I have gotten the first half of this question, but need help on the second half.

Let $\displaystyle X$ be a random variable with geometric distribution of parameter $\displaystyle p$.

a)Show that $\displaystyle X$ satisfies the memoryless property:

$\displaystyle P(X>n+k|X>n)=P(X>k)$

Here is my answer to this half:

If $\displaystyle X$ has the probability distribution function $\displaystyle P(x)=pq^{x-1}, x=1,2,...$, $\displaystyle p+q=1$

Then $\displaystyle P(X>n+k|X>n)=P(X>k)$, $\displaystyle \forall n,k>0$

Consider the following, where $\displaystyle y$ is some positive integer such that $\displaystyle X>y$

$\displaystyle P(X>y)=P(X=y+1)+P(X=y+2+...=pq^y+pq^{y+1}+pq^{y+2} +...$

$\displaystyle =pq^y(1+q+q^2+...)=\frac{pq^y}{1-q}=\frac{pq^y}{p}=q^y$

Therefore, $\displaystyle P(X>y)=q^y\Rightarrow P(X>n)=q^n, P(X>k)=q^k, P(X>n+k)=q^{n+k}$

$\displaystyle P(X>n+k|X>n)=\frac{P(X>n+k, X>n)}{P(X>n)}=\frac{P(X>n+k)}{P(X>n)}=\frac{q^{n+k }}{q^n}=q^k=P(X>k)$

(obtained from this source: memoryless property of geometric distribution)

b)Reciprocally, show that any random variable $\displaystyle Y$, having nonnegative integer values, whose distribution satisfies the property above has necessarily a geometric distribution.

I presently don't have an answer to this half of the question, so I could use a hand.