In article <[EMAIL PROTECTED]>,
Shahram Hosseini <[EMAIL PROTECTED]> wrote:
>Hi everybody,
>The discrete random process n(t), uniformly distributed in the interval
>(-0.5,05), is filtered by a first order AR system to generate the
>sequence s(t)=a*s(t-1)+n(t). What is the probability density function of
>s(t)?
>Thank you in advance for your help.
>PS: I know that the answer can be obtained by solving the following
>functional equation:
>a*f(s)=Int from -0.5 to 0.5 of [f((s-y)/a)dy]
>but I can not solve it.
I doubt that a simple form can be given for any value of a;
however, it should not be too difficult to compute it
numerically. In the tails, one should use complex variable
procedures, as indicated.
The mgf of n(t), for any t, is sinh(t/2)/(t/2). So the
mgf of the distribution sought is
m(w) = \prod_0 sinh(a^k*t/2)/(a^k*t/2).
One could use the usual inversion formula to obtain the
density or cdf by making w purely imaginary and using
m(iw) as the characteristic function.
In the tails, it is probably better to use the inversion
formula in the form
f(s) = \int m(w)*exp(-ws) dw /(2*\pi*i),
where the range of integration is from c-i\infty to c+\infty,
and c can be chosen to speed up convergence, such as using
steepest descent. Also, if c > 0, the right tail probability
can be obtained directly by dividing the integrand by w. If
c < 0, one gets the negative of the left tail probability.
The steepest descent approach for this is not at the same
point as for the density for a particular s.
Also, as the range is finite, one can use other devices in
the tails, such as restricting the values of some of the
n's and modifying the mgf appropriately.
--
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, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
[EMAIL PROTECTED] Phone: (765)494-6054 FAX: (765)494-0558
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