How to obtain the following total variation? I read in a book the following statement but can't see why it is "clear".
Suppose $a \in [0,1/2]$. Let's say we are given $(Z,T)$ which is a random variable with uniform distribution over $[0,1]\times[a/2,1-a/2]$. Set 
$$
V = T\cdot\mathbf{1}(Z\in[a/2,1-a/2]) + Z\cdot\mathbf{1}(Z\notin[a/2,1-a/2])
$$
We can show that both $Z$ and $V$ are uniformly distributed over $[0,1]$. Let $P_{(Z,V)}$, $P_Z$ and $P_V$ denote the laws of $(Z,V)$, $Z$ and $V$ respectively. The author claims that it is clear that
$$
||P_{(Z,V)} - P_Z \otimes P_V|| = 4a-2a^2
$$
where $||\cdot||$ denote the total variation of a signed measure.
I have to admit that I couldn't see why the statement is clear. Can anyone gives some hint? Many thanks!
 A: I hope my computations are correct. First of all I believe that you used the notation $$\| \mu \|_{\text{TV}} = 2 \sup_{A}|\mu (A)|$$ I will show that
$$
\sup_A |P_{(Z,V)}(A) - P_Z \otimes P_V(A)| = 2a - a^2
$$
First, the guessing part. Simply take $I_0 = [a/2, 1-a/2]$ and define
$$A_0 = \{(z,v) \ | \  \text{ either }(z,v) \in I_0 \times I_0 \text{ or } z= v\}$$ Note that this is exactly the support of the joind distribution. Hence
$$|P_{(Z,V)}(A_0) - P_Z \otimes P_V(A_0)|= 1-|I_0|^2 = 2a - a^2.$$
Now we have to prove that this is a supremum. Here many arguments can be followed. For example we can compute the density of $V$ given $Z$. Then
$$
p_{(V|Z)}(v,z) = \frac{1_{I_0}(v)}{|I_0|} 1_{I_0}(z) + \delta_z(v) 1_{I_0^c(z)}.
$$
Now we divide the proof in steps. First we show that for any $A,$ $$  \mu(A) :=P_{(Z,V)}(A) - P_Z \otimes P_V(A) \le 2a - a^2$$
Suppose $ A_0 \subset A,$ then the claim immediately follows because $A_0$ is the support of $P_{(Z,V)}.$ Also it is clear that $\mu(A) \le \mu(A \cup \{z = v\})$ so that we can restrict to considering sets of the form $A = \overline{A} \cup \{z = v\},$ with $\overline{A}\subset I_0 \times I_0.$ Now observe that the density of $p_{(Z,V)}(z,v)$ is strictly larger than the density $p_{Z}(z)p_{V}(v)$ on $I_0 \times I_0.$ This is enough to conclude that the supremum of $\mu$ is assumed in $A_0.$
Now we pass to proving that
$$
\nu(A): =  P_Z \otimes P_V(A)- P_{(Z,V)}(A) \le 2a - a^2
$$
The same considerations of before show us that the maximum for $\nu$ is assumed on the set $A_0^c$, on which we have
$$ \nu(A_0^c) = 2a - a^2.$$
This proves the claim.
