# Decomposition of probability measures with a bounded total variation distance

Fix a probability space $$(\Omega, \mathcal{F})$$. Let $$P$$ and $$Q$$ be two probability measures on $$(\Omega, \mathcal{F})$$ such that there exist $$\varepsilon$$ and $$\delta$$ that for every $$A \in \mathcal{F}$$ we have $$P(A) \leq e^\varepsilon Q(A) + \delta$$ and $$Q(A) \leq e^\varepsilon P(A) + \delta$$.

Question: Is the following claim true?

There exists probability measures $$\mu$$, $$\mu'$$, $$\nu$$, $$\nu'$$ and a constant $$C >0$$ such that the following holds:

1. $$P = (1-\delta)\mu + \delta \mu'$$
2. $$Q = (1-\delta)\nu + \delta \nu'$$
3. $$\Vert\frac{\text{d}\mu}{\text{d}\nu} \Vert_{\infty} \leq C$$ (ess-sup with respect to $$\nu$$)
4. $$\Vert \frac{\text{d}\nu}{\text{d}\mu} \Vert_{\infty} \leq C$$ (ess-sup with respect to $$\mu$$)

For the case of $$\varepsilon=0$$, this claim is true and the proof is based on the relationship between total variation distance and Wasserstein distance.

• I am interested in the proof of the claim. I edited the question.
– MMH
Commented May 19, 2023 at 14:52

Let $$\tau=1/2 P+1/2Q$$. Then both $$P$$ and $$Q$$ are absolutely continuous with respect to $$\tau$$ and admit densities $$p$$ and $$q$$, respectively, and the total variation distance between $$P$$ and $$Q$$ is given by $$\int|p-q|~\mathrm d\tau$$. Now, $$p=p\wedge q+(p-q)_+$$ and $$q=p\wedge q+(q-p)_+$$ and $$\int (p-q)_+~\mathrm d\tau=\int (q-p)_+~\mathrm d\tau=\delta/2$$ and $$\int p\wedge q~\mathrm d\tau=1-\delta/2$$. Let $$g=\frac{1}{1-\delta/2}~ p\wedge q$$ and $$h=\frac{1}{\delta/2}~ (p-q)_+.$$ Then $$p=(1-\delta/2)g+\delta/2~h=(1-\delta)g+\delta/2~g+\delta/2~h=(1-\delta)g+\delta(1/2 g + 1/2 h).$$ Note that $$g$$ and $$h$$ are densities with respect to $$\tau$$. We can then take $$\mu$$ to be the probability measure that has density $$g$$ with respect to $$\tau$$ and $$\mu'$$ be the probability measure that has density $$1/2~g+1/2~h$$ with respect to $$\tau$$.
A parallel argument works for $$Q$$ and we can even take $$\nu=\mu$$.
• Thanks Michael. I made a mistake in the statement of the claim. Your proof is correct for the case that $\varepsilon=0$.