A much more general problem. Suppose there are k samples, each from R(0,1),
each of size n. Suppose the largest value in sample i is x_i. Then the
product z of the x_i is derived by Rider (JASA, 1955). Yours is the (very)
special case in which n=1 and k=2. The general result, in TeX, is
p(z) = \frac{n^k}{(k-1)!}z^{n-1}(-\log z)^{k-1}
Also, of course, if x_i are R(0,1), i=1,...,n, then the distribution of
-2 log(x_1...x_n) is chi-square with 2n dfr.
At 3:43 +0000 07/01/2000, Gautam Sethi wrote:
>Herman Rubin <[EMAIL PROTECTED]> wrote:
>: In article <002a01bfe2f1$5548e6e0$[EMAIL PROTECTED]>,
>: David A. Heiser <[EMAIL PROTECTED]> wrote:
>
>
>:>The product and convolution are two different things. The product gives a
>:>triangular distribution. If I remember correctly, the distribution is
>:>triangular even if the two have different supports. I never tried out the
>:>convolution.
>
>: The convolution is the distribution of the sum; the product
>: has a quite different distribution.
>
>: The distribution of the sum for different ranges is a symmetric
>: trapezoid, with the central part having the density of the one
>: with larger range and length the difference of the ranges. If
>: the ranges are equal, this becomes a triangle; proof left to
>: the reader.
>
>thanks herman. i was able to derive the convolution pretty easily. however,
>the product seems a lot harder. is the product easy to figure out too? any
>tips and tricks i should know? someone suggested that the density of z = x*y
>is -log(x) if both x and y have the support [0,1]. is this correct?
>
>best,
>gautam.
>
>
>
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