# Kolmogorov's Forward equations in BUS3024S

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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:

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

+1 vote
answered Aug 24 by (2,330 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.