# 1955 Miklós Schweitzer Problem 3 - concentration inequality

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!

• If you put 아ppreciate into google you go back to this question. Jul 16, 2022 at 23:52
• ㅋㅋㅋㅋ - fixed now thanks ;) Jul 17, 2022 at 0:09
• Perhaps Chebyshev will help Jul 17, 2022 at 18:00
• 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 Jul 17, 2022 at 18:07

I think we can use the Camp-Meidell inequality (, ).

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.

References.

 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

• 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 Jul 18, 2022 at 10:49
• @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 $g(\cdot)$. Jul 18, 2022 at 13:30