# Probability (Geometric Distribution)?

• Oct 5th 2009, 06:41 PM
rangergiya
Probability (Geometric Distribution)?
Can someone help me with this question...im so lost...thanks in advance...

If X has the geometric distribution with parameter p, show that
P(X > m + n | X > m) = P(X > n).
• Oct 6th 2009, 12:30 PM
Moo
Hello,

Well, note that $P(X>k)=\sum_{j=k+1}^\infty (1-p)p^j$

And then just write that $P(X>m+n|X>m)=\frac{P(X>m+n,X>m)}{P(X>m)}=\frac{P(X >m+n)}{P(X>m)}$

because if X>m+n happens, X>m happens too.
• Oct 7th 2009, 08:48 PM
matheagle
Quote:

Originally Posted by Moo
Hello,

Well, note that $P(X>k)=\sum_{j=k+1}^\infty (1-p)p^j$

And then just write that $P(X>m+n|X>m)=\frac{P(X>m+n,X>m)}{P(X>m)}=\frac{P(X >m+n,X>m)}{P(X>m)}$

because if X>m+n happens, X>m happens too.

I believe you meant to drop that last inequality

$P(X>m+n|X>m)=\frac{P(X>m+n,X>m)}{P(X>m)}=\frac{P(X >m+n)}{P(X>m)}$
• Oct 8th 2009, 07:30 AM
qpmathelp
Quote:

Originally Posted by rangergiya
Can someone help me with this question...im so lost...thanks in advance...

If X has the geometric distribution with parameter p, show that
P(X > m + n | X > m) = P(X > n).

try the following link for some explanation memoryless property of geometric distribution

mean and variance of the geometric distribution