Upper bound on expectation given probability

Question

Let $X$ be a positive random variable. I know that $\mathbb{P}[X \leq a] \geq q$. Any hints about how to find an upper bound on $E[X]$ in terms of $q$ and $a$? Using Markov inequality, I can obtain $E[X] \geq a(1-q)$. But I need an upper bound, not a lower one. Any hints? Thanks!

Answer 1

If $X=a$ with probability $q$ and $N$ with probability $1-q$ (where $N>a$) then $EX=aq+N(1-q)$. So there are examples with $EX$ as large as you want for a given $a$ and $q$.

Answer 2

There does not exist any such upper bound. Take $\mathbf{P}(X=0)=\frac{1}{2}$ (here $a=0$ and $q=\frac{1}{2}$) and $\mathbf{P}(X=n)=\frac{1}{2}$, for $n$ integer. Then $$ \mathbf{E}(X)=\frac{n}{2}. $$ So we can make $\mathbf{E}(X)$ as large as we want. However, if we have $$ \forall a\geq 0: \quad \mathbf{P}(X\geq a) \leq Q(a), $$ then you may use the integral identity $$ \mathbf{E}(X)=\int_{0}^\infty \mathbf{P}(X\geq a) \mathrm{d}a $$ to get an upper bound.