Show Process is a Poisson Process (Check 3 conditions).

Question

Suppose $A_t$ is a Process with rate 1. Show that $B_t=A_{2t}$ is a Poisson process.

I have to check three conditions:

A Poisson process with rate $\lambda$ is a random process $N=(N_t: t\geq 0$) so that:

$1.$ $N$ is a counting process

$2.$ $N$ has independent increments, for any positive integer $m$, and any $t_0<t_1<...t_m$, the increments $N_{t_1}-N_{t_0}$, $N_{t_2}-N_{t_1}$,....$N_{t_m}-N_{t_{m-1}}$ are mutually independent.

$3.$ $N(t)-N(s)$ has Poisson($\lambda(t-s))$ for $t \geq s$.

My main problem is with step 2.

My Attempt:

1. Since $A_t$ is a counting process, $B_t=A_{2t}$ must also clearly be a counting process. (Not sure if I should give more explanation...)

I'm having trouble writing this part formally or properly, I think some of my logic is a bit confused or not precise:

2. Let $m$ be a positive integer and consider $t_0<t_1<...t_m$ and $t_m< t_{m+1}< t_{m+2},....<t_{2m}$. We have that the increments $A_{t_1}-A_{t_0}, A_{t_2}-A_{t_1},A_{t_3}-A_{t_2}, A_{t_4}-A_{t_3}...,A_{t_m}-A_{t_{m-1}}, A_{t_{m+1}}-A_{t_{m}},.....A_{t_{2m}}-A_{t_{2m-1}}$ are mutually independent since $A_t$ is a Poisson Process.

Thus the increments,

$A_{t_{2m}}-A_{t_{2m-1}}+A_{t_{2m-1}}-A_{t_{2m-2}}=A_{t_{2m}}-A_{t_{2m-2}}=B_{t_{m}}-B_{t_{m-1}}$, $A_{t_{2m-2}}-A_{t_{2m-3}}+A_{t_{2m-3}}-A_{t_{2m-4}}=A_{t_{2m-2}}-A_{t_{2m-4}}=B_{t_{m-1}}-B_{t_{m-2}}$,

.....$A_{2t_1}-A_{2t_0}=B_{t_{1}}-B_{t_{0}}$ are mutually independent.

Hopefully this is right:

3. Let $t \geq s$. We have that $B(t)-B(s)=A(2t)-A(2s)=A(2t)-A(2t-1)+A(2t-1)-A(2t-2)+....A(2s+1)-A(2s)$. We know that $A(2t)-A(2t-1), ....A(s+1)-A(s)$ are all independent Poisson random variables with rate $1$. We know that the sum of independent Poisson random variables is Poisson with the sum of the rates. There are $2(t-s)$ terms in the above sum. Thus:$A(2t)-A(2s)=A(2t)-A(2t-1)+A(2t-1)-A(2t-2)+....A(2s+1)-A(2s)$ is Poisson with rate $2(t-s)$. Hence $B(t)-B(s)$ is Poisson with rate $2(t-s)$ for $t \geq s$. Hence $B_t$ is a Poisson process with rate 2.

Answer 1

This all seems correct, both part 2 and 3. I guess you could argue in 1 that B counts as A, only twice as fast, but it's still just integers. :)