Consider a random variable X, find its PMF

Question

This question has me stumped. I'm fairly certain that the answer to part a.) is pretty simple: $P(X>n+m|X>n)=\frac{P(X>n+m)}{P(X>n)}$, because $X>n+m$ is a subset of $X>n$, which is $\frac{a^{n+m}}{a^n}=a^m$. But part b.) is where I run out of steam. What avenue should I be taking? It's the inequality that's confusing me. I can't see how to obtain $P(X=x)$ from $P(X>n)$ and $P(X>n+m)$.

Answer 1

Hint for c):

If $X$ is a random variable that only takes values in $\mathbb N\cup\{0\}$ then:

$$\mathsf EX=\sum_{n=0}^{\infty}\mathsf P(X>n)$$

This on base of:$$p_1+2p_2+3p_3+\cdots=(p_1+p_2+p_3+\dots)+(p_2+p_3+\cdots)+(p_3+\cdots)+\cdots$$

Answer 2

$$\Pr(X>0)=a^0=1 \rightarrow \Pr(X=0)=0$$

$$\Pr(X=1)=\Pr(X>0)-\Pr(X>1)=1-a $$ $$\Pr(X=2)=\Pr(X>1)-\Pr(X>2)=a-a^2 $$ $$\Pr(X=3)=\Pr(X>2)-\Pr(X>3)=a^2-a^3 $$ ... $$\Pr(X=k)=\Pr(X>k-1)-\Pr(X>k)=a^{k-1}-a^k $$