Expected value exponential inequality non-negative random variable

Question

Let $X$ be a non-negative random variable and $p \geq e$, $q > 0$ be two constant values such that $$ P [X \geq x] \leq p e^{-x^2/q^2} \quad \forall x \geq 0. $$ Prove that $$ \mathbb{E}[X] \leq q(1+\sqrt{\log p}). $$

There's my first attempt. Using the identity $$ \mathbb{E}[X] = \int_0^\infty (1-F(x)) d x - \int_{-\infty}^0 F(x) d x $$ and the fact that $X \geq 0$ I obtain: $$ \mathbb{E}[X] = \int_0^\infty (1-F(x)) d x $$ and using the hypothesis I get $$ \mathbb{E}[X] \leq \int_0^{+\infty} p e^{-x^2/q^2} d x = pq \int_0^{+\infty} e^{-z^2} dz = \frac{pq \sqrt{\pi}}{2} $$ where the last equality is given by the gaussian integral.

But then I'm stuck.

The second attempt I tried was to partition the events set in some smart way, that is using $( X \geq q )$ or $(X \geq \sqrt{\log p} )$ but I couldn't go anywhere.

Answer 1

By replacing $X$ with $X/q$, we can w.l.o.g. assume $q = 1$.

Note that the inequality $\mathbb{P}(X \ge x) \le p e^{-x^2}$ is a strong inequality for large $x$, but rather poor for small $x$. Indeed, for $x = 0$ this only yields $\mathbb{P}(X \ge x) \le p$, which by assumption is greater than $e$. Since Probabilities are bounded by $1$, this inequality is poor when $pe^{-x^2} > 1$, or equivalently $x < \sqrt{\log(p)}$. We should therefore split up the integral:

$$\mathbb{E}[X] = \int_0^\infty \mathbb{P}(X \ge x) \,dx = \int_0^\sqrt{\log(p)} \mathbb{P}(X \ge x) \, dx + \int_{\sqrt{\log(p)}}^\infty \mathbb{P}(X \ge x) \, dx.$$

Now the first term is bounded by $\sqrt{\log(p)}$. For the second term, use the inequality in the assumption, as well as $e^{-x^2} \le x e^{-x^2}$ for $x \ge 1$.

Answer 2

For $z\geq 0$

$$\Bbb E [X] = \Bbb E [X 1_{\{X< z\}}] + \Bbb E [X1_{\{X\ge z\}}] \leq z \Bbb P (X< z) + \int_0^\infty \Bbb P(X1_{\{X \ge z\}} \geq x) dx\\ =z \Bbb P (X< z) + \int_0^z \Bbb P(X \ge z) dx +\int_z^\infty \Bbb P(X \geq x) dx \\ = z + \int_z^\infty \Bbb P (X\geq x) dx = z + \int_0^\infty \Bbb P (X\geq x+z)dx\\ \leq z+\int_0^\infty e^{- \left( \frac{x^2}{q^2} + \frac{z^2}{q^2} - \log (p) \right)} e^{-\frac{2xz}{q^2}}dx\\ = z+ q \int_0^\infty e^{- \left( x^2 + \frac{z^2}{q^2} - \log (p) \right)} e^{-\frac{2xz}{q}} dx $$

Thus for $z= q \sqrt{\log (p)}$ we have

$$\Bbb E [X] \leq q \sqrt{\log (p)} + q \int_0^\infty e^{- x^2} e^{-2x \sqrt{\log (p)}} dx \leq q \sqrt{\log (p)} + q \int_0^\infty e^{- x^2} e^{-2x } dx\\ =q \sqrt{\log (p)} + q \int_0^\infty e^{- (x^2 +2x +1) +1} dx = q \sqrt{\log (p)} + q \int_0^\infty e^{- (x+1)^2 +1} dx\\ = q \sqrt{\log (p)} + q e^1 \int_1^\infty e^{- x^2} dx$$ Using this bound A bound for error function $$\int_1^\infty e^{- x^2} dx \leq \frac{e^{-1}}2$$ we get $$\Bbb E [X] \leq q \sqrt{\log (p)} + \frac q 2$$