Conference rusure::math

     To be sure that I understand you:
     
     F(x) = the probability that the random variable's value is <= x.
     
     F(0-) = 0 means the probability that the random variable's
               value is < 0 (*not* <=) is zero.  So if
               probability(x = 0) = 1/2 then F(0) = 1/2, but
               if F is continuous then F(0) = 0.
     
     lim x -> oo F(x) = 1 (the random variable probably has
                           a value :-)
     
     Since F is monotonic, it has at most countably many points
     of discontinuity, and so is a measurable function, and
     so the first integral is defined (although it may be
     infinite).
     
     1-F(x) starts [in the range zero to infinity] out being
     <= 1 and is non-increasing, declining to 0 as x -> oo.
     
     The integral of xdF(x) would be the expected value of x.
     
     But we can't assume that F has a nice derivative.
     
     We must show that both integrals diverge or both have
     the same (finite) value.
     
     Integrating the second integral by parts between 0 and
     t gives
                                       |t
          integral(0,t) xdF(x) = xF(x) |  - integral(0,t) F(x)dx
                                       |0
     
                               = tF(t) - integral(0,t) F(x)dx
     
     Evaluating the first integral between zero and t gives
     
          integral(0,t) (1-F(x))dx = t - integral(0,t) F(x)dx
     
     and this is defined since F is measurable.
     
     The integrals from 0 to oo are the limits of the integrals
     from 0 to t as t -> oo; since F(t) -> 1 as t -> oo it
     would appear that these are the same.  So this must be
     the "trivial" part that you mentioned; and the step where
     I used integration by parts would be where I assumed
     that there was a density function (the "nice derivative").

     And finally, the problem is to get the same result without
     that simplification? 
     
     Dan

    If F(x) is a bounded nondecreasing function of x (with greatest lower
    bound and least upper bound) on an open interval a < x < b then
    x is clearly a bounded nondecreasing function of F.  This would
    justify integration by parts; the "area on the graph" between
    y = 1 and y = F(x) would be the same if you transposed the y and x
    axes.  One of the purposes of Lebesgue integration is to allow this
    kind of reasoning under wide conditions.

    The solution probably consists in quoting the right result about
    Lebesgue integration and rigorously justifying the statement that F is
    a measurable function of x <=> x is a measurable function of F under
    these conditions...

    It is certainly possible for the integrals to diverge as x -> infinity.

    - Jim

    .1 puts the solution when there is a density quite nicely - you
    can attack either side by parts in fact.  The solution may even
    be correct under the weaker assumption that the non-atomic part
    of the probability measure be absolutely continuous with respect
    to Lebesgue measure.
    
    .2 is interesting but F(x) need not have an inverse.  In the case
    of the Cantor distribution function, the range set is of measure
    zero, that is F(x) is constant a.e. x but uniformly continuous!
    
    Chuck

�   .2 is interesting but F(x) need not have an inverse.  In the case
�   of the Cantor distribution function, the range set is of measure
�   zero, that is F(x) is constant a.e. x but uniformly continuous!

    I don't think this is true, given the conditions.

    The cumulative probability function of the Cantor set
    (also called a Devil's Staircase, if I recall) has an
    inverse, for example.  In fact, the length (measure) of both its
    x and y projections are equal to 1, so the total length of the
    this continuous "curve" is 2.

    - Jim

     The lengths of the x and y projections of y = x for 0 <= x <= 1
     are each 1, but that curve has length sqrt(2).
     
     Dan

     re .3
     
>>     .2 is interesting but F(x) need not have an inverse.  In the case
>>     of the Cantor distribution function, the range set is of measure
>>     zero, that is F(x) is constant a.e. x but uniformly continuous!
     
     That doesn't sound right.  I know what the Cantor set
     is, but could you please define this distribution function.
     Perhaps you mean it has a derivative defined almost
     everywhere which is constant where defined?
     
