A simple proof I'm having issues with

[Ant64]

Hi all,

The question is: Given that X follows a geometric distribution, prove that

P(X>j+k | X>k) = P(X>j).

So this is all I have:

P(X>k+j) = (1-p)^k+j, (a proof covered in the text)
= (1-p)^k • (1-p)^j,
No, what I want to say is,

since P(X>k) = (1-p)^k = 1,
then P(X>k+j) = 1 • (1-p)^j,
= P(x>j).
But I don't know if I'm allowed to separate the expression (1-p)^k • (1-p)^j into separate probabilities... Thanks in advance!

 

[Stephen Tashi]

since P(X>k) = (1-p)^k = 1,

That isn't true.



The definition of conditional probability tells you that

$$P( X > j + k | X > k) = \frac{ P( (X > j + k) \cap (X > k) )} {P(X > k)}$$

Set theory tells you that $(X > j + k) \cap (X > k) = (X > j + k)$

 

[Ant64]

Okay, thanks a ton. I can take it from there!