Difference between revisions of "Derivation of delta exponential T=isolated points"
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(Created page with "For $t>s$, find $e_p$ by computing $$\begin{array}{ll} e_p(t,s) &= \exp \left( \displaystyle\int_s^t \dfrac{1}{\mu(\tau)} \log(1+\mu(\tau)p(\tau)) \Delta...") |
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$$\begin{array}{ll} | $$\begin{array}{ll} | ||
e_p(t,s) &= \exp \left( \displaystyle\int_s^t \dfrac{1}{\mu(\tau)} \log(1+\mu(\tau)p(\tau)) \Delta \tau \right) \\ | e_p(t,s) &= \exp \left( \displaystyle\int_s^t \dfrac{1}{\mu(\tau)} \log(1+\mu(\tau)p(\tau)) \Delta \tau \right) \\ | ||
− | &= \exp \left( \displaystyle\sum_{k=\pi(s) | + | &= \exp \left( \displaystyle\sum_{k=\pi(s)}^{\pi(t)-1} \log(1+\mu(t_k)p(t_k) \right) \\ |
− | &= \displaystyle\prod_{k=\pi(s) | + | &= \displaystyle\prod_{k=\pi(s)}^{\pi(t)-1} 1+\mu(t_k)p(t_k). |
\end{array}$$ | \end{array}$$ |
Revision as of 23:27, 9 June 2015
For $t>s$, find $e_p$ by computing $$\begin{array}{ll} e_p(t,s) &= \exp \left( \displaystyle\int_s^t \dfrac{1}{\mu(\tau)} \log(1+\mu(\tau)p(\tau)) \Delta \tau \right) \\ &= \exp \left( \displaystyle\sum_{k=\pi(s)}^{\pi(t)-1} \log(1+\mu(t_k)p(t_k) \right) \\ &= \displaystyle\prod_{k=\pi(s)}^{\pi(t)-1} 1+\mu(t_k)p(t_k). \end{array}$$