Q:

Determining a Mass Function to $P(X>k+1|X>k) = (k+1)/(k+2)$

I have been struggling with this exercise for a while now and I could a push into the right direction. The exercise is the following:
Let $X$ be a random variable which may assume only positive integer values $\{1, 2, \ldots\}$. If $$P(X >k +1 | X > k) = \frac{k+1}{k+2}$$ what is the mass function of X?
After attempting different ideas to this problem, I decided to look it at the problem in terms of cases.

Let $k=1$. Then $P(X>2|X>1) = \frac{2}{3}$
Let $k=2$. Then $P(X>3|X>2)=\frac{3}{4}$

And so on. I am not sure what to make of this pattern. Any help is appreciated. Thank you.

A:

The result is: $$P(X=k)=\dfrac{1}{k(k+1)} \qquad \forall k \in \mathbb N$$
You have that $$P(X>k+1|X>k)=\dfrac{P(X>k+1 \text{ and } X>k)}{P(X>k)}=\dfrac{P(X>k+1)}{P(X>k)}=\dfrac{P(X\ge k+2)}{P(X\ge k+1)}=\dfrac{P(X\ge k+2)}{P(X\ge k+2)+P(X=k+1)}=\dfrac{k+1}{k+2}$$ Solving for $P(X=k+1)$ you have that $$P(X=k+1)=\dfrac{P(X\ge k+2)}{k+1}$$ or equivalently (since the above is true for all $k\in \mathbb N$) $$P(X=k)=\dfrac{P(X\ge k+1)}{k}$$ For $k=1$ you have that $$P(X=1)=\dfrac{P(X\ge2)}{1}=1-P(X<2)=1-P(X=1) \implies P(X=1)=\dfrac{1}{2}$$ since $X$ takes values only in $\{1,2, \ldots\}$. For $k=2$ you have that $$P(X=2)=\dfrac{P(X\ge3)}{2}=\dfrac{1-P(X=2)-P(X=1)}{2}=\dfrac{1}{6}$$ Similarly for $k=3$ you can calculate that $$P(X=3)=\dfrac{1}{12}$$ Now, observe that $$P(X=1)=\dfrac{1}{2}=\dfrac{1}{1\cdot2} \\ P(X=2)=\dfrac{1}{6}=\dfrac{1}{2\cdot3}\\P(X=3)=\dfrac{1}{12}=\dfrac{1}{3\cdot4}$$ Conjecture: $$P(X=k)=\dfrac{1}{k(k+1)}$$ and prove it with induction. (Of course, you can also check that these probabilities sum up to $1$ before proceeding with induction.)

Indeed, this holds true for $k=1,2,3$ and assume it holds for $k \in \mathbb N$. For $k+1$ you have $$P(X=k+1)=\dfrac{P(X\ge k+2)}{k+1}=\dfrac{1-\sum_{l=1}^{k+1}P(X=l)}{k+1} \implies\\ \implies (k+2)P(X=k+1)=1-\sum_{l=1}^{k}P(X=l)$$ where by the induction assumption the RHS can be written as $$1-\sum_{l=1}^{k}\dfrac{1}{l(l+1)}=1-\dfrac{k}{k+1}=\dfrac{1}{k+1}$$ Hence, $$P(X=k+1)=\dfrac{1}{(k+1)(k+2)}$$ which completes the induction.
(we have to test if these probabilities add up to $1$ and then prove it by induction).

