In article <[EMAIL PROTECTED]>,
H. J. Wang <[EMAIL PROTECTED]> wrote:
>Hi,
>Suppose X, Y are independent random variables with normal distributions.
>The means and variances are different. Assume X1 and Y1 are random
>variables with the probability distributions f(X | X>=0) and g(Y|Y>=0),
>respectively. That is, X1 and Y1 the non-negative truncations of X and
>Y, respectively. Does anyone know whether in this case Z = X1 + Y1 is
>still a truncated normal? Any reference on this? Thanks in advance!
It is not. An easy way to see this is to use the fact
that the truncated normal distribution has a strictly
positive finite density at the low end of its range.
However the convolution of two such has density zero
at the low end.
--
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
