Probability (Geometric Distribution)?

**rangergiya** · October 5th 2009, 06:41 PM

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).

**Moo** · October 6th 2009, 12:30 PM

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.

**matheagle** · October 7th 2009, 08:48 PM

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

$P(X>m+n|X>m)=\frac{P(X>m+n,X>m)}{P(X>m)}=\frac{P(X >m+n)}{P(X>m)}$

**qpmathelp** · October 8th 2009, 07:30 AM

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

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