# A problem with elementary inequality involving probabilities and Brier scoring rule

I am trying to prove certain relations between certain values of the so called Brier inaccuracy measure (Brier scoring rule).

Given a vector $p = (p_1, \ldots p_n)$, where $p_1 + \ldots p_n = 1$ and $p_i \geq 0$ for all $i \leq n$ we can compute the Brier inaccuracy of the vector $p$ (assuming we are given $n$ propositions and $p_i$'s denote our degrees of belief in the corresponding propositions if the proposition $i$, corresponding to $p_i$ is actually true):

$B_i(p) = (1-p_i)^2 + \sum_{j=1, j\neq i}^n p_j^2$.

If we find out that e.g. the proposition $n$ is false, we can conditionalize on it and obtain a posterior inaccuracy:

$Q_i(p) = (1 - \frac{p_i}{1-p_n})^2 + (\frac{1}{(1-p_n)^2})\sum_{j=1, j\neq i}^{n-1} p_j^2$.

In general, it is possible that by such conditionalization the measure of inaccuracy can increase. However, I conjecture that if $p_i = max\{p_j: j<n\}$, then the inaccuracy has to decrease (and actually, if all $p_i$'s are equal for $i <n$ then I already know that it also decreases).

That is, there is the following problem:

Assume that:

(1) $0<p_i<1$ for all $i\leq n$,

(2) $p_1 + \ldots + p_n = 1$,

(3) $p_1 \geq p_2 \geq \ldots \geq p_{n-1}$,

(4) $p_1 > p_{n-1}$.

Then consider: $B_1(p) = (1-p_1)^2 + \sum_{j=2}^n p_j^2$

and:

$Q_1(p) = (1 - \frac{p_1}{1-p_n})^2 + (\frac{1}{(1-p_n)^2})\sum_{j=2}^{n-1} p_j^2$.

What I want to demonstrate is that for any vector $p$ satisfying conditions (1)-(4) it holds that:

$B_1(p) > Q_1(p)$.

I did try to compare how much of a decrease in inaccuracy one gets by the conditionalization with how much increase one gets from it.

What I have concluded is that I need to prove the folloqing inequality:

$2p_1 + \sum_{i=1}^{n-1}p_i - (\sum_{i=1}^{n-1}p_i)^2 > (\frac{1}{1-p_n})\cdot \sum_{i=1}^{n-1}p_i^2 + \sum_{i=1}^{n-1}p_i^2$.

Is there now an elementary way to use our assumptions (1)-(4) to prove the inequality?

Clearly $\sum_{i=1}^{n-1}p_i - (\sum_{i=1}^{n-1}p_i)^2\geq 0$, so we can ignore those terms in the inequality. Since $(\frac{1}{1-p_n})\sum_{i=1}^{n-1}p_i^2 \geq \sum_{i=1}^{n-1}p_i^2$ it suffices to prove $$p_1 > (\frac{1}{1-p_n})\cdot \sum_{i=1}^{n-1}p_i^2$$
which can be written as the obvious inequality $$p_1(p_1+\cdots+p_{n-1}) > p_1^2+\dots+p_{n-1}^2$$