How to find the CDF of a random variable uniformly distributed around another random variable? I'm working on some game theory models of incomplete information (which I've posted about a few times here). I think this question is pretty straightforward though, so the actual context is unimportant.
Suppose there is a random variable uniformly distributed between $b_l$ and $b_h$, i.e. $B \sim U(b_l, b_h)$. Someone draws $b$ from $B$, and then defines another random variable $X \sim U(b - \epsilon, b + \epsilon)$. You know $b_l, b_h, \epsilon$, and the distributions of $B$ and $X$, but you do not know $b$, the actual number drawn from the $B$ distribution.
What is the CDF of $X$? My intuition is that the mass goes from $b_l - \epsilon$ to $b_h + \epsilon$ and is triangular, with the peak at $P(X < \frac{b_l + b_h}{2}) = .5$, but I'm really not sure. Thanks in advance for any help!
Edit: $b_l, b_h$, and $\epsilon$ are all predefined and fixed.
 A: Your variable $X$ is simply a sum of two randoms $$x=b+e$$, where the randoms are from uniform distributions $b\sim  U(b_l,b_h)$ and $e\sim U(-\varepsilon,\varepsilon)$
If $b_h-b_l=2\varepsilon$, then $X$ is from a triangular distribution. Otherwise, it'll be a trapezoidal distribution. It must very easy to come up with CDF.
UPDATE
Her's one way of calculating the CDF. Start with a PDF $f(x)$, which is easily defined as follows:


*

*$\frac{x-b_l+\varepsilon}{(b_h-b_l)2\varepsilon}$, when $x\in [b_l-\varepsilon,b_l+\varepsilon]$

*$\frac{1}{b_h-b_l}$, when $x\in [b_l+\varepsilon,b_h-\varepsilon]$

*$\frac{-x+b_h+\varepsilon}{(b_h-b_l)2\varepsilon}$, when $x\in [b_h-\varepsilon,b_h+\varepsilon]$


Obviously, these are when $2\varepsilon\le b_h-b_l$.
Now, to get the CDF simply take the integral $\int_{b_l-\varepsilon}^{b_h+\varepsilon}f(x)dx$, which is easy but too long for me to type
A: First recall that
$$
f(X|X \in [b-\epsilon,b+\epsilon])=  \frac{1}{b+\epsilon - (b-\epsilon) } =\frac{1}{2\epsilon}
$$
and
$$
f(X) = f(X|X \in [b-\epsilon,b+\epsilon]) \times Pr(X \in [b-\epsilon,b+\epsilon])+ f(X|X \not\in [b-\epsilon,b+\epsilon]) \times Pr(X \not\in [b-\epsilon,b+\epsilon])
$$
but since $f(X|X \not\in [b-\epsilon,b+\epsilon]) =0$ the above simplifies to 
$$
f(X) = f(X|X \in [b-\epsilon,b+\epsilon]) \times Pr(X \in [b-\epsilon,b+\epsilon])
$$
Note $ X \in [b-\epsilon,b+\epsilon] \equiv b \in [X-\epsilon,X+\epsilon]$ and that
$$
Pr(b \in [X-\epsilon,X+\epsilon]) = \int_{\max(b_l,X-\epsilon)}^{\min(b_h,X+\epsilon)}\frac{1}{b_h-b_l}db=\frac{\min(b_h,X+\epsilon)-\max(b_l,X-\epsilon)}{b_h-b_l}
$$
Thus;
$$
f(X)=\frac{\min(b_h,X+\epsilon)-\max(b_l,X-\epsilon)}{2\epsilon(b_h-b_l)}
$$
The above can be broken up into a piece-wise function differently depending on the values of $\epsilon,b_l,$ and $b_h$.
Simulation in R
To validate this answer I also ran a simulation where $\epsilon=2,b_l=0,$ and $b_h=1$.  This results in the following pdf
$$
f(X) = \begin{cases}
\frac{X+\epsilon-b_l}{2\epsilon(b_h-b_l)} = \frac{X+2}{4}\;\;\mathrm{if}\;\;X \in [-2,-1)\\
\frac{b_h-b_l}{2\epsilon(b_h-b_l)} = \frac{1}{4}\;\;\mathrm{if}\;\;X \in [-1,2]\\
\frac{b_h+\epsilon-X}{2\epsilon(b_h-b_l)} = \frac{3-X}{4}\;\;\mathrm{if}\;\;X \in (2,3]\\
0\;\;\;\;\mathrm{otherwise}
\end{cases}
$$
The trapezoidal shape of $f(X)$ is validated with simulation and a kernel density estimator which I plot below

The code in R for the above plot is; 
e=2
bl=0;
bh=1;
b=runif(1e5,bl,bh)
x=runif(1e5,min=b-e,max=b+e)
plot(density(x,from=-2,to=3),main="Density Estimate (KDE)",lwd=2)
abline(h=.25,col=2,lty=2,lwd=2)
abline(v=-1,col=4,lty=4,lwd=2)
abline(v=2,col=4,lty=4,lwd=2)

