99
Lv 4
99 asked in 科學數學 · 1 decade ago

分配函數與不連續點

Show that a distribution function has at most countably many discontinuities.

2 Answers

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  • 1 decade ago
    Favorite Answer

    Since distribution function is montone decreasing

    so it has at most countably many discontiouous points

    2008-03-17 00:51:43 補充:

    Let f be the distribution function defined on an interval I

    Let S={c€I:f is discontinuous at c}

    then we have lim(x->c+)f(x)<lim(x->c-)f(x)

    so we can choose a rational number c(r) such that

    lim(x->c+)f(x)<c(r)<lim(x->c-)f(x)

    If c ,d in S so that c<d

    then r(d)<lim(x->d-)f(x)<=r[(c+d)]/2<=lim(x->c+)f(x)<r(c)

    =>r(c)≠r(d)

    so r:S->Q be a one to one mapping

    Since Q is countable, so is its subset

    =>S is countable

    2008-03-17 00:54:01 補充:

    應該是r(c)才

    2008-03-18 12:48:46 補充:

    應該是f((c+d)/2)

    Source(s): me
  • 99
    Lv 4
    1 decade ago

    分配函數是nondecreasing吧?!

    怎麼從 "nondecreasing" 推得 "at most countably many discontiouous points" 呢?

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