Probability question about geometric distribution Let X have a geometric distribution. Show that <math xmlns="http://www.w3.org/1998/Math/MathML"> <mi>P</mi> <mo stretchy="false">(</mo> <mi>X</mi> <mo>&gt;</mo> <mi>k</mi> <mo>+</mo> <mi>j</mi> <mrow class="MJX-TeXAtom-ORD"> <mo stretchy="false">|</mo> </mrow> <mi>X</mi> <mo>&gt;</mo> <mi>k</mi> <mo stretchy="false">)</mo> <mo>=</mo> <mi>P</mi> <mo stretchy="false">(</mo> <mi>X</mi> <mo>&gt;</mo> <mi>j</mi> <mo stretchy="false">)</mo> </math>, where k and j are nonnegative integers.

Raiden Barr 2022-10-19 Answered
Probability question about geometric distribution
Let X have a geometric distribution. Show that P ( X > k + j | X > k ) = P ( X > j ), where k and j are nonnegative integers.
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Answers (1)

Kyle Delacruz
Answered 2022-10-20 Author has 21 answers
Step 1
I will use the formula of conditional probability: P ( A | B ) = P ( A B ) P ( B ) .
Thus P ( X > k + j | X > k ) =
P ( X > ( k + j )   and   X > k ) / P ( X > k ) = P ( X > ( k + j ) ) / P ( X > k )
Step 2
As we are working with a geometric distribution, P ( X > ( k + j ) ) = q k + j and P ( X > k ) = q k .
Then P ( X > ( k + j ) ) / P ( X > k ) = q k + j / q k = q j = P ( X > j )
= P ( X > j ) when q is the probability of failure.
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