Characteristic Function of a Compound Poisson Process

mikhairu
Messages
2
Reaction score
0
Hello,

I am trying to find a characteristic function (CF) of a Compound Poisson Process (CPP) and I am stuck :(.

I have a CPP defined as X(t) = SIGMA[from j=1 to Nt]{Yj}. Yj's are independent and are Normally distributed.

So, in trying to find the CF of X I do the following:
(Notation: CFy = Characteristic Function of y (y is subscript))

CF(X) = E[exp{i*u*X}]
= E[ (E[exp{i*u*Y}])^N ]
= E[ (CFy(u))^N ]
= E[ (exp{ ln[CFy(u)] })^N ]
which is really just a moment generating function Phi_N (Phi subscript N):
Phi_N( ln[CFy(u)] ).

I don't know how to go from here... Y's are Normally distributed but the entire process is Poisson.. so I'm not sure how to combine these. Please help!

Thank you.
 
on Phys.org
mikhairu said:
Hello,

I am trying to find a characteristic function (CF) of a Compound Poisson Process (CPP) and I am stuck :(.

I have a CPP defined as X(t) = SIGMA[from j=1 to Nt]{Yj}. Yj's are independent and are Normally distributed.

So, in trying to find the CF of X I do the following:
(Notation: CFy = Characteristic Function of y (y is subscript))

CF(X) = E[exp{i*u*X}]
= E[ (E[exp{i*u*Y}])^N ]
= E[ (CFy(u))^N ]
= E[ (exp{ ln[CFy(u)] })^N ]
which is really just a moment generating function Phi_N (Phi subscript N):
Phi_N( ln[CFy(u)] ).

I don't know how to go from here... Y's are Normally distributed but the entire process is Poisson.. so I'm not sure how to combine these. Please help!

Thank you.

After line 3 it should be possible to evaluate expressions of the form E[z^N] using P[N=n]=exp(-L*t)*(L*t)^n/n! where L is the rate of the Poisson process.
 

Similar threads

Replies
8
Views
3K
Replies
8
Views
3K
  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 36 ·
2
Replies
36
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
8K
  • · Replies 3 ·
Replies
3
Views
2K