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Kolmogorov Forward differential equation

I

Indy007

Member
Hi!

I'm looking at Chapter 5 section 4 of the notes (page 20), and am having difficulty following the logic behind the calculation of the Kolmogorov Forward equation.

Specifically, I'm having problem with the first step where:

Sum[Pik(t)Pkj(h)] = Pij(t) +h*Sum[Pik(t) *MUkj +o(h)]

I can't seem to figure out how this has been determined. I've tried using the relationship as specified on page 20, but end up with the following:

Sum[Pik(t)Pkj(h)] = Pij(t)(1+MUjj+oh) +h*Sum[Pik(t) *MUkj +o(h)]

Apologies if I've overlooked something obvious, but have been staring at the notes for a while now and can't seem to wrap my head round it!:confused:

I'd appreciate any help you can give me.

~Indy
 
My guess is, you should not substitute them directly, instead write it like this:

= Sum P_ik(t) { P_kj(h) + h.Mu_kj + o(h)}

where if k = j, then the {} part becomes 1 + h.Mu_jj + o(h)
and if k =/ j, then {} becomes h.Mu_kj + (o)h
just like how 5.7 is.

Then, breaking up the summation:

= Sum_k=j P_ik(t).P_kj(h) + ... (the part you have correctly)
but the first part is really P_ij(t), so there you have it.

Can someone please confirm my derivation, please?

Thanks.
 
I think you're on the right track Indy, just need one more step.

You've got

Sum[Pik(t)Pkj(h)] = Pij(t)(1+MUjj+oh) +h*Sum [Pik(t) *MUkj +o(h)]
= Pij(t) + Pij(t)(MUjj + oh) + h*Sum [Pik(t) *MUkj +o(h)]

where the summation is for values of k not equal to j.
By changing the summation to take all values of k, the Pij(t)(MUjj + oh) gets included as well, so you're left with

Pij(t) +h*Sum[Pik(t) *MUkj +o(h)]

as required.
 
I think you're on the right track Indy, just need one more step.

You've got

Sum[Pik(t)Pkj(h)] = Pij(t)(1+MUjj+oh) +h*Sum [Pik(t) *MUkj +o(h)]
= Pij(t) + Pij(t)(MUjj + oh) + h*Sum [Pik(t) *MUkj +o(h)]

where the summation is for values of k not equal to j.
By changing the summation to take all values of k, the Pij(t)(MUjj + oh) gets included as well, so you're left with

Pij(t) +h*Sum[Pik(t) *MUkj +o(h)]

as required.

Hi Michael

How do you combine the 2nd and 3rd term together when one has "h" and the other does not?
 
Last edited by a moderator:
Hrm... Er, clearly you can't. :) Guess I should have looked a little more closely before posting...

The equation that Indy posted:

Sum[Pik(t)Pkj(h)] = Pij(t)(1+MUjj+oh) +h*Sum[Pik(t) *MUkj +o(h)]

isn't quite correct, I think it should be

Sum kES[Pik(t)Pkj(h)]
= Pij(t)Pjj(h) + Sum k!=j[Pik(t)Pkj(h)]
= Pij(t)[1+h*MUjj+o(h)] + Sum k!=j [Pik(t) [h*MUkj +o(h)]]

Now you can take the Pij(t) * (hMUjj + o(h)) into the summation.
 
For the FDE of P3,7(t), is it:

[row 4 of P]*[column 8 of A]

i.e. [P30(t),P31(t),...]*[Generator matrix values for column 8]
 
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