I am trying to approximate the pdf of the random variable

$Y = X + \frac{1}{b} \lceil b(Z-X) \rceil$

where $b$ is constant and $b \in \mathbb{N}$

$X$ is uniform random variable


and $Z$ is a normally distributed random variable independent of $X$


From experimentation I'm able to verify with a high level of certainty that I've correctly calculated the distribution of $\frac{1}{b} \lceil b(Z-X) \rceil$ but I'm not sure what to do from there. I originally thought convolving the distribution with the distribution of $X$ would produce the results I desired but soon realized that this was incorrect because $X$ is part of both parts of the sum.

What operation would I need to perform in order to estimate the distribution?


3 Answers 3



Consider $P(Y \geq y)$

$$ \begin{align*} P(Y \geq y) &= P\big((X+\frac{1}{b}\lceil{b(Z-X)\rceil)} \geq y\big)\\ &=P(\frac{1}{b}\lceil{b(Z-X)\rceil} \geq y-X\big)\\ &=P(\lceil{b(Z-X)\rceil} \geq by-bX\big)\\ &=P({b(Z-X)} > by-bX-1\big)\\ &=P(Z > y-\frac{1}{b}) \end{align*} $$

and support of $Y$ is $(-\infty, +\infty)$


Proof for $\lceil x \rceil \geq y \implies x > y-1$

$$ \begin{align} \lceil x \rceil &= min\{n \in {R} | \ n \geq x \}\\ &x-1 \leq \lceil x \rceil< x+1 \implies x+1 >y \implies x > y-1 \end{align} $$

Does the reverse hold? i.e. Given $x > y-1$ Is $\lceil x \rceil \geq y $

$ x > y-1 \implies x+1 >y$ and $x+1 > \lceil x \rceil$ But that does NOT imply $\lceil x \rceil \geq y$

  • $\begingroup$ Thank you for response, I'm a bit confused about the step where you removed the ceiling function. It appears that your saying the inequality $\lceil x \rceil \geq y$ is equivalent to $x \geq y - 1$ however I don't think that this is necessarily the case since ${x,y} = {0.9,1.1}$ evaluates to false for the first inequality and true for the second. It's possible that I'm unaware of some property of the probability function. $\endgroup$
    – jodag
    Sep 25, 2015 at 23:46
  • $\begingroup$ @jodag Indeed, I have edited my post. $\endgroup$ Sep 25, 2015 at 23:49
  • $\begingroup$ Thank you, I realized after I posted the comment that I had a typo and was probability in the midst of correcting it when you responded. It appears that even with the latest edit the fact still holds that $\lceil x \rceil \geq y$ implies $x > y-1$ but the reverse does not hold true, i.e. $x > y-1$ does not imply $\lceil x \rceil \geq y$ example $x = 0.9$ and $y=1.1$ for example. This leads me to believe that $P(\lceil b(Z-X) \rceil \geq by-bX) \neq P(b(Z-X) > by-bX-1)$ $\endgroup$
    – jodag
    Sep 25, 2015 at 23:56
  • $\begingroup$ True, the reverse does not hold true and it should be $\lceil x \rceil \geq y-1$. I am using the formal inequality, which does hold true by the very definition of ceiling function. $\endgroup$ Sep 26, 2015 at 0:12
  • $\begingroup$ Interesting, I don't understand but I'll take your word for it. The result of your derivation indicates that $Z$ is a normal distribution with the same variance as $Y$ and a mean of which has been shifted by $\frac{1}{b}$. $\endgroup$
    – jodag
    Sep 26, 2015 at 0:19

Since $Y=\frac{1}{b}(bX+\lceil bZ-bX \rceil)$ and $bX\sim U(0,1)$, $$Y|Z\sim \frac{1}{b}U(bZ,bZ+1)\sim U(Z,Z+\frac{1}{b})$$

Integrating over Z yields $$f_Y(y)=\int_{y-\frac{1}{b}}^y bf_Z(z)dz=b\big(F_Z(y) - F_Z(y-\frac{1}{b})\big)$$

  • $\begingroup$ Yes this certainly appears to be the correct solution thank you! $\endgroup$
    – jodag
    Sep 28, 2015 at 7:32

Thanks to @rightskewed I was able to come up with what I believe is the answer using a similar approach

Considering $P(Y = y)$

$$ \begin{align*} P(Y = y) &= P\big((X+\frac{1}{b}\lceil{b(Z-X)\rceil)} = y\big)\\ &=P\big(\lceil{b(Z-X)\rceil} = yb - Xb\big)\\ &=P(b(Z-X) \leq yb-Xb < b(Z-X)+1\big)\\ &=P(yb-Xb < b(Z-X)+1\big) - P(yb-Xb < b(Z-X)\big)\\ &=P(y-X < Z-X+\frac{1}{b}\big) - P(y-X < Z-X\big)\\ &=P(y-\frac{1}{b} < Z\big) - P(y < Z\big)\\ &=1-P(Z \leq y-\frac{1}{b}\big) - (1-P(Z \leq y)\big)\\ &=P(Z \leq y) - P(Z \leq y-\frac{1}{b}\big) \end{align*} $$

So that seems to indicate that

$$ \begin{align*} f_Y(y) = F_Z(y) - F_Z(y-\frac{1}{b}) \end{align*} $$

  • 1
    $\begingroup$ $P(Y=y)=0$ for a continuous r.v Y $\endgroup$ Sep 26, 2015 at 3:33
  • $\begingroup$ Of course it does >_< $\endgroup$
    – jodag
    Sep 26, 2015 at 5:44
  • $\begingroup$ My answer seems correct, btw. $\endgroup$ Sep 27, 2015 at 20:14

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