endgroup$
    – user672552
    Commented Jul 17, 2022 at 18:07

1 Answer 1


I think we can use the Camp-Meidell inequality ([1], [2]).

If $\lambda < \frac{2}{\sqrt 3}$, we have $$P(|\xi | \ge \lambda D ) \le 1 - \frac{\lambda}{\sqrt 3} \le \frac{1}{1 + \lambda^2}.$$

If $\lambda \ge \frac{2}{\sqrt 3}$, we have $$P(|\xi | \ge \lambda D ) \le \frac{4}{9\lambda^2} \le \frac{1}{1 + \lambda^2}.$$

We are done.


[1] https://vrs20.amsi.org.au/wp-content/uploads/sites/75/2020/01/hanna_matthew_vrs-report.pdf

[2] Thomas M. Sellke and Sarah H. Sellke, “Chebyshev Inequalities for Unimodal Distributions,” The American Statistician Vol. 51, No. 1 (Feb., 1997), pp. 34-40 (7 pages). https://www.jstor.org/page-scan-delivery/get-page-scan/2684690/0

  • $\begingroup$ Wow… beautiful thanks - I had to run through the proof as I haven’t heard of it before - it is definitely the way to expand my initial lazy approach $\endgroup$
    – user672552
    Commented Jul 18, 2022 at 10:49
  • $\begingroup$ @pSrIoGcNeAsLs You are welcome. I guess the official solution (if any) does not apply the Camp-Meidell inequality. Perhaps there is some approach to use $$P\Big(g(|\xi|/D) \ge g(\lambda)\Big) \le \frac{\mathbb{E}(g(|\xi|/D))}{g(\lambda)}$$ for some $g(\cdot)$. $\endgroup$
    – River Li
    Commented Jul 18, 2022 at 13:30

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