1

I'm attempting to prove that the set $\Delta X = \{π ∈ \mathbb{R}^n |\sum \pi_i=1, \pi_i \ge 0, \forall i\}$ is a convex subset of the vector space $\mathbb{R}^n$. It is specified that $X$ is a finite set of size $n$ and $\Delta X$ is the space of probabilities on $X$.

That $\Delta X$ needs to be convex is not immediately clear to me, as I do not see what contradiction is made by letting $\pi_0 = \alpha (\pi_1) + (1-\alpha)\pi_2 \not \in \Delta X$. If the probability space is of size $n=2$, it could be that $ X = \{x_1, x_2\}$ where $P(x_1)= \frac{1}{3}$ and $P(x_2)= \frac{2}{3}$. I know that $\{\frac{1}{3}, \frac{2}{3}\}$ is not convex.

I do know this is trivially true for $n=1$, as $ X = \{x'\}$ where $\Delta X=\{1\}$, but I am unsure of how to expand this to the nontrivial case.

This may be as simple as me misunderstanding the question, any tips are appreciated.

Corran Horn
  • 61
  • 1
    You are confusing a set of probability vectors being convex with any particular vector being convex. The former means, basically checking that the vector average of two prob vectors is again a prob vector. The latter is nonsense. – kimchi lover Sep 01 '18 at 18:24
  • @kimchilover so my $\Delta X$ will not itself be of size $n$, and I'm showing that $\Delta X$ will contain any probability that $X$ provides us? – Corran Horn Sep 01 '18 at 18:27
  • 2
    Take the case where $n=2$. What is $\Delta X$? It is the line segment in the plane, of all points $(x,y)$ for which $x\ge0, y\ge0, x+y=1$, connecting the points $(0,1)$ to $ (1,0)$. $\Delta X$ is not a finite set. You job is to show that it is convex, – kimchi lover Sep 01 '18 at 18:31
  • What is your definition of a convex subset of $\mathbb{R}^n$ ? – coffeemath Sep 01 '18 at 18:51

0 Answers0