What type of distribution is $f(x; \theta)=\theta e^{-x}+2(1-\theta)e^{-2x}$?

Question

Does anyone know what type of distribution this is?

$$f(x;\theta)=\theta e^{-x}+ 2(1-\theta)e^{-2x}$$

Where $\theta \in [0,1]$ and for $x \ge 0$.

I am trying to show that $f(x;\theta)$ is a density function.

Answer 1

Hint: It is a mixture (sum of nonnegative weights is $1$) of two exponential distribution densities. For fixed $\theta$, one needs to check that $f(x;\theta ) >0$ for all $x$ and that $\int_{-\infty}^\infty f(x; \theta) dx = 1$

Answer 2

A density function $f:\mathcal{X}\to \mathbb{R}$ must satisfy two properties:

1: $f(x)\geq 0 ~ \forall x\in\mathcal{X}$

2: $\int_{\mathcal{X}} f(x)\mathrm{d}x=1$.

Let's now investigate $g(x|\theta)=\theta e^{-x} + 2(1-\theta)e^{-2x}$. Well, if $\theta \in [0,1]$, it's pretty clear $g(x|\theta) \geq 0$. And, $$\int_{0}^{\infty}\theta e^{-x} + 2(1-\theta)e^{-2x} ~\mathrm{d}x=(-\theta e^{-x} + (-1/2)\cdot 2(1-\theta)e^{-2x})\bigg|^{\infty}_{0}$$ $$=(-\theta e^{-x} + (\theta-1)e^{-2x})\bigg|^{\infty}_{0}$$ $$=-\theta(0-1)+(\theta-1)(0-1)$$ $$=\theta+1-\theta=1.$$

Answer 3

Does anyone know what type of distribution this is?

It's a

Hyperexponential Distribution