How to express combined discrete-continuous RVs in one pdf?

Question

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?

Answer 1

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.