Let $X_1, X_2, \ldots$ be independent and identically distributed (iid) Bernoulli random variables. For each $k,$ $$P\{X=k\} = p, \text{ } 0 <p<1.$$ Set $N_0 = 0$ and $t=1,2, \ldots,$ set $$N=\sum_{k=1}^{t} X_k.$$

Find the joint distribution of $$P\{N_{t_1} = i_1, \ldots, N_{t_d} =i_d \}$$

  • $\begingroup$ Repost of math.stackexchange.com/q/3733385/321264. $\endgroup$ – StubbornAtom Jun 25 '20 at 22:04
  • $\begingroup$ @StubbornAtom. That one contained errors and it was closed for that reason. The one here is the updated version $\endgroup$ – Ab2020 Jun 25 '20 at 22:26
  • $\begingroup$ You are supposed to edit the previous post to add context. It will then be reopened. $\endgroup$ – StubbornAtom Jun 26 '20 at 5:59
  • $\begingroup$ I'm sorry for that $\endgroup$ – Ab2020 Jun 26 '20 at 13:46

Assuming $t_1<t_2<\ldots<t_d$, one can write $P(N_{t_1}=i_1,\ldots,N_{t_d}=i_d)=P(N_{t_1}=i_1,N_{t_2}-N_{t_1}=i_2-i_1,\ldots)$. Each RV of the form $N_{t_k}-N_{t_{k-1}}$ is $Bin(t_k-t_{k-1},p)$ independent of the others.

To conclude, the probability is $\prod {t_k-t_{k-1} \choose i_k-i_{k-1}}p^{i_k-i_{k-1}}(1-p)^{t_k-t_{k-1}-(i_k-i_{k-1})}=p^{i_d}(1-p)^{t_d-i_d}\prod {t_k-t_{k-1} \choose i_k-i_{k-1}}$

  • $\begingroup$ Is it possible to also find $P\{ N_{t_1}-i_1, \ldots, N_{t_d} \}$? $\endgroup$ – Ab2020 Jun 25 '20 at 22:15
  • $\begingroup$ I don't understand this notation. $\endgroup$ – YJT Jun 25 '20 at 22:56
  • $\begingroup$ I still don't follow. The last expression is exactly what is calculated in my answer. $\endgroup$ – YJT Jun 26 '20 at 5:07

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