Let $X$ be a random variable such that $P(X < t + c) \ge 1-e^{-t^2/2}$ for $t > 0$ and $c > 0$ is some known constant.
I am interested in deriving upper bounds for $E[|X|^k]$ for $k \in \mathbb{Z}$.
My approach is as follows:
$E[|X|^k]^{1/k} = E[|X-c+c|^k]^{1/k} \le E[|X-c|^k]^{1/k} + c$ by the triangle inequality.
Hence, it only remains to bound $E[|X-c|^k]^{1/k}$.
I'm going to exploit the fact that $E[|X-c|^k]^{1/k} = \int_0^{\infty} P(|X-c|^k \ge u)\,du = \int_0^{\infty} ku^{k-1} P(|X-c| \ge u)\,du$.
However, since $P(|X-c| \ge u) \ge P(X-c\ge u)$, I could not invoke the tail probability that I have for $X$.
Any suggestions on how to proceed? If I have results on $P(d – t < X) \ge 1-e^{-t^2/2}$ for some constant $d$, $t>0$, will that be needed here? Thanks!
Best Answer
You can use the following approach
$$ E[|X|^k]=\int_{0}^{\infty} t^{k-1}P(|X|>t)\,dt = \int_{0}^c t^{k-1}P(|X|>t)\,dt + \int_{c}^{\infty} t^{k-1}P(|X|>t)\,dt \\ \leq c^{k-1} + \int_{0}^{\infty} (z+c)^{k-1}P(|X|>z+c)\,dz \leq c^{k-1} + \int_{0}^{\infty} (z+c)^{k-1}e^{-\frac{z^2}{2}}\,dz. $$ In the last inequality, we have used $P(|X|>z+c) \leq e^{-\frac{z^2}{2}}$.
You will need some assumptions on the lower tail. Otherwise, you can construct a random variable whose upper tail is Gaussian and lower tail is like Cauchy, and therefore you will not be able to prove any meaningful upper bound even for $E|X|$. Let me know if this clarifies your doubts.