perhaps i have to do this
1/2 D^2(x) T''(x) + M(x)T'(x) = 0
h = T'(x)
h'/h = 2M(x) / D^2(x)
maybe? whats the solution of this?
cheers all
hi all
so this question is quite tough to answer.
the fundamental solution of the kolmogorov backward equation is:
dp/dt = Lp = 1/2 D^2(x) d2p/dx2 + M(x)dp/dx
where D is the variance and M is the mean.
given a stationary distribution Lp = 0 so if we have a function T(x)
1/2 D^2(x) T''(x) + M(x)T'(x) = 0
a solution to this is the harmonic function
now i want to know how is
T'(x) = C exp(-int_xo^x 2M(s)/D^2(s) ds )
and
T(X) = int^x exp(-int_xo^x 2M(y)/D^2(y) dy ) du
My kingdom for latex..... i hope someone can help...