Lower bound for left tail probability

Question

If we consider $X$ to be a non-negative random variable, then can we find a lower bound on the left tail probability, that is $\mathbb{P}(X \leq t) \geq C$ where t and C are two positive constants?

Answer 1

This is an extremely broad question, and has trivial answers (e.g. take $t>0$ large enough that $C:=\mathbb P(X\le t)>0$). I'm going to interpret your question in the following way:

Given $t>0$, does there exist a positive constant $C=C(t,m)$ such that $$\mathbb P(X\le t)\ge C$$ for every non-negative random variable $X$ such that $\mathbb E[X]=m$?

This is a natural question, since if you reverse the inequalities you get Markov's inequality for $C=\frac{m}t$. (If you ask for higher moments instead of $\mathbb E[X]$, clearly we also have tighter bounds such as Chebyshev or Chernoff; more control of the tails obviously gives you more control in the inequality. I'm going to stick to the first moment.) In fact, as BGM points out in the comments, this immediately gives you $$\mathbb P(X\le t)\ge1-\frac{m}t.$$ However, this is not very interesting - we are interested in "tail" probabilities after all, and so it makes sense to restrict to the case where $t\le m$ (in which case the above bound is useless).

Unfortunately, it turns out that with such a restriction the answer to the above question is "no". To see this, assume without loss of generality that $m>1$ and let $X_n$ have distribution $$\mathbb P(X_n=m+a_n)=1-a_n,\quad\mathbb P(X_n=m+1)=\frac{a_n(m-1+a_n)}{m+1},\quad\mathbb P(X_n=0)=\frac{a_n}{m+1}$$ where we assume $1>a_n\to0$. Then $\mathbb E[X_n]=m$ and, for any $t\le m$, $\mathbb P(X\le t)=\frac{a_n}{m+1}\to0$ as $n\to\infty$, so there is no uniform lower bound.

You can probably use similar tricks for the analogous question where $C$ can depend on the second moment or even higher, I haven't checked this.