# total variation distance between 2 distributions decreases?

Say we have 2 probability distributions $$\pi_0, \alpha_0$$ on the same state space $$S$$ with some transition matrix $$P$$. Define $$\pi_1 = \pi_0 \cdot P$$, $$\alpha_1 = \alpha_0 \cdot P$$. The goal is to show that $$\vert\vert \pi_1 - \alpha_1\vert\vert_{TV} \leq \vert\vert \pi_0 - \alpha_0\vert\vert_{TV}$$
where $$\vert\vert\cdot\vert\vert_{TV}$$ denotes the total variation distance, which can be defined as $$sup_{E \in S}\vert\pi(E)-\alpha(E)|$$.

Here is my attempt:

$$\vert\vert \pi_1 - \alpha_1\vert\vert_{TV} = \vert\vert \pi_0P - \alpha_0P\vert\vert_{TV} = sup_{E \in S}\vert(\pi_0P)(E)-(\alpha_0P)(E)|$$

This is such a silly question, but how do I get that matrix $$P$$ out of there, and how do I use the fact that it's specifically a probability transition matrix (ie. all entries between $$0$$ and $$1$$) to reach the conclusion?

An equivalent definition of total variation distance is: $$||x-y||_{TV} = \frac{1}{2} \sum_i |x_i-y_i|.$$ Therefore: $$||\pi_1 - \alpha_1||_{TV} = ||(\pi_0-\alpha_0)P||_{TV} \leq \frac{1}{2} \sum_i \sum_j P_{j,i} | \pi_0(j) - \alpha_0(j)|.$$ After changing the order of summation we obtain: $$||\pi_1 - \alpha_1||_{TV} \leq \frac{1}{2} \sum_j | \pi_0(j) - \alpha_0(j)| \sum_i P_{j,i}.$$ Since $$P$$ is stochastic this is equal to $$\frac{1}{2} \sum_j | \pi_0(j) - \alpha_0(j)| = ||\pi_0 - \alpha_0||_{TV}.$$