# Thread: Geometric distribution of parameter p: memoryless property

1. ## Geometric distribution of parameter p: memoryless property

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

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

a) Show that $X$ satisfies the memoryless property:
$P(X>n+k|X>n)=P(X>k)$

Here is my answer to this half:

If $X$ has the probability distribution function $P(x)=pq^{x-1}, x=1,2,...$, $p+q=1$
Then $P(X>n+k|X>n)=P(X>k)$, $\forall n,k>0$
Consider the following, where $y$ is some positive integer such that $X>y$
$P(X>y)=P(X=y+1)+P(X=y+2+...=pq^y+pq^{y+1}+pq^{y+2} +...$
$=pq^y(1+q+q^2+...)=\frac{pq^y}{1-q}=\frac{pq^y}{p}=q^y$
Therefore, $P(X>y)=q^y\Rightarrow P(X>n)=q^n, P(X>k)=q^k, P(X>n+k)=q^{n+k}$
$P(X>n+k|X>n)=\frac{P(X>n+k, X>n)}{P(X>n)}=\frac{P(X>n+k)}{P(X>n)}=\frac{q^{n+k }}{q^n}=q^k=P(X>k)$
(obtained from this source: memoryless property of geometric distribution)

b) Reciprocally, show that any random variable $Y$, 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.

2. I was just playing around with it and, this seems to work

let $a=P(X>1)$

then by the definition of memoryless we have $P(X>2|X>1)=P(X>1)=a$

But $P(X>2|X>1)={P(X>2)\over P(X>1)}$ so $P(X>2)=a^2$

so you should continue in this manner and derive $P(X>k)=a^k$

Then $P(X=k)=P(X>k-1)-P(X>k)=a^{k-1}-a^k=a^{k-1}(1-a)$

3. Originally Posted by matheagle
I was just playing around with it and, this seems to work

let $a=P(X>1)$

then by the definition of memoryless we have $P(X>2|X>1)=P(X>1)=a$

But $P(X>2|X>1)={P(X>2)\over P(X>1)}$ so $P(X>2)=a^2$

so you should continue in this manner and derive $P(X>k)=a^k$
I see how that works (proof by induction), but I'm not sure if that's what's expected. Here's a hint I was given by my Prof.

You may consider $G(n)=P(Y>n)$ and translate the memoryless property into a relation between $G(n+k)$, $G(n)$, and $G(k)$, to obtain a recursive formula in $n$ leading to the pmf of $Y$.

Maybe I'm reading it wrong, but I'm having trouble figuring out how to do it like that.

4. why don't you show him what I did, I finished it as well.

I'm not sure what to say, but I can mimick what I just did with G.

G(n+k)=G(n)G(k) by the memoryless property

So G(n+1)=G(n)G(1) and G(n+2)=G(n+1)G(1)=G(n)[G(1)]^2.....

5. Your method seems like it'd be a good alternative to what I think is expected of us.

If possible, though I'd also like to obtain the answer in terms of what my Prof. gave to me. I'll list his hint below (if no one can answer it this way, I'll use matheagle's method).

Relating to the first hint (see above), when you get relationships between $G(n+k)$, $G(n)$, and $G(k)$, take $k=1$ to get $G(n+1)$ from $G(n)$. So recursively you will get $G(n)$ as a function of $G(1)$. Take $G(1)=1-p$, then write $G(n)$ in terms of $p$, and use the fact that $P(Y=k)=P(Y>k-1)-P(Y>k)$, to get the form of $P(Y=k)$ as a geometric distribution with parameter $p$.

It's probably easy, though I'm slightly confused by the wording.

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### geometric distribution memoryless property

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