In article <[EMAIL PROTECTED]>,
ZHANG Yan <[EMAIL PROTECTED]> wrote:
>> P(sum(T_i) < a | max(T_i) < b) P{max(T_i) < b}
>> Or perhaps it is easy somehow to compute the n-fold convolution of the
>> conditional distribution?
It depends on what one means by "easy", and this depends
on the precise distribution. If m(z) = E(exp(z*T_k)|T_k < b},
then P(sum(T_k) > a | max(T_k) < b) is given by
(1/2*\pi*i)*\int_{c-i*\infty}^{c+i*\infty) m(z)^n*exp(-za)/z dz,
where c > 0. Frequently a good approximation can be made
by steepest descent.
>Thanks for your kind help.
>It seems for me that the conditional probability
>P(sum(T_i) < a | max(T_i) < b)
>is difficult to compute. Can you plz give more information? Thanks.
If you want "closed form", these are generally unavailable.
The easiest case is that of the uniform; it is just the
probability that \sum(U_k) < a/b, where the U_k are
uniform (0,1), and this is in a "closed form" given in
the 19th century. Probably the next easiest is that of
a gamma distribution, where the distribution of the T's
given the sum is independent of the sum, but this is
still a little nasty, although studied.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
[EMAIL PROTECTED] Phone: (765)494-6054 FAX: (765)494-0558
.
.
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