Hi, it's me a gain.
Probably, my question is very trivial. But I'm not very familiar with conditional distributions and a little bit confused right now.
How can I compute the density functions of the following conditional distributions dependent on \(\displaystyle n\)?
(I) \(\displaystyle Pr(B_n-B_{n-1}\leq x | B_N-B_{N-1}=b)\)
(II) \(\displaystyle Pr((B_n-B_{n-1}) + Y \leq x | B_N-B_{N-1}=b, Y=y)\)
Where \(\displaystyle B\) is a Brownian motion and for the time points it holds: \(\displaystyle n \leq N\)
\(\displaystyle Y\) is a normally distributed r.v. and independent of the Brownian motion
For (I) and \(\displaystyle n<N\) it's just the density function for the Brownian increments due to their independence.
and for \(\displaystyle n=N\) is it just 0 or 1?
Thanks in advance.
Probably, my question is very trivial. But I'm not very familiar with conditional distributions and a little bit confused right now.
How can I compute the density functions of the following conditional distributions dependent on \(\displaystyle n\)?
(I) \(\displaystyle Pr(B_n-B_{n-1}\leq x | B_N-B_{N-1}=b)\)
(II) \(\displaystyle Pr((B_n-B_{n-1}) + Y \leq x | B_N-B_{N-1}=b, Y=y)\)
Where \(\displaystyle B\) is a Brownian motion and for the time points it holds: \(\displaystyle n \leq N\)
\(\displaystyle Y\) is a normally distributed r.v. and independent of the Brownian motion
For (I) and \(\displaystyle n<N\) it's just the density function for the Brownian increments due to their independence.
and for \(\displaystyle n=N\) is it just 0 or 1?
Thanks in advance.