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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?

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  • $\begingroup$ Your notation is difficult to decipher. What exactly is $n$? How can you hope to define it with an inequality? I suspect that once you are able to write down a clear mathematical definition of $n$, most of the demonstration you seek will be complete. $\endgroup$ – whuber Jul 14 '17 at 14:16
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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 $\lceil k \rceil = 2 =n-1$.

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  • $\begingroup$ Thank you for the answer! It makes sense. And I was inaccurate in my definitions. In order for the original equality $\lceil k \rceil = n$ to hold the definition of quantiles should be different: $P( \lceil x \rceil \leq n) \geq 1 - \alpha$. Your proof makes sense, thanks! How should I refer to it in a paper? $\endgroup$ – Ivan Svetunkov Jul 14 '17 at 18:55
  • $\begingroup$ And yes, you are quite right that $n$ is the smallest integer number. $\endgroup$ – Ivan Svetunkov Jul 14 '17 at 19:10
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    $\begingroup$ This is simple enough that you ought to be fine just asserting the result. If you want to cite this thread, you could emulate the example in the question at stats.meta.stackexchange.com/questions/2900/…. $\endgroup$ – whuber Jul 14 '17 at 19:21

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