Timeline for In general is $\frac{d\,\mu_1}{d\,\mu_2}\circ T = \frac{d\,T\mu_1}{d\,T\mu_2}$?
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| Jun 1, 2022 at 14:38 | comment | added | Sanae Kochiya | @Steve Thank you for your answers! For your question, I believe at least I need $T$ to have a right inverse to have the correct equation stand even without assuming the system is ergodic. However, now if I assume the equation holds (or the equivalent one stated by leo), then I will have $\mu_1 = T\mu_1$ (?) When the system is ergodic and this equation holds, then all characteristic functions will be the constant one (?) | |
| Jun 1, 2022 at 14:27 | vote | accept | Sanae Kochiya | ||
| Jun 1, 2022 at 11:32 | comment | added | Steve | Thanks @leomonsaingeon for your remark. If $T$ is not invertible, which property of $T$ is the necessary one to obtain the result you write? If $T$ is for instance constant, then the result you write cannot hold of course. | |
| Jun 1, 2022 at 10:06 | comment | added | leo monsaingeon | Yes, and in fact an even better way to phrase the correct statement is $\frac{dT\mu_1}{dT\mu_2}\circ T=\frac{d\mu_1}{d\mu_2}$, which also holds true in case $T$ is not invertible. | |
| Jun 1, 2022 at 8:43 | history | answered | Steve | CC BY-SA 4.0 |