combining Gamma distributions with different k parameters
FangQ
My research recently involved the modeling of photon migration in
highly scattering random media. I need some input to understand the
path-lengths of the photons (particles under random walk) exiting from
the domain.
Photon scattering length inside a random medium is modeled by a random
variable (RV) of exponential distribution. As a consequence, I believe
that when the photon exits the domain, the total path-length is a sum
of all exponential RVs. If there are N scattering events before the
exit, the path-length is supposed to be a RV with Gamma (or Erlang)
distribution with k=N. However, the photon does not have to scatter N
times, N can be any number from 0 to inf, so, that means the path-
length of a exit photon should be a "combined distribution" of all
different Gammas with k from 0~inf. I am not entirely sure about the
function form of k(n).
Here are my questions:
1. what kind form do you expect if one want to "combine" these Gamma
distributions with different k parameters? do you think there will be
an analytical form for that? is integrate(Gamma(k(n),theta),n,0,inf) a
reasonable form? or it should be convolution type?
2. I know that adding two RVs with Gamma(k1,theta) and Gamma(k2,theta)
is a Gamma(k1+k2,theta), will this simplify the integrate(Gamma(k
(n),theta),n,inf) to Gamma(integrate(k(n),n,0,inf),theta)?
3. is there a known distribution to describe how many scattering
events happened before a random particle exit a domain at a given
point? (which is k(n) as we mentioned above)
I am great appreciated if someone can point me to the references that
are related to this problem. Any hint or suggestion is also welcome!
thanks
Qianqian
Ray Koopman
That sounds like N should be treated as a random variable.
Then fX(x) = sum_n fXn(x|n) * fN(n).
Is there theory that suggests a particular distribution for N?
Ray Koopman
Make that fX(x) = c * sum_n fXn(x|n) * fN(n),
where c is chosen so that fX(x) integrates to 1 over (0,infinity).
Ray Koopman
> [...]
> Make that fX(x) = c * sum_n fXn(x|n) * fN(n),
> where c is chosen so that fX(x) integrates to 1 over (0,infinity).
If every fXn integrates to 1, and fN sums to 1, then c = 1
and drops out. Looks like it's one of those days :(
FangQ
> That sounds like N should be treated as a random variable.
> Then fX(x) = sum_n fXn(x|n) * fN(n).
> Is there theory that suggests a particular distribution for N?
thanks for your comments on my questions.
I think (not entirely sure) that the distribution of N is a Poisson
distribution with a mean of x*mus, i.e. fN(n)=(x*mus)^n*exp(-x*mus)/
(n!), where mus is the scattering coefficient of the medium (constant,
it has a unit of 1/length, denoting the probability of hitting a
scattering event per unit length, x is the total path length in the
medium).
Now I don't know if this complicates the situation: fN(n) is now a
function of x; but it does look promising. I will do more derivations
from here.
danh...@yahoo.com
The answer to part 1 is yes, there is an analytic form that gives a
closed form expression. A random sum of exponentials (which is what
you have) is an exponential when the number of summands
has a geometric distribution. I.e., the probability that there are n
summands has probability
(1-p)p^(n-1), n=1,2,3,... for 0<p<1.
Ray Koopman
If X|n ~ Gamma[n+1] and N ~ Poisson[m] then Mathematica gets
fX[x,m] = Exp[-(m+x)] * Hypergeometric0F1Regularized[1,m*x],
for which the mean is m+1. (Note that I've taken the scale parameter
-- your 'theta' -- to be 1, so everything will need to be scaled.)
Note also that the number of path segments is n+1, not n. Using just
n would mean that a pathlength of 0 has a nonzero probability.