Given is the set of probability density functions defined as $P:=\left \{ p(x)\mid p(x)\, is\ a\ probability \ density \ function \right \}$

Is $P$ a convex set?

I am not sure that here i have to use the classical definition for the convex set. In the lecture we have seen that for the general case, suppose $p:\mathbb{R}^{n}\rightarrow \mathbb{R}$ satisfies $p(x) \geq 0$ for all $x\in C$ and $\int _{C} p(x)dx=1$ where $C\subseteq \mathbb{R}^{n}$ is convex. Then $\int_{C}p(x)xdx\in C$ if the integral exists. I guess i have to use this definition to see if $P$ is convex but i have no idea where to start... Can anybody help me with this problem, please? Thank you in advance!


1 Answer 1


Take the definition of convex set (e.g. chapter 2.1.4 from Boyd & Vandenberghe).

Can you say that $p(x) = \theta \, p_1(x) + (1-\theta) \, p_2(x) \in P$ for any $0 \leq \theta \leq 1$ and any $p_1(x),\,p_2(x) \in P$?

Yes, $p(x)$ is a valid pdf, it is non-negative and integrates to 1. BTW, the resulting density is called a mixture density distribution (to draw a sample from this distribution, you can imagine that you first flip a loaded coin with heads probability $\theta$, then you sample from $p_1$ if the outcome is a head, from $p_2$ if a tail).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .