- #1

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found k=1-p

[tex] E(X)=\sum kxp^x =p/(1-p)[/tex]

[tex]P(X>=x) = 1-\sum_{x'=0}^{x} kp^x' = p^{x+1}[/tex]

P(X=y|X>=x) =?

- Thread starter Gregg
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- #1

- 459

- 0

found k=1-p

[tex] E(X)=\sum kxp^x =p/(1-p)[/tex]

[tex]P(X>=x) = 1-\sum_{x'=0}^{x} kp^x' = p^{x+1}[/tex]

P(X=y|X>=x) =?

- #2

vela

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- #3

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is A∩B={y>=x}?

- #4

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SO is it [tex] p^{y-x} [/tex]

- #5

vela

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No, they're just normal sets. For instance, if y=5 and x=3, you'd have A={5} and B={3, 4, 5, 6, 7, ...} and their intersection would be A∩B={5}.

is A∩B={y>=x}?

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