Convolution of two non-independent probability distributions (Exponential, Uniform)

Question

I want to describe a probability distribution which — what I believe — is a convolution between two dependent (rather than two independent) probability distributions.

In many ways this question is very similar to mine, although my situation is more generic/general: I have an exponential distribution with rate $\frac{1}{\lambda}$ (i.e., mean $\lambda$) and a uniform distribution on $[a,b]$ (where $a>0$ and $a<b$), and I am trying to find the combined general distribution. However, my problem is the exponential distribution should be dependent on the result of the uniform distribution... at least, that's what I believe is needed to describe the situation where I generate a uniform random variable and sum it with a exponential random variable (I may not be describing this correctly; I am trying to define the general distribution describing the total time between two "back-to-back"/sequential, stochastic events whose rates can be described by the distributions above).

In more mathematical terms, I am dealing with the functions:

$\mathbf{\\ f_X(x)=\begin{cases} \frac{1}{\lambda} e^{-\frac{x}{\lambda}} & x\geq 0\\ 0 & \text{otherwise} \end{cases}} \ \hspace{20pt} \ \mathbf{f_Y(x)=\begin{cases} \frac{1}{b-a} & a\leq x\leq b\\ 0 & \text{otherwise} \end{cases}}$

...and I am looking for the result $(f_X*f_Y)(x)$ without knowledge of these functions being independent.

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My understanding about convolutions is extremely elementary: I roughly understand what they are, what purpose(s) they serve, and to some degree how they are built, but I struggle to define them with respect to functions with complex domains — especially for situations involving dependent probability distributions. As in, I know that:

$(f_X*f_Y)(x) = f_Z(z) = \int_{x = -\infty}^\infty f_{XY}(x,z-x) \, dx$ if $Z=X+Y$

...but we should not be able to apply the standard approach of $f_Z(z) = \int_{x = -\infty}^\infty f_X(x) \, f_Y(z-x) \, dx$ since we may not be able to say $f_{XY}(x,z-x)=f_X(x) \, f_Y(z-x)$ (I only say 'may' because I truly have no idea how independence or dependence affects the convolution of probability distributions).

I have some idea that the domain of $f_Z(z)$ is $[a,\infty)$ provided $a>0$ (or, at least, should be — based on what I understand to be true about the general distribution needed to describe my situation), but I have no understanding on how to go any further with — or without — this information.

Answer 1

Note: I'm not too sure if this is correct since it is somewhat "convoluted" (pun intended) and contrived, but this is the best I could scrap together with my understanding.

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I tried to take advantage of the properties of the Laplace transform to derive a "backwards approach" at solving the convolution problem via inversion.

Namely, let it be said that if $f_X, f_Y$ have well-defined Laplace transforms $\mathcal{L}[f_X],\mathcal{L}[f_Y]$, then

$ (1)\, \mathcal{L}[f_X*f_Y]=\mathcal{L}[f_X]\mathcal{L}[f_Y]. $

...so a good first step is to work out the Laplace transforms of the individual functions, then perform a simple product.

For $f_X$,
$ (2)\, \mathcal{L}[f_X](s)=\int_{0}^{\infty} e^{-st}f_X(t) \, dt=\frac{1}{\lambda s + 1}. $

...and for $f_Y$,
$ (3)\, \mathcal{L}[f_Y](s)=\int_{a}^{b} e^{-st}f_Y(t) \, dt=\frac{e^{-as}-e^{-bs}}{(b-a)s}. $*

Now, it's only a matter of finding the product, which is rather easy:

$ (4)\, \mathcal{L}[f_X]\mathcal{L}[f_Y]=\frac{e^{-as}-e^{-bs}}{(b-a)s(\lambda s + 1)}. $

But, $(4)$ isn't $f_X*f_Y$; it's $\mathcal{L}[f_X*f_Y]$ (according to $(1)$). So, how do we get $f_X*f_Y$ from $(4)$? Using the inverse Laplace transform! Using indicator functions where needed, we have:

$ (5)\, \mathcal{L}_s^{-1}[\frac{e^{-as}-e^{-bs}}{(b-a)s(\lambda s + 1)}](x)=\frac{1}{b-a}[\mathbb{1}(x-a\geq0)(1-e^{(a-x)/\lambda})+\mathbb{1}(x-b\geq0)(e^{(b-x)/\lambda}-1)]. $

Taking it all together,

$ (f_X*f_Y)(x)= \begin{cases} \frac{1}{b-a}[1-e^{(a-x)/{\lambda}}] & a \leq x \leq b & \\ \frac{1}{b-a}e^{(a-x)/\lambda}[e^{(b-a)/\lambda}-1] & x > b & \\ 0 & \text{otherwise} & \end{cases} $

...given $b>a$ and $\lambda> 0$.

This seems to agree with the expected solution given in the question I referred to and strongly agrees with the domain I need for $f_Z(x)$ (i.e., $x\in[a,\infty)$), which suggests that I was wrong to assume that $f_X$ and $f_Y$ were non-independent/dependent (this was certainly due to my poor understanding on the criteria necessary for dependence between probability distributions).

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*_{This is where it feels like I'm hand-waving a certain detail about Laplace transforms, so I'm a bit unsure if the limits are practical; it makes sense for the domain of $f_Y$, but I'm not entirely certain that I can arbitrarily restrict the limits of integration to this domain when deriving Laplace transforms.}