Let $\Omega$ be a set of $n$ elements.
I need to do the following two things:
- Describe the set $\mathcal{P}$ of all probability distributions supported on $\Omega$.
- Show that $\mathcal{P}$ is convex.
Now, the second part is probably very easy once I've figured out the first part. For the first part, I was given the hint: "represent a distribution on $\Omega$ by a vector and describe its properties to obtain an analytic description of $\mathcal{P}$".
What I'm confused about is, are we talking about cumulative distribution functions? Are we talking about probability density functions? Because $\Omega$ consists of $n$ elements, are we talking about discrete probability distributions? Or do we just mean that for any given distribution, the probabilities of elements are between $0$ and $1$ and add up to (or integrate to, if it's continuous) to $1$?
The wording of Part 1 is very confusing to me and I don't understand how to represent a distribution on $\Omega$ by a vector. Also, what does it mean that the probability distribution is "supported" on $\Omega$? Usually, the support of a function is the set of all $x$ such that the function value is nonzero, so does this mean that no point in $\Omega$ gives us a zero probability?
I apologize for my absolute cluelessness, but I really need help understanding this! Could somebody please help me?
Thank you.
WLOG assume that $\Omega = \{1,2,\ldots,n\}$. Let $\mathcal F=2^\Omega$. A probability distribution $P\in\mathcal P$ is a map $P:\mathcal F\to\mathbb R$ such that
The set $\mathcal P$ then is just the collection of measures on $\Omega$ which assign the value $1$ to $\Omega$. Now let $P,Q\in\mathcal P$ and $\lambda\in(0,1)$. It is clear that $\lambda P(E) + (1-\lambda)Q(E)\geqslant 0$ for each $E\in\mathcal F$ and that $\lambda P(\Omega) + (1-\lambda)Q(\Omega) = 1$. If $A,B\in\mathcal F$ are disjoint, then $$ \lambda P(A\cup B) + (1-\lambda)Q(A\cup B) = \lambda(P(A)+P(B)) + (1-\lambda)(Q(A)+Q(B)), $$ and so $\lambda P + (1-\lambda)Q\in\mathcal P$. It follows that $\mathcal P$ is convex.