# 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 '17 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? – Gabriel Jul 28 '17 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 '17 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? – Gabriel Jul 28 '17 at 16:00
• Yes, that would work. – whuber Jul 28 '17 at 16:36

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