# Maximal “distance” of two probability measures?

Consider the two coins (i.e., probability measures on the discrete set ${0,1}$) $C_{0.9}$ and $C_{0.99}$, where $C_{x}$ is the coin having probability of turning head equal to $x$.

Let $\mu_{0.9}$ and $\mu_{0.99}$ be the probability measures on $2^{\omega}$ (Markov chains) obtained by flipping infinitely often the two coin $C_{0.9}$ and $C_{0.99}$ respectively. So that, for instance, the probability assigned by $\mu_{0.9}$ to the (open) set of sequences starting with two zeros is $0.81$.

Question: I would like to calculate $D(\mu_{0.9}, \mu_{0.99})$ defined as:

$D(\mu_{0.9}, \mu_{0.99})$ = $\displaystyle \bigsqcup_{B} \{ |\mu_{0.9}(B) - \mu_{0.99}(B)| \ \$ ; $B$ a Borel subset of $2^{\omega} \}$.

I don't know how to approach this problem. I have the feeling that the above supremum should be strictly less than $1$ and that the maximizing event should be the set of sequences starting with about 25 ones. The number 25 comes from the function $f : \mathbb{R}\rightarrow \mathbb{R}$ defined as $f(x)= {0.99}^x - {0.9}^x$ which attains its maximum around 25.

Question 2: More generally, is the function $D$ (mapping pairs of probability measures to the unit interval) as defined above well known?

Thank you in advance for any comment!

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Regarding Question 2: Total variation distance. –  martini Mar 27 '12 at 13:30
@martini: Thanks! –  MM83 Mar 27 '12 at 14:09

Call $B$ the event that the asymptotic proportion of heads exists and is 0.9, then the law of large numbers shows that $\mu_{0.9}(B)=1$ and $\mu_{0.99}(B)=0$. Hence $D(\mu_{0.9},\mu_{0.99})=1$. One sees that in fact the measures $\mu_{0.9}$ and $\mu_{0.99}$ are mutually singular.