- #1

Ant64

- 4

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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!

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!

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