I am trying to figure out limits of integration, to find the probability distribution of $P(X<Y-Z)$.
Please verify and correct if I can write like below.
$$P(X<Y-Z)=\int_{y=0}^{\infty}\int_{z=0}^{z=y}\int_{x=0}^{x=y-z}f_{X}(x)\,f_{Y}(y)\,f_{Z}(z)\,dx\,dz\,dy,$$
where $X=M+S$, $\,\,M\sim Exp(\gamma), \,\,S \sim Exp(\mu), \,\,Y \sim Exp(\xi)$ and $\,\,Z \sim U(0,Y)$.
We can see $Z$ is dependent on $Y$. I have taken independent above while writing the densities of $X,Y$ and $Z$.
$$P(X<Y-Z)=\int_{y=0}^{\infty}\int_{z=0}^{z=y}\int_{x=0}^{x=y-z}(f_{M}(x)*f_{S}(x))\,\frac{1}{y} \,\xi e^{- \xi y } \,dx\,dz\,dy.$$
Correct me if i am wrong if I can write above, Please suggest on limits of integration and dependence of $Z=Y$.
$*$ represents the convolution operator. $\,\,Exp()$ represents the exponentially distributed random variable, and $U()$ represents the uniformly distributed random variable.
Indeed. Rather, $~Z$ is conditionally dependent on $Y$, which is more correctly indicated by: $Z\mid Y~\sim~\mathcal{U}(0,Y)$
Now let $W=Y-Z$, then by symmetry: $W\mid Y~\sim~\mathcal{U}(0,Y)$, thus:
$$\begin{align}f_{\small W}(w) &= \int_w^\infty f_{\small Y}(y)\,f_{\small W\mid Y}(w\mid y)\,\mathrm d y\end{align}$$
So
$$\begin{align}\mathsf P(X<Y-Z) &= \mathsf P(X<W)\\[1ex]&=\int_0^\infty f_{\small X}(x)\,\mathsf P(x<W)\,\mathrm d x\\[1ex]&= \int_0^\infty f_{\small X}(x)\,\int_x^\infty f_{\small W}(w)\,\mathrm d w\,\mathrm d x\\[1ex]&=\int_0^\infty f_{\small X}(x)\,\int_x^\infty\!\int_w^\infty\! f_{\small Y}(y)\,f_{\small W\mid Y}(w\mid y)\,\mathrm d y\,\mathrm d w\,\mathrm d x\\[1ex]&=\int_0^\infty [f_{\small S}{\star}f_{\small M}](x)\int_x^\infty\!\int_w^\infty\!\dfrac 1y\,\xi\,\mathsf e^{-\xi y}\,\mathrm d y\,\mathrm d w\,\mathrm d x\\[1ex]&~~\vdots \end{align}$$