- #1
TeenieBopper
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Statistics: geometric distribution "proof" problem
If Y has a geometric distribution with success probability p, show that:
P(Y = an odd integer) = [itex]\frac{p}{1-q^{2}}[/itex]
p(y)=p(q)[itex]^{2}[/itex]
p(1)=pq^0
p(3)=pq^2
p(5)=pq^4
.
.
p(2k+1)=pq^2k
I also know the sum of a geometric series is basically [itex]\frac{first-next}{1-ratio}[/itex]
Basically, I'm stuck on how to something set up so I can do some manipulation to eventually lead to P(Y = an odd integer) = [itex]\frac{p}{1-q^{2}}[/itex]
Homework Statement
If Y has a geometric distribution with success probability p, show that:
P(Y = an odd integer) = [itex]\frac{p}{1-q^{2}}[/itex]
Homework Equations
p(y)=p(q)[itex]^{2}[/itex]
The Attempt at a Solution
p(1)=pq^0
p(3)=pq^2
p(5)=pq^4
.
.
p(2k+1)=pq^2k
I also know the sum of a geometric series is basically [itex]\frac{first-next}{1-ratio}[/itex]
Basically, I'm stuck on how to something set up so I can do some manipulation to eventually lead to P(Y = an odd integer) = [itex]\frac{p}{1-q^{2}}[/itex]