Determine all probability distributions

Question

Let $(\mathbb{N}_0,\mathcal{P}(\mathbb{N}_0),\mathbb{P})$ be a probability space and $A_k:=\left\{k,k+1,k+2,\ldots\right\}$ for $k\in\mathbb{N}_0$. Assume that $$ \mathbb{P}(A_{k+l}|A_k)=\mathbb{P}(A_l)~~~\text{for all}~~~k,l\in\mathbb{N}_0. $$ Determine all probability distributions $\mathbb{P}$ on $\mathcal{P}(\mathbb{N}_0)$ with this property.

I did not find a real understanding to this task. Can you give me some help how to start this task?

Of course I made some efforts, which lead me to: $$ \mathbb{P}(A_{k+l}|A_k)=\frac{\mathbb{P}(A_{k+l}\cap A_k)}{\mathbb{P}(A_k)}=\mathbb{P}(A_l)~~~\Leftrightarrow~~~\mathbb{P}(A_{k+l}\cap A_k)=\mathbb{P}(A_k)\mathbb{P}(A_l).~~~~~(*) $$

For $l=0$ it is $\mathbb{P}(A_{k+l}\cap A_k)=\mathbb{P}(A_k)$, so that $(*)$ says that $\mathbb{P}(A_l)=1$.

For $l>0$ it is $\mathbb{P}(A_{k+l}\cap A_k)=\mathbb{P}(A_{k+l})$, so that $(*)$ says that $\mathbb{P}(A_{k+l})=\mathbb{P}(A_k)\mathbb{P}(A_l)$.

Furthermore, the usual properties of a probability measure have to be fullfilled, i.e. $\mathbb{P}(\emptyset)=0$, $\mathbb{P}(\mathbb{N}_0)=1$, $\mathbb{P}\left(\biguplus_n B_n\right)=\sum_n \mathbb{P}(B_n)$ for disjoint $B_n\in\mathcal{P}(\mathbb{N}_0)$.

But to be honest, I do not know how to use this... or if this is the right beginning...

Answer 1

Let $a=P(A_1)$, then $a$ is in $[0,1]$ and $P(A_{k+1}\mid A_k)=a$ for every $k$, which, since $A_{k+1}\subset A_k$, implies $P(A_{k+1})=aP(A_k)$ for every $k$. An easy induction shows that $P(A_k)=a^k$ for every $k\geqslant0$, hence $P(\{k\})=(1-a)a^k$ for every $k\geqslant0$ since $\{k\}=A_k\setminus A_{k+1}$. The sequence $(A_k)$ is nonincreasing with intersection $\varnothing$ hence $P(A_k)\to P(\varnothing)=0$ when $k\to\infty$, that is, $a$ is in $[0,1)$. Finally:

If $P(A_1)=0$, $P(\{0\})=1$ hence $P$ is the Dirac measure at $0$. If $P(A_1)$ is in $(0,1)$, $P$ is the geometric distribution with parameter $a$. And $P(A_1)=1$ is impossible.

Answer 2

Let write $A_k$ as $(X\geq k)$, since that's a more normal way of writing it. Then the statement we're given is

$$P(X \geq k + \ell \; |\; X \geq k) = P(X \geq \ell).$$

We can rewrite this as

$$P(X \geq k + \ell) = P(x \geq \ell)\,P(X \geq k)$$

Write $p_i$ for $P(X = i)$. Then

$$P(X \geq k) = 1 - p_0 - p_1 - \cdots - p_{k-1}$$

so the previous expression becomes

$$(1 - p_0 - \cdots - p_{k+\ell-1}) = (1 - p_0 - \cdots - p_{k-1})(1 - p_0 - \cdots - p_{\ell-1})$$

Take $\ell = 1$:

$$(1 - p_0 - \cdots - p_{k}) = (1 - p_0 - \cdots - p_{k-1})(1 - p_0)$$

You should be able to see what to do from there.