Suppose that 0 ≤ p n ≤ 1 and put α n = min { p n , 1 − p n } .Show that if ∑ α n converges, then there exist on some discrete probability space events A n satisfying P ( A n ) = α n .In this section, it has already proved that for any sequence of measures { μ n } on the class of all subsets of R with finite support, there exist on some probability space a sequence of independent simple random variables { X n } whose distribution is μ n .But the probability space constructed in that proof is just the unit interval and I am unable to modify it to meet the requirement of this problem.

Question

Suppose that   0  ≤      p    n    ≤  1 and put       α    n    =  min      {          p      n        ,        1    −          p      n        }  .Show that if   ∑      α    n   converges, then there exist on some discrete probability space events       A    n   satisfying   P      (          A      n        )    =      α    n  .In this section, it has already proved that for any sequence of measures       {          μ      n        }   on the class of all subsets of       R   with finite support, there exist on some probability space a sequence of independent simple random variables       {          X      n        }   whose distribution is       μ    n  .But the probability space constructed in that proof is just the unit interval and I am unable to modify it to meet the requirement of this problem.

Gillian Kelly · Accepted Answer

First, we do this with a continuous probability space, then modify it to be discrete. Let   (  Ω  ,      F    ,  P  ) be the unit interval [0,1] with Lebesgue measure, and define      A    n    =      {                            [          0          ,                      α            n                    ]                          if                       p            n                    &amp;lt;          1                      /                    2                                                          [                      α            n                    ,          1          ]                          if                       p            n                    ≥          1                      /                    2                        You can check that indeed   P  (      A    n    )  =      p    n   for all   n  ≥  1.To make this discrete, consider taking the unit interval and cutting it at every point       α    n  . I claim that this divides the unit interval into at most countably many pieces. You can then collapse all of those pieces into a single outcome in a new probability space, where the probability of that outcome is the length of the piece.It is nontrivial to show the number of pieces is countable. For example, if the sequence   (      α    1    ,      α    2    ,  …  ) enumerates the rational numbers in (0,1/2), then snipping the unit interval at these points would mean there would be one &quot;piece&quot; for every irrational number in (0,1/2). However, since       ∑    n        α    n   converges, nothing like this can happen, as convergence implies the set   {      α    n    ∣  n  ∈      N    } each       α    n    &amp;gt;  0 has a neighborhood which avoids all of the other       α    m   for   m  ≠  n.In fact,       ∑    n        α    n    &amp;lt;  ∞is a much stronger assumption than is needed to do this construction. It suffices to assume   {      α    n    ∣  n  ∈      N    } has at most countably many accumulation points.

Suppose that 0 ≤ p n </msub> ≤ 1 a

Answered question

Answer & Explanation

New Questions in Pre-Algebra