Martingale Problem and Quadratic Family
Fuente:
arXiv
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| Format: | Preprint |
| Veröffentlicht: |
2025
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| _version_ | 1866916938127507456 |
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| author | Wang, Haoming |
| author_facet | Wang, Haoming |
| contents | Assuming uniqueness of the martingale problem for Markov processes of generators $q_t$ in a quadratic family like \[q_t(i,j) = a_t(i) q_0(i,j)^2 + b_t(i) q_0(i,j) - \frac{a_t(i)}{N} \sum_k q_0(i,k)^2,\] where $a_t(i),b_t(i)$ are predictable processes, $N$ is the number of states, and $q_0$ represents the generator of a stationary reference Markov process which satisfies $q_0(i,j)>0$ for all $i,j$, we obtain the sufficient and necessary conditions for the Girsanov transformation. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2509_06016 |
| institution | arXiv |
| publishDate | 2025 |
| record_format | arxiv |
| spellingShingle | Martingale Problem and Quadratic Family Wang, Haoming Probability 60G46 (primary), 60J10, 60J28, 60J35 (secondary) Assuming uniqueness of the martingale problem for Markov processes of generators $q_t$ in a quadratic family like \[q_t(i,j) = a_t(i) q_0(i,j)^2 + b_t(i) q_0(i,j) - \frac{a_t(i)}{N} \sum_k q_0(i,k)^2,\] where $a_t(i),b_t(i)$ are predictable processes, $N$ is the number of states, and $q_0$ represents the generator of a stationary reference Markov process which satisfies $q_0(i,j)>0$ for all $i,j$, we obtain the sufficient and necessary conditions for the Girsanov transformation. |
| title | Martingale Problem and Quadratic Family |
| topic | Probability 60G46 (primary), 60J10, 60J28, 60J35 (secondary) |
| url | https://arxiv.org/abs/2509.06016 |