Timeline for The disintegration of the convolution of two probability measures
Current License: CC BY-SA 3.0
6 events
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| Apr 4, 2018 at 4:01 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Apr 3, 2018 at 22:46 | comment | added | Iosif Pinelis | Previous comment continued: Let $\mu_t$ and $\nu_t$ be the probability distributions of $D_{\sigma,\rho,t}$ and $B_t$, respectively. Now, to get $\hat\mu_t$ from $\mu_t$, vary $\rho$ and $\sigma$ so that $(1-\rho^2)\sigma^2$ stay constant; cf. e.g. page 4 of bit.ly/2GAqbaJ . I think very likely this will give an example when $(\mu_t)_h=(\hat\mu_t)_h$ for $p_* \mu_t$-almost all $h \in H$ and, equivalently, for $p_* \hat\mu_t$-almost all $h \in H$ (and for all $t\ge0$), but $(\mu_t * \nu_t)_h\ne(\hat\mu_t * \nu_t)_h$. | |
| Apr 3, 2018 at 22:40 | comment | added | Iosif Pinelis | Well, Alex, you did not mention semigroups in your original question, and now your question is quite a bit different. I think you can get a counterexample for entire semigroups, though, but you'd have to work somewhat harder. I actually think about any example will be a counterexample. I suggest you try the following. Let $G$ and $H$ be as in my example. Let $(B_t)$ and $(C_t)$ be two independent standard Brownian motions. Let $D_{\sigma,\rho,t}:=\sigma(\rho B_t+\sqrt{1-\rho^2}C_t)$ for $\sigma>0$ and $\rho\in(0,1)$. | |
| Apr 3, 2018 at 21:18 | comment | added | Alex M. | Does the edit (last paragraph) that I have performed on my question clarify what I am looking for? | |
| Apr 3, 2018 at 20:52 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Apr 3, 2018 at 20:43 | history | answered | Iosif Pinelis | CC BY-SA 3.0 |