Variation Distance: Explaining Finite Cases

[boboYO]

I was doing some reading and I came across this:

http://en.wikipedia.org/wiki/Total_variation_distance_of_probability_measures

So apparently for the finite case,

$$\max_{x} ( \left| P(x) - Q(x) \right|)\quad \mbox{ is equivalent to}\quad \frac{1}{2} \sum_x {\left| P(x)-Q(x)\right|}$$


but isn't this is a counterexample?

x         1        2        3        4
P(x)    0.25     0.25     0.25      0.25

Q(x)    0.10     0.20     0.35      0.35

|P-Q|   0.15     0.05     0.10      0.10

sum(|P-Q|)/2= 0.2

max(|P-Q|)=0.15



So I was thinking, maybe they meant 'equivalent' in a different sense? Could somebody please explain?

 

[rochfor1]

I believe they might mean that those two metrics define the same topology on the set of probability measures.

 

[HallsofIvy]

Rochfor1 is correct. "Equivalent" simply means that they will give the same results in any probability measures, not that they are equal.

 

[boboYO]

Thanks guys.