How to express combined discrete-continuous RVs in one pdf? Let's say we have a random variable $X$ that behaves in two different ways where $X\sim$Bernoulli(1/3) AND $X\sim U(0,1)$. $X$ follows the Bernoulli distribution 25% of the time and the uniform distribution 75% of the time. How do I express this into a single probability density function? Do I condition? Do I keep them as distinctly separate portions of the pdf?
 A: Since the Bernoulli distribution is discrete, this distribution does not have a density in the most usual sense.  It does have a cumulative distribution function:
$$
F(x) = \Pr(X\le x) = \begin{cases}
0 & \text{if } x<0, \\
1/6 + 3x/4 & \text{if } 0\le x<1, \\
1 & \text{if }x\ge 1.
\end{cases}
$$
In senses other than the most usual one, you could say that
$$F'(x) = f(x) = \frac 1 6 \delta (x) + \frac1{12}\delta(x-1)+ \begin{cases} \frac 3 4 & \text{if }0< x< 1 \\[6pt] 0 & \text{if }x<0\text{ or }x>1,  \end{cases}$$
where $\delta$ is Dirac's delta function.
One could also speak of a density with respect to a measure other than Lebesgue measure, putting point masses at $0$ and $1$.
At any rate, these issues that complicate the concept of a probability density function do not afflict the cumulative distribution function.
Postscript in response to a comment:
$$M_X(t) = \mathbb E(e^{tX}) = \mathbb E(\mathbb E(e^{tX}\mid Y))$$
where $Y = 1\text{ or }0$ according as $X$ is at one of the endpoints or in the interior.  So this is
\begin{align}
& \mathbb E(e^{tX}\mid Y=1)\Pr(Y=1)+\mathbb E(e^{tX}\mid Y=0)\Pr(Y=0) \\[10pt]
= {} & \mathbb E(e^{tX}\mid Y=1)\cdot\frac 1 4 + \mathbb E(e^{tX}\mid Y=0)\cdot\frac 3 4 \\[10pt]
= {} & \left( e^{0t}\frac23 + e^{1t}\frac 13 \right)\cdot\frac 1 4 + \frac 3 4 \int_0^1 e^{tx}\,dx = \cdots\cdots
\end{align}
So it's just the appropriate weighted average of the two moment-generating functions.
