Welcome to the hotseat. We've prepared a guide if you'd like to read more about how it works.

Kolmogorov's Forward equations in BUS3024S

+1 vote
84 views
asked Aug 22 in BUS 3024S - Contingencies by anonymous

Hi guys, I am a bit confused about the Kolmogorov forward equations that we derive in Contingencies:


$$\frac{d}{dt}\  _tp_x = \Sigma_{k \neq j} (_tp_x^{ik}\mu_{x+t}^{kj} - _tp_x^{ij}\mu_{x+t}^{jk}) $$ 

Because they look a bit different to those I have derived before in other stats courses and Models, which are:

$$ \frac{d}{dt} \  _tp_x = \Sigma_{k \neq j } (_tp_x^{ik}\mu_{x+t}^{kj}) $$


Are they equivalent? Are they different? Please help.



1 Answer

+2 votes
answered Aug 24 by Njabulo.Dube (2,950 points)

Taking Kolmogorov's forward DE as presented in contingencies, we can simplify the RHS as follows:

$$ \sum_{k \neq j} (_tp_x^{ik}\mu_{x+t}^{kj} - _tp_x^{ij}\mu_{x+t}^{jk}) = \sum_{k \neq j} (_tp_x^{ik} \mu_{x+t}^{kj}) -  \sum_{k \neq j} (_tp_x^{ij} \mu_{x+t}^{jk})$$ 

$$ \sum_{k \neq j} (_tp_x^{ik}\mu_{x+t}^{kj} - _tp_x^{ij}\mu_{x+t}^{jk}) = \sum_{k \neq j} (_tp_x^{ik} \mu_{x+t}^{kj}) - _tp_x^{ij}  \sum_{k \neq j} \mu_{x+t}^{jk}$$ 

$$ \sum_{k \neq j} (_tp_x^{ik}\mu_{x+t}^{kj} - _tp_x^{ij}\mu_{x+t}^{jk}) = \sum_{k \neq j} (_tp_x^{ik} \mu_{x+t}^{kj}) + _tp_x^{ij}\mu_{x+t}^{jj}$$  since \(   \mu_{x+t}^{jj} = -  \sum_{k \neq j} \mu_{x+t}^{jk} \). Therefore:

$$ \sum_{k \neq j} (_tp_x^{ik}\mu_{x+t}^{kj} - _tp_x^{ij}\mu_{x+t}^{jk}) = \sum_{k} (_tp_x^{ik} \mu_{x+t}^{kj}) = \frac{d}{dt} \  _tp_x^{ij} $$ 

I will leave it to you to verify the DE as derived in your statistics classes as see why the last line above should hold in general.

...