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).

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- Oct 5th 2009, 06:41 PMrangergiyaProbability (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 PMMoo
Hello,

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

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

because if X>m+n happens, X>m happens too. - Oct 7th 2009, 08:48 PMmatheagle
- Oct 8th 2009, 07:30 AMqpmathelp

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

mean and variance of the geometric distribution