CDF and MGF of a Sum of a discrete and continuous random variable I am currently dealing with the following exercise:
Given the random variables $X \sim Be(p), Y \sim Exp(\lambda)$, and assume they are independent. Set $Z:= X + Y$.


*

*Compute the Moment Generating Function (MGF) $M_{z}(t)$ of $Z$

*Compute the Distribution Function (CDF) $F_{z}$ of $Z$. Is $Z$ an absolutely continuous random variable?


From the theory I have studied, I have some questions regarding how to deal with this kind of exercises:


*

*When it comes to sum to sum two independent random variables (discrete or continuous) and then calculating their CDF, does this always mean using convolution? Does this differ from calculating their joint PDF or CDF?

*If X depended on Y, the joint CDF would be obtainable by using the law of total probability?

*Since they are independent, the joint MGF is simply the product of the two MGFs?

 A: I will provide some advice about question (1) because (as indicated in the comments) it's the heart of the matter.
Because there are many ways to calculate with distributions, it's not always apparent how to proceed.  When you're stuck, it's often a good idea to resort to the basic definition; namely, for all real numbers $z$ the CDF of $X+Y$ is given by the probability that $X+Y \le z:$
$$F_{X+Y}(z) = {\Pr}_{(X,Y)}(X+Y \le z).$$
It can help to focus on the simplest aspect of a problem.  In this case, the Bernoulli variable looks particularly simple to me, because it can attain only two values.  This invites us to break the problem into two parts, one for each possible value of $X.$  Thus:


*

*$X=0$ with probability $1-p.$  In this case $X+Y=Y$ and therefore $$\Pr(X+Y\le z) = \Pr(Y \le z) = F_Y(z).$$

*$X=1$ with probability $p.$  Now $X+Y=1+Y$ and therefore $$\Pr(X+Y\le z) = \Pr(1+Y\le z) = \Pr(Y \le z-1) = F_Y(z-1).$$
The foregoing are mutually exclusive events: the axioms of probability assert we may add their chances.  Technically, we're working with conditional probabilities and we are computing
$$\eqalign{
\Pr(X+Y \le z) &= &\Pr(X+Y\le z\mid X=0)\Pr(X=0) \\&&+ \Pr(X+Y\le z\mid X=1)\Pr(X=1) \\
&= &F_Y(z)(1-p) + F_Y(z-1)p.}$$
I'm not supposing you would have formulated the problem like this at the outset; but by the time you have done the foregoing calculations, it's helpful to express what they reveal to you in this more rigorous form.
You should be able to proceed to an explicit answer from here by using whatever formula you like for the exponential CDF $F_Y.$  To determine whether this is a continuous variable, I recommend plotting $F_{X+Y}$ for typical values of $p$ and $\lambda,$ as shown by the thick curve in this figure (where $p=1/2$ and $\lambda=1$):

The dashed line sketches $F_Y$ itself.  The amount by which the curve departs from the dashed line represents the contribution of $pF_Y(z-1),$ whose graph has the same shape as that of $(1-p)F_Y,$ but shifted one unit right and rescaled by $p/(1-p).$  (As always, the graph continues infinitely far to the right and left, equal to $0$ for all negative values of $z$ and reaching $1$ asymptotically for large positive values of $z.$)  
The question of continuity is resolved by determining whether the graph of $F_{X+Y}$ has any vertical breaks in its rise from $0$ to $1.$  I hope the picture makes it clear that this depends on whether $F_Y$ itself is continuous; I'll leave the answer at that.
