Assume that $F_X$ is the CDF of the random variable X and $Q_X$ its quantile function.Prove that $Q_X(F_X(t)) \le t $ and $F_X(Q_X(p)) \ge p $.
I substituted $F_X(t)=P[X \le t]$ and then substituted that in the quantile function $Q_X(p)=min(x:F_X(x) \ge p)$, in this case p is the CDF but i am confused how to go further,seems pretty straightforward but i am missing something. Can some one please help
My first guess is that you might have missed the property that CDF is a non-decreasing function of $x$. Hence: $$ \forall x(x : F_X(x)≥p) \implies F_X(x)≥p $$ It is almost obviously true for a specific $x$ to satisfy the above derivation: $$ x = min(x : F_X(x)≥p) \implies F_X(x)≥p $$ Together with the definition of quantile function, we have: $$ F_X(Q_X(p)) \geqslant p $$ For the first inequality, I would consider a intuitive answer: If $F_X(x)$ is strictly increasing in a neighborhood around point $x = t$, then the equality is attained. The only way to violate the equality is that $x = t$ appears as the right endpoint of a flat region of CDF. In this way, the quantile is the left endpoint of this flat region, which is obviously less than $t$ by their relative position