James from the MacMillan Academy was the
first to crack this problem - well done James!
Steve says
The essence of this problem lies in the fact that a cdf F(x) is
non-decreasing and satisfies 0\leq F(x) \leq 1, whereas a pdf is
a non-negative function which integrates to 1 between -\infty and
\infty. The areas under the curves are the key points to
consider. A cdf F(x) can either reach the value 1 for a finite
value of x_1 or tend to 1 as x\to \infty. In the first case,
provided that the area under the cdf to the left of x_1 is 1 then
this will work as a pdf for a random variable which is 0 if
x> x_1. In the second case, the area diverges and, therefore,
cannot be used as a pdf.
To get the curves to match exactly for values of x< x_1 we
will need to solve the equation F'(x) = f(x).
Here is James's solution
Part 1
A CDF is always an increasing function which goes to 0 as X
goes to -\infty and 1 as x goes to \infty and a pdf is a
function that is never negative and has an area under it of
1.
The CDF can reach 1 at any point; let us only consider ones that
reach it at 0 as all others either can be generalised from these by
moving them left or right i.e. considering F(x-a) where a is
the point where it reaches one, or they don't reach 1 until
infinity which I shall deal with later.
As the PDF can do what it likes when x is positive (so long as it
stays non-negative and doesn't let the area exceed 1) it can make
up difference between the area to the left of x=0 and 1 and so the
area to the left of x=0 need only be between 0 and 1. So the only rules controlling the CDF
are that to the left of where it reaches F(x)=1 the area underneath
it must be less than or equal to 1 which is true for any of a
multitude of CDF's
Regarding the ones that don't reach 1 until \infty these would
have to be a solution to PDF=CDF for x< \infty, consider any
finite value of x if F(x)> 0. At that point then it must be
that value or greater until x=\infty giving area under it \geq
F(x)\times \infty which is absurd and definitely larger than 1
unless F(x) is infinitesimal for all x in which case F(x)
wouldn't be tending to 1 and so such a CDF cannot possibly be a
PDF.
Part 2
The integral of a CDF from -\infty to \infty is always
\infty. Thus a CDF cannot be used as a pdf.
Part 2
As f(x)=F'(x) for f(x)=F(x) we would need F(x)=F'(x). Thus,
writing F(x)=y, the CDF must be a solution to the differential
equation