# Use expected bound to derive high probability bound

Suppose that for some integer $p\geq 2$, we have

$$(\mathbb{E}[\lVert \mathbf{X} \rVert^p])^{\frac{1}{p}}\leq C$$ where $C>0$ is a constant as upper bound.

Now, my question is how to convert this into a high probability result? Specifically, I want to have the result like

with probability at least $1-\delta$, $$\lVert \mathbf{X} \rVert \leq D$$ Here, $D = f(C,\delta)$. Besides, hope the order of $\delta$ is $log$-order.

A simple method is use Markov inequality, then $D=C/\delta^{1/p}$, so is it possible to be tighter for a high probability bound?

Let $X$ be a real random variable such that $\mathbb{P}(X=0)=1-\delta$ and $\mathbb{P}(X=D)=\delta$ where $D=C/\delta^{1/p}$. Then we have $(\mathbb{E}(X^p))^{1/p}=C$ so one cannot hope to get a better bound than $C/\delta^{1/p}$ without any other assumptions.