Talk:Indecomposable distribution: Difference between revisions

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Maybe Typo ???

{\begin{matrix}V=\left\{{\begin{matrix}1&\mathrm {with} \ \mathrm {probability} &1,\\0&\mathrm {with} \ \mathrm {probability} &1-q,\end{matrix}}\right.\end{matrix}}

I feel this should be

{\begin{matrix}V=\left\{{\begin{matrix}1&\mathrm {with} \ \mathrm {probability} &q,\\0&\mathrm {with} \ \mathrm {probability} &1-q,\end{matrix}}\right.\end{matrix}}

-hgkamath

Yes. The "q" and "1" keys are close together. Thanks. I've fixed it. Michael Hardy 01:30, 28 August 2006 (UTC)[reply]

Suppose a random variable Y has a geometric distribution

\Pr(Y=n)=(1-p)^{n}p\,

... now let D_n be the nth binary digit of Y ... then the Ds are independent and

Y=\sum _{n=1}^{\infty }{D_{n} \over 2^{n}},

@@ Line 28: / Line 28: @@
 Yes.  The "q" and "1" keys are close together.  Thanks.  I've fixed it. [[User:Michael Hardy|Michael Hardy]] 01:30, 28 August 2006 (UTC)
+== Don't understand example ==
+:''Suppose a random variable Y has a [[geometric distribution]]''
+:::<math>\Pr(Y = n) = (1-p)^n p\, </math>
+:''... now let D<sub>n</sub> be the nth binary digit of Y ... then the Ds are independent and''
+:::<math> Y = \sum_{n=1}^\infty {D_n \over 2^n}, </math>
+Umm. What?? I don't know how to fix this. --[[User:God made the integers|God made the integers]] ([[User talk:God made the integers|talk]]) 20:29, 13 January 2017 (UTC)