0

Given two absolutely continuous probability measures $\mu,\sigma \in \mathcal P_2(\mathbb R^n)$ and two maps $T_1, T_2$ such that $$(T_1 \circ T_2)_\#\sigma =\mu$$ where $(\cdot)_{\#}$ denotes the pushforward operator. I saw that it is a general property that

$$(T_1 \circ T_2)_\#\sigma={T_1}_{\#}({T_2}_{\#}\sigma)$$

How can one make sense of this or see this?

Tesla
  • 1,380

1 Answers1

2

If I'm not wrong, for a probability $\mu$ and a map $T$, one defines $$T_\# \mu(A) := \mu(T^{-1}(A)),$$ for a measurable set $A$, right?

Then, in your notations: $$\begin{align*} (T_1 \circ T_2)_\#\sigma(A) & = \sigma((T_1\circ T_2)^{-1}(A)) \\ & = \sigma(T_2^{-1}(T_1^{-1}(A))) \\ & = (T_{2\#}\sigma)(T_1^{-1}(A)) \\ & = T_{1\#}(T_{2\#}\sigma)(A). \end{align*}$$ That is, $$(T_1 \circ T_2)_\#\sigma = T_{1\#}(T_{2\#}\sigma).$$

fcz
  • 278
  • And then, formally, wouldn't it be

    $${T_2}{#}\sigma(T_1^{-1}(A))$$ instead of $$({T_2}{#}\sigma)(T_1^{-1}(A))$$, since $\sigma$ is applied onto the measurable set?

     – Tesla Nov 04 '21 at 09:36
  • However, we would then get $${T_2}#{T_1}#\sigma(A)$$, so I guess I must have got something wrong – Tesla Nov 04 '21 at 09:38
  • 1
    I don't quite get you. Note that, given a probability $\sigma$ and a map $T$, then $T_#\sigma$ is a new probability. So $(T_{2#}\sigma)(T_1^{-1}(A))$ can be read as the $T_{2#}\sigma$-probability of the measurable set $T_1^{-1}(A)$. What would it mean, for you, the expression $T_{2#}\sigma(T^{-1}(A))$? To me, they mean the same: a probability acting on a measurable set. I just added the parenthesis to clarify. – fcz Nov 04 '21 at 09:42