     Dan

    OK will post Cantor distribution this PM.  About the derivative.
    F' exists except on the Cantor set and where it exists, F'=0.
    Since the value of a function on a set of measure zero has no effect
    on the value of the integral, the integral of F' is equal to zero.
    This is an example where differentiation followed by integration
    does not get you back where you started!
    
    Chuck

     Okay, it seemed like you were say F was constant a.e.,
     not F' or the integral of F'.
     
     Dan

�  The lengths of the x and y projections of y = x for 0 <= x <= 1
�  are each 1, but that curve has length sqrt(2).

   True, but it is also true that the length of the Devil's Staircase
   is 2 and *not* sqrt(2)!  In spite of its being continuous.  This is
   what's amazing and counterintuitive about this transfinite counting
   stuff...

   - Jim

     re .9
     
     Your wording in .4 suggested that because each projection
     was 1 the total length must be 2; .5 was just a
     counterexample.  I'm not familiar with this staircase
     so I'm not saying that its length isn't 2, just that
     .4 didn't establish that it was.
     
     Dan

This topic seems to have gone after the Cantor distribution function -
not what I had in mind but maybe just as interesting.  As promised, I
will present the construction along the lines in Chung (see .0).

For completeness, I construct the Cantor ternary set.  Consider the
closed interval [0,1] in the reals.  For statements about topology, the
usual topology is assumed.  Let U1,1 denote the open middle third of
[0,1] that is (1/3,2/3).  Let U1,2 and U3,2 be the open middle thirds
of the intervals obtained by removing U1,1.  Now at this stage, let
the open middle third of [0,1] be U2,2.  Continue in this fashion
constructing Uk,n.  At each step we get 2^n-1 open intervals, that is,
k ranges from 1 to 2^n-1.  Let U be the union of all Uk,n for every
k and n.  Note that U is open and by adding up the lengths of all the
unique intervals, it is clear that U has measure exactly 1.  Let C denote
the relative complement of U in [0,1] i.e. C=[0,1]\U.  C is called the
Cantor ternary set.  It is closed, completely disconnected and of
measure zero.  It is also called fractal.

Define the function G on U in the following way.  Let ck,n=k/2^n
for every k and n.  G(x)=ck,n for x an element Uk,n.  This is consistent
but I won't prove it.  Note that G(x) is constant on each Uk,n and it
can therefore be differentiated on U (U is open!) giving G'=0 on U.
Let F be defined on the real line with F(x)=0 for x<=0, F(x)=1 for x>=1
and F(x)=G(x) on U.  F is the Cantor distribution function.  It is
uniformly continuous and increasing.  It has no inverse since it is
constant on intervals and therefore inverse images of points are sets
(actually the Uk,n).  Note that the measure associated with F has no
mass on U - all of the mass is in C, a set of Lebesgue measure zero.

re .3 I did not say what I meant in this one, the above is correct.

About the length of F in [0,1].  Looks like it is 2 since the length
in the intervals Uk,n is obviously 1 and for the rest, it must rise
continuously on a set of measure zero (very loose argument) by 1.

Devil's staircase seems an odd name for a function with no steps.

    Chuck

    I think I said something stupid.  If we define Gn in the obvious
    way on U, for any fixed n the length of Gn is clearly one.  The
    limit of the length of Gn as n goes to infinity must be one.  The
    total variation of G is probably 2 as advertised but the length
    seems to be one or have I missed something.
    
    Chuck

    Should think before writing!  There is a very simple proof that
    the length of arc of the Cantor distribution function is 2.  Just
    construct a sequence of functions of known arc length coverging
    to it and the result is immediate.
    
    In thinking about the problem again, I am happy with the proof of
    .0 for absolutely continuous (w.r.t. Lebesgue measure) distributions
    which uses integration by parts (absolute continuity justifies this).
    It is also easy to prove for distributions having a countable number
    of step discontinuities (I have an argument for one discontinuity
    which will extend).  By linearity, any distribution which is absolutely
    continuous except for a countable number of step discontinuities will
    satisfy the equality in .0.  The problem for me hinges on finding
    a way to handle distributions which are not absolutely continuous
    (like the Cantor distribution which is only uniformly continuous).