Re: Repeated Bernoulli test in which the success probability is itself binomially distributed
- From: Ray Vickson <RGVickson@xxxxxxx>
- Date: Wed, 12 Aug 2009 22:54:01 -0700 (PDT)
On Aug 12, 5:43 pm, Adan Mithrillion <diablodestruct...@xxxxxxxxx>
wrote:
Sorry. Previously I got it wrong by mixing the two kinds of tests up.
The sub-tests should not be Bernoulli tests. The result of a single
test A[i,j] in a bigger test A[i] is a random variable r[i,j]
distributed in [0,1]. We repeat this test for, say, 20 times. A bigger
test A[i] is a success if the sum of all these random variables in A
[i] is greater than a given number, say, 15. Then we repeat the bigger
test A[i] as a Bernoulli test, which will be simple when the success
probability is obtained. I am asking for the probability when the
distribution of r[i,j] is uniform and when the probability density
describing the distribution of r[i,j] is linear(a linear function
defined on [0,1], you can give an arbitrary example).
So, if I now understand correctly (eliminating the i, j, r[i,j] and A
[i] and just describing what actually occurs): we perform some test n
= 20 (independent) times; these individual tests give iid U(0,1)
values. If the sum of these n values is > 15 we declare "success".
You
want the probability of success. Is that the problem? Well, we can
develop the cdf of the sum of n iid U(0,1) random variables, using
Laplace transforms (LT). If f is the U(0,1) density, its LT is g(s) =
int{x=0..1} exp(-x*s) dx = (1/s)[1-exp(-s)]. So, the density f_n(x)
of
the sum of n iid copies has LT g_n(s) = g(s)^n, while the LT of the
cumulative F_n(x) = int{y=0..x} f_n(y) dy is G_n(s) = g_n(s)/s. Using
the binomial expansion we get G_n(s) = sum{j=0..n} (-1)^j*C(n,j)*exp
(-
j*s)/s^(n+1). To invert this, use the following: 1/s^(n+1) --> 1{x >
0}
*x^n/n! for n >= 0. Also, if h(s) <--> r(x), then exp(-a*s)*h(s) <-->
r
(x-a); thus, exp(-j*s)/s^(n+1) <--> 1{x > j}*(x-j)^n/n! for n >= 0.
Thus, we have F_n(x) = sum_{j=0..n} (-1)^j*C(n,j)*1{x >j}*(x-j)^n/n!
=
sum_{j=0..n} (-1)^j * 1{x > j}*(x-j)^n/[j!*(n-j)!] = Pr{U1+...+Un <=
x}.
For the specific case of n = 20, Maple gives P{sum <= 15} = .
9999695678
using 30 digits to do the computations---due to the possibility of
serious
cancellation errors. Thus, P{success} = .304322e-4 .
For the case of a linear density f(x) = (1+b*x)/(a + b/2) for x in
[0,1], we can use a similar Laplace transform method to get the
density (and cdf) of 20 iid r's with density f. However, the results
are messy. The density f20(x) of the sum has 3,311 terms, of which
the
first three are -1/29256054573178125*1/(2*a+b)^20*a^3*b^17*Heaviside
(x-19)*(x-19)^29, -1/7114537504341867600000*1/(2*a+b)
^20*b^19*a*Heaviside(x-11)*(x-11)^33 and -17/17929782016990593750*1/
(2*a+b)^20*a^6*b^14*Heaviside(x-19)*(x-19)^30. There are 3,308
additional terms like that. It requires 200 pages to print out the
complete formula. Of course, numerical computations for given a, b
and
x may require using very high precision, because of serious and
significant cancellation errors. Maple can handle it just fine, but
less sophisticated programs might very well fail or give highly
suspect results.
R.G. Vickson
.
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- From: Adan Mithrillion
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