Hallo everybody,
I've problems computing the following conditional distribution:
Let \(\displaystyle B=(B_t)\) be a Brownian motion.
How can I compute the following conditional distribution function:
\(\displaystyle Q^Z_{t_n}(\cdot \mid \mu=\theta,B_{t_N}=i)\)?
= condtional distribtuin of \(\displaystyle Z_{t_n}\) given \(\displaystyle \mu=\theta,B_{t_N}=i\)
where
\(\displaystyle Z_{t_n}:=\mu + \sigma(B_{t_n}-B_{t_{n-1}}) \)
\(\displaystyle \sigma>0\) a constant
\(\displaystyle t_{n-1}<t_n<t_N\)
\(\displaystyle \mu\) is independent of \(\displaystyle B\) and normally distributed
Does there exist a density function \(\displaystyle q_{t_n}\) with respect to a \(\displaystyle \sigma-\)finite measure \(\displaystyle \lambda\), i.e.
\(\displaystyle Q_{t_n}^{Z}(dz\mid \theta,i)=q_{t_n}^{Z}(z\mid \theta,i)\lambda(dz)\)?
Can anybody help me?
Thanks in advance
I've problems computing the following conditional distribution:
Let \(\displaystyle B=(B_t)\) be a Brownian motion.
How can I compute the following conditional distribution function:
\(\displaystyle Q^Z_{t_n}(\cdot \mid \mu=\theta,B_{t_N}=i)\)?
= condtional distribtuin of \(\displaystyle Z_{t_n}\) given \(\displaystyle \mu=\theta,B_{t_N}=i\)
where
\(\displaystyle Z_{t_n}:=\mu + \sigma(B_{t_n}-B_{t_{n-1}}) \)
\(\displaystyle \sigma>0\) a constant
\(\displaystyle t_{n-1}<t_n<t_N\)
\(\displaystyle \mu\) is independent of \(\displaystyle B\) and normally distributed
Does there exist a density function \(\displaystyle q_{t_n}\) with respect to a \(\displaystyle \sigma-\)finite measure \(\displaystyle \lambda\), i.e.
\(\displaystyle Q_{t_n}^{Z}(dz\mid \theta,i)=q_{t_n}^{Z}(z\mid \theta,i)\lambda(dz)\)?
Can anybody help me?
Thanks in advance
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