<p>This uses the fact that the L1-norm between two probability distributions is equal to twice their total variation distance.</p>

<p>See <a href="https://en.wikipedia.org/wiki/Total_variation_distance_of_probability_measures" rel="nofollow">https://en.wikipedia.org/wiki/Total_variation_distance_of_probability_measures</a></p>

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

CC BY-SA 3.0
2016-08-29T14:22:22.037Z
2016-08-29T14:22:22.037Z
nichehole
<p>I'm reading this paper <a href="http://eccc.hpi-web.de/report/2016/044/download/" rel="nofollow">http://eccc.hpi-web.de/report/2016/044/download/</a>, at page 7 you can see this equation:</p>

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

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

<p>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?</p>

CC BY-SA 3.0
2016-08-27T18:00:21.797Z
Presumably, summing over $t$ of both $p_t$ and $p_{t|w,w}$ yields 1 so they cancel out.
Aryeh
CC BY-SA 3.0
2016-08-27T19:02:01.527Z
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
CC BY-SA 3.0
2016-08-28T18:13:10.560Z
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
CC BY-SA 3.0
2016-08-29T20:07:21.143Z
I suggest closing the question. It's not research level and I believe the answer is resolved in the comments.
Aryeh
<pr.probability>
Issue with inequalities involving probablities
Marco
# Issue with inequalities involving probablities

(Asked by: username_0 on 2016-08-27)

I'm reading this paper <http://eccc.hpi-web.de/report/2016/044/download/>, at page 7 you can see this equation:

$$ \sum\_t \mathbb{E}\_{w \in W} [|p\_t - p\_{t|w,w}|] = 2 \sum\_t \mathbb{E}\_{w \in W} [\max(0,p\_t - p\_{t|w,w})]$$

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?

**Tags:** pr.probability

> **Comments:**
> **username_1**: Presumably, summing over $t$ of both $p_t$ and $p_{t|w,w}$ yields 1 so they cancel out.  
> **username_0**: 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?  
> **username_1**: 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.  
> **username_1**: I suggest closing the question. It's not research level and I believe the answer is resolved in the comments.  

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## 1 answers

### Answer by username_2 on 2016-08-29. Score: 2

This uses the fact that the L1-norm between two probability distributions is equal to twice their total variation distance.

See <https://en.wikipedia.org/wiki/Total_variation_distance_of_probability_measures>

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


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