3

Suppose I have a random variable $X$, and a high probability bound of the form: $$P(X\geq f(\delta))<\delta$$ where $f$ is a positive function in $\delta>0$. In general, how can I use this to bound the expected value of $X$?

We know that \begin{align*} E[X]\leq \int_{0}^{\infty} P(X\geq x) dx =\int_{0}^{f(\delta)} P(X\geq x) dx+\int_{f(\delta)}^{\infty} P(X\geq x) dx\\ \leq f(\delta)+\int_{f(\delta)}^{\infty} P(X\geq x) dx \end{align*} However, this is as far as I have gotten....

Any ideas would be much appreciated!

Merry
  • 998

1 Answers1

0

Without saying anything about $f$, I believe this is not possible.

But suppose you have $f(\delta) = B + c\ln^m(1/\delta)$ for some positive reals $B, c > 0$ and some positive integer $m > 0$ for all $\delta \in (0,1)$. Then by writing the expected value of a nonnegative random variable into an integral and using $\delta = e^{-(t/c)^{1/m}}$ for any $t > 0$ you get $$ \mathbb{E}[X] - B \le \mathbb{E}\big[\max\{X-B, 0\}\big] = \int_0^\infty\mathbb{P}\{X > B+t\} dt \le \int_0^\infty e^{-(t/c)^{1/m}} dt = m!\,c , $$ hence $\mathbb{E}[X] \le B + m!\,c$.

The formula I showed above is mentioned in (2) in this paper in the context of statistical estimation. As pointed out here, this paper also explores the topic in the context of statistical estimation.

The integral on the right hand side in my derivation might not be that obvious. I used the formula $\int_0^\infty e^{-ax^b}dx = (1/b)a^{-1/b}\Gamma(1/b)$ with $a = (1/c)^{1/m}$ and $b = 1/m$ where $\Gamma(\cdot)$ is the Gamma function satisfying $\Gamma(m) = (m-1)!$ for positive integer $m > 0$.

gabalz
  • 371