Linking error probability based on total variation Consider probability measure $\mu_{XY}$ defined on $\mathbb{R}^d \times \{1,2,3\}$, and sub-probability measures $\mu_1$, $\mu_2$, and $\mu_3$ as $\mu_1(A):=P(X\in A, Y=0)$ and $\mu_2(A):=P(X\in A, Y=2)$ and $\mu_3(A):=P(X\in A, Y=3)$ for Borel sets $A$, so that $\mu:=\mu_1+\mu_2+\mu_3$ is the probability distribution of $X$. Let 
\begin{equation*}
 p:=P(Y\ne f(X)). 
\end{equation*}
where 
$$ f(x):=
\left\{\begin{matrix}
1 & \mbox{if}~~ \rho_1(x) > \rho_2(x), \rho_3(x) \\ 
2 & \mbox{if}~~ \rho_2(x) \ge \rho_1(x), \rho_2(x) > \rho_3(x) \\ 
3 & \mbox{if}~~ \rho_3(x) \ge \rho_1(x), \rho_2(x)  \:, 
\end{matrix}\right.
$$
for all $x\in \mathbb{R}^d$ and for each $i=1,2,3$ the function $\rho_i$ is the density of $\mu_i$ with respect to the measure $\mu$. 
Can we write $p$ as a function of $\delta(\mu_1, \mu_2)$, $\delta(\mu_1, \mu_3)$, and $\delta(\mu_2, \mu_3)$, where $\delta(.,.)$ is the total variation? 
For the case of having only two sub-probability measures, Proposition 4.2 of Levin'08 implies that $2p = 1- \delta(\mu_1, \mu_2)$.  
 A: The answer is no. E.g., let $(A_1,A_2,A_3)$ be any Borel partition of $\mathbb{R}^d$ such that $A_j\ne\emptyset$ for each $j$. 
Let $\nu$ be any measure on $\mathbb{R}^d$ such that $\nu(A_j)=1$ for each $j$. 
For each $i$ and $j$ in $\{1,2,3\}$, let the density of the measure $\mu_i(\cdot)=P(X\in \cdot,Y=i)$ with respect to $\nu$ equal the constant $\mu_{i,j}$ on the set $A_j$, where 
the $3\times3$ matrix $(\mu_{i,j})$ is
$$
\frac1{10}\,\begin{bmatrix}
 2 & 1 & 0 \\
 1 & 2 & 1 \\
 0 & 1 & 2 \\
\end{bmatrix}. 
$$ 
Then ($\mu_1+\mu_2+\mu_3$ is indeed a probability measure,)
$\delta(\mu_1, \mu_2)=\frac1{10}(|2-1|+|1-2|+|0-1|)=\frac3{10}=\delta(\mu_2, \mu_3)$, $\delta(\mu_1, \mu_3)=\frac25$, $f(x)=j$ if $x\in A_j$ (the comparisons between the densities $\rho_i$ do not depend on the measure with respect to which the densities are taken -- as long as such densities exist), 
and $p=\frac25$ (in this case, the sum of the off-diagonal entries of the matrix). 
Replacing here the previous matrix by 
$$
\frac1{60}\,\begin{bmatrix}
 21 & 7 & 0 \\
 9 & 13 & 0 \\
 0 & 7 & 3 \\
 \end{bmatrix},
$$ 
we get the same values of $\delta(\mu_1, \mu_2)$, $\delta(\mu_2, \mu_3)$, and $\delta(\mu_1, \mu_3)$, but $p=\frac{23}{60}\ne\frac25$. 
So, $p$ is not a function of $\delta(\mu_1, \mu_2)$, $\delta(\mu_2, \mu_3)$, and $\delta(\mu_1, \mu_3)$. 
Added: However, one always has the following lower bound on the misclassification probability $p$ in terms of the total variation norms $\delta(\mu_i,\mu_j)=\|\mu_i-\mu_j\|$:
\begin{equation*}
 p\ge\frac12-\frac1{2(k-1)}\,\sum_{1\le i<j\le k}\|\mu_i-\mu_j\|. \tag{1}
\end{equation*}
Here we consider the more general setting, where $k\in\{2,3,\dots\}$, $Y$ takes values in the set $[k]:=\{1,\dots,k\}$, 
$f(x)=i$ if $x\in\bigcap_{j\in J_i}A_{ij}$, $J_i:=[k]\setminus\{i\}$, 
\begin{equation*}
 A_{ij}:=\begin{cases}
 \{x\colon\rho_i(x)>\rho_j(x)\}&\text{ if }i<j,\\
\{x\colon\rho_i(x)\ge\rho_j(x)\}&\text{ if }i>j,
\end{cases}
\end{equation*}
so that $\bigcup_{i\in[k]}\bigcap_{j\in J_i}A_{ij}=\mathbb R^d$. Here everywhere $i$ and $j$ are in $[k]$. 
Thus, 
\begin{multline*}
 1-p=P(f(X)=Y)=\sum_{i\in[k]}P\Big(Y=i,X\in\bigcap_{j\in J_i}A_{ij}\Big)
 =\sum_{i\in[k]}\mu_i\Big(\bigcap_{j\in J_i}A_{ij}\Big) \\ 
 \le\sum_{i\in[k]}\min_{j\in J_i}\mu_i(A_{ij})
 \le\sum_{i\in[k]}\frac1{k-1}\sum_{j\in J_i}\mu_i(A_{ij})
 =\frac1{k-1}\sum_{1\le i<j\le k}[\mu_i(A_{ij})+\mu_j(A_{ji})] \\ 
 =\frac1{k-1}\sum_{1\le i<j\le k}\int(\rho_i\vee\rho_j)\,d\mu
 =\frac1{k-1}\sum_{1\le i<j\le k}\int\frac{\rho_i+\rho_j+|\rho_i-\rho_j|}{2}\,d\mu \\ 
 =\frac1{k-1}\sum_{1\le i<j\le k}\frac{\|\mu_i\|+\|\mu_j\|+\|\mu_i-\mu_j\|}{2}  
 =\frac12\,+\frac1{2(k-1)}\sum_{1\le i<j\le k}\|\mu_i-\mu_j\|;
\end{multline*}
here we used the fact that $\sum_i\mu_i=\mu$, a probability measure, so that $\sum_i\|\mu_i\|=1$. Thus, we have (1). 
