# Issue with inequalities involving probablities

$$\sum_t \mathbb{E}_{w \in W} [|p_t - p_{t|w,w}|] = 2 \sum_t \mathbb{E}_{w \in W} [\max(0,p_t - p_{t|w,w})]$$

I think what we're using here is this: $$a+b+|a-b| = 2 \max(a,b)$$

but with $a=0$ and $b=p_t-p_{t|w,w}$ we get: $$p_t - p_{t|w,w} + |p_t - p_{t|w,w}| = 2\max(0,p_t - p_{t|w,w})$$ I know it's a bit of a silly question but I just can't figure it out, maybe I'm missing some fact about probabilities?

• Presumably, summing over $t$ of both $p_t$ and $p_{t|w,w}$ yields 1 so they cancel out. – Aryeh Aug 27 '16 at 18:00
• using $b=p_t - p_{t|w,w}$ i need to substract from the right-hand side $p_{t|w,w}-p_t$, so we would have: $$\sum_t \mathbb{E}_{w \in W} (p_{t|w,w}-p_t)$$, i can split this in two sums, the first giving $$\sum_t \mathbb{E}_{w \in W} p_{t|w,w} = \sum_t p_t$$ and the other give the same so they cancel out, is this correct? – Marco Aug 27 '16 at 19:02
• I didn't read the paper but if $t$ is indeed the "index" of the distribution (i.e., summing over $t$ yields 1) then that's correct. – Aryeh Aug 28 '16 at 18:13
• I suggest closing the question. It's not research level and I believe the answer is resolved in the comments. – Aryeh Aug 29 '16 at 20:07

To prove the equality you take the subset A in the definition to be the set of points where $$p_t > p_{t|w,w}.$$