Quantiles of rounded up values and rounded up quantiles I am currently working with a rounded up values of continuous distribution and I am trying to find out some of its general properties. I have the following model:
$y = \lceil x \rceil$
where $x$ has continuous distribution and $\lceil x \rceil$ is a rounded up value of $x$. I am mainly interested in quantiles of distribution of $\lceil x \rceil$. While the quantiles of $x$ are straight forward, the rounded up version seems trickier. I could do simulations, but I have a feeling that there is a much simpler and neater way of working with its quantiles. So if we define quantiles $n$ and $k$ the following way:
$k: P(x < k) = 1 - \alpha,$
$n: P(\lceil x \rceil < n) \geq 1 - \alpha,$
then I have a very strong feeling that the following is true:
$\lceil k \rceil = n$
This means that the quantiles of rounded up $x$ are equal to rounded up quantiles of $x$. The simulations I've done in R so far support this finding and I cannot find any case, where this would be violated. However I need an appropriate mathematical proof for this, and I can't seem to find one. And, by the way, in case this is not always true, I need to know the conditions when this equation does not hold.
Does anyone have any references to the proof or any ideas of how to prove this?
 A: These are unusual definitions of quantiles, so let's be careful with them and with the distinctions between inequalities and equalities.  It makes a difference, because the answer is not what you think.
The analysis rests on the equivalence of these two statements whenever $n$ is an integer:
$$x \le n-1 \iff\lceil x \rceil \lt n.$$
I take your definition of "$n$" to mean that $n$ is the smallest value for which $\Pr(\lceil x \rceil < n)$ equals or exceeds $\alpha$.  Thus, it must be true that both
$$\Pr(x \le n-1) = \Pr(\lceil x \rceil \lt n) \ge 1-\alpha$$
and
$$\Pr(x \le n-2) = \Pr(\lceil x \rceil \lt n-1) \lt 1-\alpha.$$
These inequalities place $k$ within the interval $(n-2, n-1]$, because the continuity of the distribution assures $$\Pr(x \le k) = \Pr(x \lt k) = 1-\alpha$$ and, from what we have already deduced from the definition of $n$,
$$\Pr(x \le n-2) \lt 1-\alpha = \Pr(x \le k) = 1-\alpha \le \Pr(x\lt n-1).$$
Therefore 
$$\lceil k \rceil = n-1.$$

As an example, suppose the distribution is uniform on $[0, 3]$.  There are $1/3$ chances each that $\lceil x \rceil$ is $1, 2,$ or $3$, corresponding to the events $(0,1]$, $(1,2]$, and $(2,3]$, respectively.    Suppose $\alpha=1/2$.  Since $$\Pr( \lceil x \rceil \lt 2) = \Pr( \lceil x \rceil \le 1) = 1/3\lt 1-\alpha$$ and $$\Pr( \lceil x \rceil \lt 3) = 2/3 \ge 1-\alpha,$$we find $n=3$.  Nevertheless $k=3/2$ and $\lceil k \rceil = 2 =n-1$.
