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](http://pages.uoregon.edu/dlevin/MARKOV/markovmixing.pdf) implies that $2p = 1- \delta(\mu\_1, \mu\_2)$. 



Linking error probability based on total variation
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). 


