Why does P(r,t) equal δ_r0 δ_t0 + sum of F(r,t')P(0,t-t')?

[WiFO215]

Let P(r,t) define the occupation probability, the probability that a particle emulating a random walk will find itself at position r at time t if starts from the origin at time zero.

Let F(r,t) define the first passage probability, the probability that a particle emulating a random walk will find itself at position r at time t FOR THE FIRST TIME if starts from the origin at time zero.

I was reading a book which says this :
"For a random walk to be at position r at time t, the walk must first reach r at some earlier time step t' and then return to r after t-t' additional steps. This connection between F(r,t) and P(r,t) can thus be expressed by the equation

$$P(r,t) = \delta_{r0} \delta_{t0} + \sum_{t'\leqt}F(r,t')P(0,t-t')$$

"

Can someone please explain how this is possible?

 

[WiFO215]

For some reason, I cannot edit the above. It is supposed to read t' $$\leq$$ t below the summation sign.

 

[WiFO215]

Can someone move this to the Classical Physics subforum? I might probably get more replies there.