probabilityprobability distributionsprobability theoryrandom variables
Let $Y$ be a random variable with distribution function
$$
F_Y(x) = \begin{cases}
0 &\quad x<0\\
p &\quad x=0\\
p + (1-p)F_X(x) &\quad x>0
\end{cases}
$$
where $X$ is a continuous random variable.
My first question is if $Y$ has a density, despite the mass at 0? Moreover, is there a way to express $Y$ as a linear combination of other random variables, like a mixture? For instance
$$
Y = (1-B)X + B\delta
$$
where $B$ is Bernoulli and $\delta$ is a degenerate r.v. at zero.
The map $g:(x,y) \mapsto (x-2y,x)$ is a differentiable and invertible function between $(0,\infty)\times (0,\infty)$ and $R=\{(z,w) | z< w \text{ and } w>0\}$, so first of all we get that the support for $(Z,W)=(X-2Y,X)$ must be $R$.
The transformation theorem for probability densities states that:
$$f_{Z,W}(z,w) = f_{X,Y}(g^{-1}(z,w)) |det(\frac{dg^{-1}}{d(z,w)}(z,w))|,$$ where $\frac{dg^{-1}}{d(z,w)}(z,w)$ is the jacobian of $g^{-1}$.
(see https://en.wikipedia.org/wiki/Probability_density_function#Vector_to_vector)
We first compute $g^{-1}(z,w)= (w,\frac{w-z}{2})$ and the jacobian $$ \frac{dg^{-1}}{d(z,w)}(z,w) = \begin{pmatrix}0 & 1 \\ -\frac12 & \frac12 \end{pmatrix},$$ which has determinant $\frac12$ for all $z,w$. We now plugin, and get $$ f_{Z,W}(z,w) = \frac12 f_{X,Y} ((w,\frac{w-z}{2})) = e^{-w}e^{-2\frac{w-z}{2}}=e^{z-2w}.$$ for all $(z,w) \in \{(z,w) | z< w \text{ and } w>0\}$. Just to verify, that this is in fact a valid density we compute $$ \int_0^\infty \int_{-\infty}^w e^{z-2w} dzdw = 1$$
You can rewrite $F_X$ as $$F_X(x) = (x-1) \mathbf{1}_{(1,2)}(x) + \mathbf{1}_{[2,\infty)}(x)\tag{$x\in\mathbb{R}$}$$ and $F_Y$ as $$F_Y(x) = \frac{x^2}{9} \mathbf{1}_{(0,3)}(x) + \mathbf{1}_{[3,\infty)}\tag{$x\in\mathbb{R}$}$$ to see how they are both defined on $\mathbb{R}$. Then you can multiply them and simplify the expression, using that $\mathbf{1}_A\cdot \mathbf{1}_B = \mathbf{1}_{A\cap B}$.
Here is an illustration (plot made in Mathematica):
Best Answer
As @angryavian and @Oliver Díaz mentioned, indeed the distribution does not have a density due to the atom at zero. Moreover, the representation $$ Y= (1-B)X + B\delta_{\{0\}} = (1-B)X $$ correctly establishes the distribution $F_Y$ provided, assuming $X$ is a positive r.v. (which is not assumed on the question but should). Indeed $$ \begin{aligned} F_Y(y) &= P(Y\leq y)\\ &= P(Y\leq y | B = 0)P(B=0) + P(Y\leq y | B = 1)P(B = 1) \\ &= P(X\leq y)(1-p) + P(0 \leq y)p\\ &= F_X(y)(1-p) + \pmb{1}_{\{y\geq 0\}}p \end{aligned} $$ which is the same c.d.f. as the one in the question.