sum of Random variable and function of that random variable 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
$X$~$U(0,\frac{1}{b})$
and $Z$ is a normally distributed random variable independent of $X$
$Z$~$N(\mu,\sigma^2)$
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?
 A: Hint:
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 $(-\infty, +\infty)$
Edit:
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 $\lceil x \rceil \geq y$
A: 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 $$f_Y(y)=\int_{y-\frac{1}{b}}^y bf_Z(z)dz=b\big(F_Z(y) - F_Z(y-\frac{1}{b})\big)$$
A: 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
$$
\begin{align*}
f_Y(y) = F_Z(y) - F_Z(y-\frac{1}{b})
\end{align*}
$$
