# Mixture distribution PDF with discrete values

I am having problems while defining the PDF expression of a mixture distribution when some of its values are discrete. For example, imagine that a given random variable $\mathbb{X}$ takes values as follows:

$$\mathbb{X} = \begin{cases} exp(1/\lambda),\quad \text{with probability}\,\, p\\ 0, \quad \text{with probability}\,\, (1-p) \end{cases}$$

So, my guess for the expression of the PDF of $\mathbb{X}$ is:

$$f(x) = (1-p)\cdot \delta(x) + p\cdot \lambda e^{-\lambda\,x}$$

Is that correct? I am not sure about the $\delta(x)$.

Thank you very much in advance.

• By definition, a discrete variable does not have a PDF. Therefore no mixture involving a discrete variable has a PDF, either. When you extend your concept of a "density" to include measures that are singular with respect to Lebesgue measure, such as the Dirac $\delta$, you can indeed express the distribution in this form--but most authorities seem to avoid calling this a "density." For an extended example of what $\delta$ is and how to work with it, see my answer at stats.stackexchange.com/a/73626/919.
– whuber
Jul 28, 2017 at 13:41
• Ok, thanks for the info, I will read it carefully. Anyway, shall I understand that I can express the distribution in that way? Jul 28, 2017 at 13:58
• Pretty close. To be clear and rigorous, it is essential that you multiply the second term by the indicator function of its intended support.
– whuber
Jul 28, 2017 at 14:01
• Something like this? $f(x) = (1-p)\cdot \delta(x) + p\cdot \lambda e^{-\lambda\,x}\cdot H(x)$ where H(x) is the Heaviside step function? Jul 28, 2017 at 16:00
• @Stubborn Yes, it can. One way to see this is to view the present mixture as a censoring of some continuous random variable supported on the reals, such as an asymmetric Laplace variable. If you're comfortable with writing likelihoods for censored variables, then you won't have any difficulty.
– whuber
Jun 11, 2020 at 14:11

$$f(x) = (1-p)\cdot \delta(x) + p\cdot \lambda e^{-\lambda\,x} \cdot H(x)$$