Re: a problem with functions of binomials.


claudia.giovannoni@xxxxxxxxx wrote:
Reef Fish wrote:
Are you trying to find an approximation for the left tail of a binomial
by the right tail of a half normal?

If we let Fb( s; p, n) denote sum[x from 0 to s] B(x;n,p), then

Fb( s; p, n) = Ff ((s+1)(1-p)/(p(n-s)); 2(n-s) , 2(s+1)).

Thank you for trying to help, but this is not what I was looking for
and the subject line (which is wrong) must have been confusing. I'll
try again to explain, and since you obviously have a much better grasp
of the literature, if I make myself clear there is a good chance you
might be able to help. So, please bear with me.

The function H(y) is the integral of a CUMULATIVE of a normal from 0 to
y where, as you guessed, the normal is itself the approximation of a
binomial. However, this is not where my problem lies.

I did misunderstand it. You have the cdf Z(.) of a unit normal, say,
so your Z is defined everywhere and Z(0) = .5 and Z -->1 as x
increases;

For z = 0 (.001) .004, Z would be .5, .50399, .50798 , .51197,
..51595.
reaching Z(1) = .84134.

Your H(y) is the INTEGRAL of this Z function from 0 to y, y>0.

so that for y = 1, H(1) would be the integral from 0 to 1 of Z(t) or
..68437.


Given the function H, a strictly positive number t and an integer n, I
can define a sequence of n terms

H(0), H(t/n),H(2t/n),.....,H(tn/n)=H(t). Thus the jth element of this
sequence is H(tj/n).

That would be (n+1) terms whose (j+1)th element is H(tj/n) wouldn't it?


Similarly, I also have a sequence of (n+1) terms

H(0),H(t/(n+1)),H(2t/(n+1)),....,H(t(n+1)/(n+1))=H(t)

In this sequence, you have (n+2) terms, from H(0) to H(t).


Not sure what you're trying to do the rest of it yet, but check and see
if I understood your Z and H functions and the sequences of H, before
I proceed any further.

-- Reef Fish Bob.

Now let a(j,k,p) = Pr (X=j) if X is a binomial with parameters k and p

I want to prove that for any strictly positive t,

Sum [j from 0 to n] a(j,n,p) x H(tj/n) > Sum [j from 0 to n+1]
a(j,n+1,p) x H(tj/n+1)

In other words, we have on the a left-hand side the expected value of a
r.v. with generic realization H(tj/n) for which the corresponding
probability is that of the realization j in a Binomial(n,p). Similarly
on the right-hand side where we have the expected value of a r.v. with
generic realization H(tj/(n+1)) for which the corresponding probability
is that of the realization j in a Binomial(n+1,p).

Now, if t=0 the difference between the two terms is trivially zero and
I can also show that this difference, as a function of the expected
value t has first derivative equal to zero and positive second
derivative when it is evaluated at zero. Thus, since the function is
continuous in t, to show the above it suffices to show that there is no
strictly positive t for which the difference is zero. This may be
easier to prove. Also note that the derivative with respect to t of the
difference between the two functions is NOT monotone so it won't help
to use that.

Again, any help is much appreciated. If you prefer, I can email you a
pdf file where the notation is much more clear.

.

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