1955 Miklós Schweitzer Problem 3 - concentration inequality

Question

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!

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.

References.

[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