Proving the finiteness of Expectation given a geometric condition

Question

For a non-negative integer-valued random variable $X$, it is given that $P(X > kN) \le p^k$ for any $k \ge 1$ with some fixed $0 < p < 1$ and fixed $N > 0$, then show that $E[X] < \infty$.

My approach :

We know that for an integer valued RV which is non-negative, $E[X]=\sum_{i=0}^{\infty}P[X>i]$.

My plan was to take $k=\frac{i}{N}$ and thus $E[X]$ is bounded above by a geometric series. But the only caveat is that $k$ in my case might not be $\ge 1$. I am stuck here! Anyone with other ideas?

Answer 1

Note that $\{kN: k \in \mathbb{N}_0\} = N\mathbb{N}_0 = \{0, N, 2N, 3N, \cdots\}$. Note that \begin{align*} P(X > i) \le P(X > kN) \quad \text{whenever } kN \le i < (k+1)N \end{align*} and so \begin{align*} E[X] = \sum_{i=0}^{\infty} P(X > i) \le N \sum_{k=0}^{\infty}P(X > kN) \le N\sum_{k=0}^{\infty} p^k = \frac{N}{1-p} < \infty \end{align*}

Answer 2

Let $Y=X/N$. Since $P(Y>0)\leq 1$ and $P(Y>i)\leq p^i$ for all $i\geq 1$, we have: $$E(Y)=\sum_{i=0}^\infty P(Y>i)\leq 1+\sum_{i=1}^\infty p^i = \frac{1}{1-p} < \infty.$$ The result follows (and $E(X)\leq N/(1-p)$).