Talk:Indecomposable distribution

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)

Don't understand example

 * Suppose a random variable Y has a geometric distribution


 * $$\Pr(Y = n) = (1-p)^n p\, $$


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


 * $$ Y = \sum_{n=1}^\infty {D_n \over 2^n}, $$

Umm. What?? I don't know how to fix this. --God made the integers (talk) 20:29, 13 January 2017 (UTC)


 * It makes no sense. The second Y is not even an integer -- probably the 2^n needs to be in the numerator. But no reference is given. So we can't correct the logic, it is too far gone.


 * In addition, the point of the example is not clearly explained, although the example is interesting and probably has a point. It reminds me of the fact that an abstract L^2 space has both a discrete decomposition and a continuous decomposition, but they are not compatible.


 * I fixed the 2^n and the wording but there still are gaps. 2001:171B:2274:7C21:B13D:6306:7B8E:4356 (talk) 21:56, 12 April 2022 (UTC)